DBMS Lab Mannual

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DataBaseManagementSystems LAB MANUAL A Helpful Hand DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING CSEROCKZ 0 WWW.CSEROCKZ.COM

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This is fully interactive lab manual for practising database and in LAB and at home as well.....You can download it and use it for your personal use as well.

Transcript of DBMS Lab Mannual

Page 1: DBMS Lab Mannual

DataBaseManagementSystems LAB MANUAL

A Helpful Hand

DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING

CSEROCKZ

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How to Write and execute sql, pl/sql commands/programs:

1). Open your oracle application by the following navigationStart->all programs->Oracle Database 10g Express Edition->Run SQL Command Line

2). You will be asked for user name, password.You have to enter user name, pass word.

3). Upon successful login you will get SQL prompt (SQL>).In two ways you can write your programs:a) directly at SQL prompt (or)b) in sql editor.

If you type your programs at sql prompt then screen will look like follow:SQL> SELECT ename,empno,2 sal from3 emp;where 2 and 3 are the line numbers and rest is the command/program……

to execute above program/command you have to press ‘/’ then enter.

Here editing the program is somewhat difficult; if you want to edit the previous command then you have to open sql editor (by default it displays the sql buffer contents). By giving ‘ed’ at sql prompt.(this is what I mentioned as a second method to type/enter the program).in the sql editor you can do all the formatting/editing/file opera-tions directly by selecting menu options provided by it.

To execute the program which saved; do the followingSQL> @ programname.sql (or)

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SQL> Run programname.sqlThen press ‘\’ key and enter.To save the day`s session ;do the followingSQL>commit;

This how we can write, edit and execute the sql command andprograms.

Always you have to save your programs in your own logins.

Background Theory

Oracle workgroup or server is the largest selling RDBMS product.it isestimated that the combined sales of both these oracle database productaccount for aroud 80% of the RDBMSsystems sold worldwide.These products are constantly undergoing change and evolving. Thenatural language of this RDBMS product is ANSI SQL,PL/SQL a supersetof ANSI SQL.oracle 8i and 9i also under stand SQLJ.

Oracle corp has also incorporated a full-fledged java virtual machineinto its database engine.since both executable share the same memoryspace the JVM can communicate With the database engine with easeand has direct access to oracle tables and their data.

SQL is structure query language.SQL contains different data types thoseare1. char(size)2. varchar(size)3. varchar2(size)4. date5. number(p,s) //** P-PRECISION S-SCALE **//6. number(size)7. raw(size)8. raw/long raw(size)

Different types of commands in SQL:

A).DDL commands: - To create a database objectsB).DML commands: - To manipulate data of a database objectsC).DQL command: - To retrieve the data from a database.D).DCL/DTL commands: - To control the data of a database…

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DDL commands:

1. The Create Table Command: - it defines each column of the tableuniquely. Each column has minimum of three attributes, a name , datatype and size.

Syntax:

Create table <table name> (<col1> <datatype>(<size>),<col2><datatype><size>));

Ex:create table emp(empno number(4) primary key, ename char(10));

2. Modifying the structure of tables.a)add new columns

Syntax:

Alter table <tablename> add(<new col><datatype(size),<newcol>datatype(size));

Ex:alter table emp add(sal number(7,2));

3. Dropping a column from a table.

Syntax:Alter table <tablename> drop column <col>;

Ex:alter table emp drop column sal;

4. Modifying existing columns.

Syntax:Alter table <tablename> modify(<col><newdatatype>(<newsize>));

Ex:alter table emp modify(ename varchar2(15));

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5. Renaming the tables

Syntax:

Rename <oldtable> to <new table>;

Ex:rename emp to emp1;

6. truncating the tables.

Syntax:

Truncate table <tablename>;

Ex:

trunc table emp1;

7. Destroying tables.

Syntax:

Drop table <tablename>;

Ex:

drop table emp;

DML commands:

8. Inserting Data into Tables: - once a table is created the mostnatural thing to do is load this table with data to be manipulated later.

Syntax 1:

insert into <tablename> (<col1>,<col2>…..<col n>) values(<val 1>,<val 2>…….<val n>);

Syntax 2:

insert into <tablename> values(&<col1>,&<col2>……,&<col n>);

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Syntax 3:

insert into <tablename> values(<val 1>,<val 2>…….,<val n>);

Ex 1:

Insert into skc (sname,rollno,class,dob,fee_paid)values(‘sri’,’104B’,’cse’,’27-feb-05’,10000.00);

Ex 2:

insert into skc values(&sname,&roll no,&class);enter sname:’sri’enter roll no:’104B’enter class:’cse’1 row created.

Ex 3:

insert into skc values(‘sri’,’104B’,cse’,’27-feb-05’,10000.00);

9. Delete operations.

a) remove all rows

Syntax:

delete from <tablename>;

b) removal of a specified row/s

Syntax:

delete from <tablename> where <condition>;

10. Updating the contents of a table.

a) updating all rows

Syntax:Update <tablename> set <col>=<exp>,<col>=<exp>;

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b) updating seleted records.

Syntax:Update <tablename> set <col>=<exp>,<col>=<exp>where <condition>;

11. Types of data constrains.

a) not null constraint at column level.

Syntax:

<col><datatype>(size)not null

b) unique constraint

Syntax:

Unique constraint at column level.<col><datatype>(size)unique;

c) unique constraint at table level:

Syntax:

Create tabletablename(col=format,col=format,unique(<col1>,<col2>);

d) primary key constraint at column level

Syntax:

<col><datatype>(size)primary key;

e) primary key constraint at table level.

Syntax:

Create table tablename(col=format,col=formatprimary key(col1>,<col2>);

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f) foreign key constraint at column level.

Syntax:

<col><datatype>(size>) references <tablename>[<col>];

g) foreign key constraint at table level

Syntax:

foreign key(<col>[,<col>]) references<tablename>[(<col>,<col>)

h) check constraint

check constraint constraint at column level.

Syntax: <col><datatype>(size) check(<logical expression>)

i) check constraint constraint at table level.

Syntax: check(<logical expression>)

DQL Commands:

12. Viewing data in the tables: - once data has been inserted into atable, the next most logical operation would be to view what has beeninserted.

a) all rows and all columns

Syntax:Select <col> to <col n> from tablename;Select * from tablename;

13. Filtering table data: - while viewing data from a table, it is rarethat all the data from table will be required each time. Hence, sql mustgive us a method of filtering out data that is not required data.

a) Selected columns and all rows:Syntax:select <col1>,<col2> from <tablename>;

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b) selected rows and all columns:Syntax:select * from <tablename> where <condition>;

c) selected columns and selected rowsSyntax:select <col1>,<col2> from <tablename> where<condition>;

14. Sorting data in a table.

Syntax:Select * from <tablename> order by <col1>,<col2> <[sortorder]>;

DCL commands:

Oracle provides extensive feature in order to safeguard informationstored in its tables from unauthoraised viewing and damage.The rightsthat allow the user of some or all oracle resources on the server arecalled privileges.

a) Grant privileges using the GRANT statementThe grant statement provides various types of access to databaseobjects such as tables,views and sequences and so on.

Syntax:GRANT <object privileges>ON <objectname>TO<username>[WITH GRANT OPTION];

b) Reoke permissions using the REVOKE statement:The REVOKE statement is used to deny the Grant given on an object.

Syntax:REVOKE<object privilege>ONFROM<user name>;

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WEEK-1

CREATING,ALTERING AND DROPPING TABLES AND INSERTING ROWS INTO A TABLE (USE CONSTRAINTS WHILE CREATING TABLES) EXAMPLES USING SELECT COMMAND .

EXAMPLE 1:

CREATING A STUDENT RELATION TABLE WITH ALL DATATYPES:

SQL> create table student252(sid number(5),sname varchar(20),sbranch char(5),dob date,spercent number(3,2));

Table created.

RELATIONAL SCHEMA FOR STUDENT RELATION :SQL> desc student252;Name Null? Type----------------------------------------- -------- ----------------------------SID NUMBER(5)SNAME VARCHAR2(20)SBRANCH CHAR(5)DOB DATESPERCENT NUMBER(5,2)

INSERT THE RECORDS INTO STUDENT RELATION:

METHOD 1:SQL>Insert intoStudent252(sid,sname,sbranch,dob,spercent) values(104,‘sri’,,’cse’,’27-feb-05’,70);1 row created.

METHOD 2:SQL>Insert intoStudent252 values(104,‘sri’,,’cse’,’27-feb-05’,70);1 row created.

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METHOD 3:SQL>Insert intoStudent252(sid,sname,sbranch,dob,spercent)values(&sid, &sname,&sbranch,&dob,&spercent);1 row created.

METHOD 4:SQL>Insert intoStudent252(sid,sname,sbranch,dob,spercent)values(&sid, ‘&sname’,’&sbranch’,’&dob’,&spercent);1 row created.

QUERY THE TABLE VALUES:

ALL ROWS AND ALL COLUMNS:

SQL> select * from student252;SID SNAME SBRANCH DOB SPERCENT------ --------------- --------------------- --------------- --------------------130 ravi it 30-1-95 60131 teja cse 21-07-87 55129 kiran mech 12-05-92 60104 sri cse 30-07-90 70133 sajith eee 12-06-89 55137 ram ece 07-07-85 40

WEEK 2 (cont…1)1) Creation, altering and dropping tables and inserting rows into a table (use constraints while creating tables) examples using SELECT command.

MODIFYING THE STRUCTURE OF TABLE

ADDING A NEW COLUMN

SQL> ALTER TABLE Emp252 ADD (age number(3),

phno number(10));

Table altered.

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MODIFYING EXISTING COLUMN

SQL> ALTER TABLE Emp252 MODIFY (phno varchar(20));

Table altered.

DROPING A COLUMN

SQL> ALTER TABLE Emp252 DROP COLUMN phno;

Table altered.

QUERY FOR THE TABLE VALUESSQL> SELECT *

FROM Emp252;

ENO ENAME ESAL DEPTNO AGE ----- -------------------- ---------- ---------- ---------- 30 ravi 51000 3 31 teja 31000 2 29 kiran 31200 1 45 allen 41000 3 33 sajith 51000 4 46 geetha 11000 4 90 veena 16000 3 85 pragna 61000 1 84 harsha 91000 3 40 sanjeev 1500 13

10 rows selected.

UPDATING ENTIRE COLUMN

SQL> UPDATE Emp252 SET age=18;

10 rows updated.

QUERY THE TABLE VALUESSQL> SELECT *

FROM Emp252;11

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ENO ENAME ESAL DEPTNO AGE ----- -------------------- ---------- ---------- ---------- 30 ravi 51000 3 18 31 teja 31000 2 18 29 kiran 31200 1 18 45 allen 41000 3 18 33 sajith 51000 4 18 46 geetha 11000 4 18 90 veena 16000 3 18 85 pragna 61000 1 18 84 harsha 91000 3 18 40 sanjeev 1500 13 18

10 rows selected.

RENAMING THE TABLE:

SQL> RENAME Emp252 TO Emp1252;

Table renamed.

SELECTING THE TABLE VALUESSQL> SELECT *

FROM Emp1252;

Example 3

CREATING A DEPARTMENT RELATION TABLE

CREATING A DEPARTMENT TABLE

SQL> CREATE TABLE Dept252(dname VARCHAR(10),dno CHAR(5),dloc VARCHAR(25));

Table created.

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DESCRIBE A STUDENT TABLE

SQL> desc Dept252; Name Null? Type ----------------------------------------- -------- ---------------------------- DNAME VARCHAR2(10) DNO CHAR(5) DLOC VARCHAR2(25)

DROPING THE TABLESQL> DROP TABLE Dept252;

Table dropped.

WEEK 3 (cont…1)

1) Creation, altering and dropping tables and inserting rows into a table (use constraints while creating tables) examples using SELECT command.

CREATING A TABLE WITH KEY CONSTRAINTS

Example 1

CREATING A TABLE WITH ‘UNIQUE ‘, ‘NOT NULL’, ‘CHECK’ AND ‘DEFAULT’ CONSTRAINT:

SQL> CREATE TABLE emp252(eid NUMBER(5) UNIQUE, ename VARCHAR(10) DEFAULT(‘UNKNOWN’), age NUMBER(3) NOT NULL, esal NUMBER(7) CHECK(esal > 1000));

Table created.

INSERTING RECORDS INTO TABLE:

SQL> INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal);13

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Enter value for eid: 1Enter value for ename: 'ravi'Enter value for age: 18Enter value for esal: 10000old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)new 1: INSERT INTO emp252 VALUES (1, 'ravi', 18, 10000)

1 row created.

SQL> /Enter value for eid: 2Enter value for ename: 'teja'Enter value for age: 18Enter value for esal: 20000old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)new 1: INSERT INTO emp252 VALUES (2, 'teja', 18, 20000)

1 row created.

SQL> /Enter value for eid: 3Enter value for ename: 'kiran'Enter value for age: 19Enter value for esal: 25000old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)new 1: INSERT INTO emp252 VALUES (3, 'kiran', 19, 25000)

1 row created.

SQL> /Enter value for eid: 4Enter value for ename: 'srinivas'Enter value for age: 19Enter value for esal: 30000old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)new 1: INSERT INTO emp252 VALUES (4, 'srinivas', 19, 30000)

1 row created.

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SQL> /Enter value for eid: 1Enter value for ename: 'alan'Enter value for age: 19Enter value for esal: 29000old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)new 1: INSERT INTO emp252 VALUES (1, 'alan', 19, 29000)INSERT INTO emp252 VALUES (1, 'alan', 19, 29000)[SHOWING AN ERROR WHILE VIOLATING UNIQUE KEY CONSTRAINT]*ERROR at line 1:ORA-00001: unique constraint (SYSTEM.SYS_C003875) violated SQL> /Enter value for eid: 7Enter value for ename: 'dravid'Enter value for age: nullEnter value for esal: 100000old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)new 1: INSERT INTO emp252 VALUES (7, 'dravid', null, 100000)INSERT INTO emp252 VALUES (7, 'dravid', null, 100000) [SHOWING AN ERROR AS NOT NULL KEY CONSTRAINT IS VIOLATED] *ERROR at line 1:ORA-01400: cannot insert NULL into ("SYSTEM"."EMP230"."AGE")

SQL> /Enter value for eid: 8Enter value for ename: 'sachin'Enter value for age: 35Enter value for esal: 100old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)new 1: INSERT INTO emp252 VALUES (8, 'sachin', 35, 100)INSERT INTO emp252 VALUES (8, 'sachin', 35, 100)*[NOT ALLOWING AS IT VOILATES CHECK CONSTRAINT FOR esal > 1000 VALUE]

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ERROR at line 1:ORA-02290: check constraint (SYSTEM.SYS_C003874) violated

Example 2

CREATING A TABLE WITH ‘PRIMARY KEY’ CONSTRAINT:

SQL> CREATE TABLE mdept252(dno NUMBER(5), dname CHAR(10), dloc VARCHAR(10), PRIMARY KEY (dno));

Table created.

SQL> desc mdept252; Name Null? Type ----------------------------------------- -------- ---------------------------- DNO NOT NULL NUMBER(5) DNAME CHAR(10) DLOC VARCHAR2(10)

INSERTING RECORDS INTO MASTER DEPARTMENT TABLE:

SQL> INSERT INTO mdept252 VALUES (&dno, &dname, &dloc);Enter value for dno: 1Enter value for dname: 'ravi'Enter value for dloc: 'hyd'old 1: INSERT INTO mdept252 VALUES (&dno, &dname, &dloc)new 1: INSERT INTO mdept252 VALUES (1, 'ravi', 'hyd')

1 row created.

SQL> /Enter value for dno: 1Enter value for dname: 'teja'

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Enter value for dloc: 'sec'old 1: INSERT INTO mdept252 VALUES (&dno, &dname, &dloc)new 1: INSERT INTO mdept252 VALUES (1, 'teja', 'sec')INSERT INTO mdept252 VALUES (1, 'teja', 'sec')*ERROR at line 1:ORA-00001: unique constraint (SYSTEM.SYS_C003876) violated

SQL> /Enter value for dno: nullEnter value for dname: 'sajithulhuq'Enter value for dloc: 'kmm'old 1: INSERT INTO mdept252 VALUES (&dno, &dname, &dloc)new 1: INSERT INTO mdept252 VALUES (null, 'sajithulhuq', 'kmm')INSERT INTO mdept252 VALUES (null, 'sajithulhuq', 'kmm') *ERROR at line 1:ORA-01400: cannot insert NULL into ("SYSTEM"."MDEPT230"."DNO")

ADDING A PRIMARY KEY TO AN EXISTING TABLE:

SQL> ALTER TABLE student252 ADD PRIMARY KEY (sid); Table altered.

SQL> ALTER TABLE emp252 ADD PRIMARY KEY (eid);ALTER TABLE emp252 ADD PRIMARY KEY (eid) *[ GIVING AN ERROR AS ONE TABLE CAN HAVE A SINGLE PRIMARY KEY AT COLUMN LAVEL]ERROR at line 1:ORA-02261: such unique or primary key already exists in the table

Example 3

CREATING A TABLE WITH ‘FORIEGN KEY’ CONSTRAINT:

SQL> CREATE TABLE detailemp252

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(eid NUMBER(5) REFERENCES mdept230 (dno), ename VARCHAR(10), esal NUMBER(7));

Table created.

INSERING RECORDS INTO DETAIL EMPLOYEE TABLE:

SQL> INSERT INTO detailemp252 VALUES (2, 'ravi', 50000);INSERT INTO detailemp252 VALUES (2, 'ravi', 50000)*ERROR at line 1:ORA-02291: integrity constraint (SYSTEM.SYS_C003877) violated - parent key not found

SQL> INSERT INTO detailemp252 VALUES (1, 'teja', 60000);

1 row created.

SQL> DELETE FROM mdept252 where dno=1;DELETE FROM mdept252 where dno=1*ERROR at line 1:ORA-02292: integrity constraint (SYSTEM.SYS_C003877) violated - child record found

SQL> SELECT * FROM detailemp252;

EID ENAME ESAL ---------- ---------- ---------- 1 teja 60000

SQL> SELECT * FROM mdept252;

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DNO DNAME DLOC ---------- ---------- ---------- 1 ravi hyd

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Exercise CREATING A CUSTOMER TABLE USING CONSTRAINTS :

SQL> CREATE TABLE cust252(cnum NUMBER(5), cname VARCHAR(10), state VARCHAR(10) DEFAULT ('ap'), phno NUMBER(5), CONSTRAINT cnum_pkkey PRIMARY KEY (cnum));

Table created.

SQL> INSERT INTO cust252 VALUES (&cnum, &cname, &state, &phno);Enter value for cnum: 1Enter value for cname: 'ravi'Enter value for state: 'bihar'Enter value for phno: 001old 2: (&cnum, &cname, &state, &phno)new 2: (1, 'ravi', 'bihar', 001)

1 row created.

SQL> /Enter value for cnum: 2Enter value for cname: 'teja'Enter value for state: 'up'Enter value for phno: 007old 2: (&cnum, &cname, &state, &phno)new 2: (2, 'teja', 'up', 007)

1 row created.

SQL> /Enter value for cnum: 2Enter value for cname: 'yama'Enter value for state: 'ap'Enter value for phno: 006old 2: (&cnum, &cname, &state, &phno)

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new 2: (2, 'yama', 'ap', 006)INSERT INTO cust252 VALUES *ERROR at line 1:ORA-00001: unique constraint (SYSTEM.CNUM_PKKEY) violated

SQL> /Enter value for cnum: 4Enter value for cname: 'huu'Enter value for state: 'ap'Enter value for phno: 101old 2: (&cnum, &cname, &state, &phno)new 2: (4, 'huu', 'ap', 101)

1 row created.

SQL> SELECT * FROM cust252;

CNUM CNAME STATE PHNO ---------- ---------- ---------- ---------- 1 ravi bihar 1 2 teja up 7 4 huu ap 101

CREATING AN ITEM TABLE USING CONSTRAINTS:

SQL> CREATE TABLE itm252(ino NUMBER(3), iname VARCHAR(10), iprice NUMBER(4,3), qtyonhand VARCHAR(5), CONSTRAINT itm252_ino_pkkey PRIMARY KEY (ino), CONSTRAINT itm230_qtyoh_chk CHECK (qtyonhand>1));

Table created.

SQL> INSERT INTO itm252 VALUES (&ino, &iname, &iprice, &qtyonhand);Enter value for ino: 1

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Enter value for iname: 'rubber'Enter value for iprice: 3.50Enter value for qtyonhand: 3old 1: INSERT INTO itm252 VALUES (&ino, &iname, &iprice, &qtyonhand)new 1: INSERT INTO itm252 VALUES (1, 'rubber', 3.50, 3)

1 row created.

SQL> /Enter value for ino: 1Enter value for iname: 'pencil'Enter value for iprice: 1.00Enter value for qtyonhand: 3old 1: INSERT INTO itm252 VALUES (&ino, &iname, &iprice, &qtyonhand)new 1: INSERT INTO itm252 VALUES (1, 'pencil', 1.00, 3)INSERT INTO itm252 VALUES (1, 'pencil', 1.00, 3)*ERROR at line 1:ORA-00001: unique constraint (SYSTEM.ITM230_INO_PKKEY) violated

SQL> /Enter value for ino: 2Enter value for iname: 'powder'Enter value for iprice: 3.00Enter value for qtyonhand: 0old 1: INSERT INTO itm252 VALUES (&ino, &iname, &iprice, &qtyonhand)new 1: INSERT INTO itm252 VALUES (2, 'powder', 3.00, 0)INSERT INTO itm252 VALUES (2, 'powder', 3.00, 0)*ERROR at line 1:ORA-02290: check constraint (SYSTEM.ITM230_QTYOH_CHK) violated

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1* CREATE TABLE invoice252(ivnno NUMBER(5), itemno NUMBER(5), qty NOT NULL, CONSTRAINT invoice252_ivnno_pkkey PRIMARY KEY(ivnno), CONSTRAINT FOREIGN KEY(itemno) REFERENCES cust252)

SQL> desc cust252; Name Null? Type ----------------------------------------- -------- ---------------------------- CNUM NOT NULL NUMBER(5) CNAME VARCHAR2(10) STATE VARCHAR2(10) PHNO NUMBER(5)

CREATING A INVOICE TABLE USING CONSTRAINTS:

SQL> CREATE TABLE invoice252(ivnno NUMBER(5), itemno NUMBER(5), qty NUMBER(5) NOT NULL, CONSTRAINT invoice252_ivnno_pkkey PRIMARY KEY (ivnno), CONSTRAINT fk_inv252 FOREIGN KEY (itemno) REFERENCES

cust252 (cnum)) Table created.

SQL> CREATE TABLE invitm252(invno NUMBER(5), itmno NUMBER(5), qty NUMBER(5) NOT NULL, CONSTRAINT invitm252_invno_itmno_pkkey PRIMARY KEY

(invno, itmno));

Table created.

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WEEK 4

2) Queries (along with subqueries) using ANY, ALL, IN, EXISTS, NOT EXISTS, UNIQUE, INTERSECT, Constraints.Example: select the rollno and name of the student who secured 4 th rank in the class

TABLE DEFINITIONS

SQL> CREATE TABLE Customer (cust_no NUMBER(4) PRIMARY KEY,last_name VARCHAR2(20),first_name VARCJHAR2(20) NOT NULL,address1 VARCHAR2(20), address2 VARCHAR2(20), city VARCHAR2(3), state VARCHAR2(20), pin VARCHAR2(6), birth_date DATE, status VARCHAR2(1),

CHECH (status IN (‘V’, ‘I’, ‘A’)));

Table created.

Insert the following data:

1 row created.

CUST NO

LAST NAME

FIRST NAME

ADDRESS1 ADDRESS2 CITY STATE PIN BIRTH DATE

STATUS

1001 UDUPI RAJ UPENDRABAUG NEAR KALPANA

UDPP KARNARATA 576101 12-DEC-62

A

1002 KUMAR RAJ A1003 BAHADUR RAJ SHANTHI VILLA NEAR

MALLIKAUDP KARNATAKA 576101 1-AUG-

70V

1004 SIMON FELIX M-J-56 ALTOBETIM PJM GOA 403002 12-FEB-71

A

1005 KUTTY RAJAN A1 TRADERS NEAR RLY STATION

KNR KERALA 67001 9-JUN-71

A

1006 PAI SHILPA 12/4B POLICE QUARTERS

MNG KARNATAKA 574154 11-DEC-70

I

1007 JAIN RAKSHIT BOSCO R.K PLAZA BNG KARNATAKA 576201 1-JAN-71

A

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QUERIES

1) To list all the fields from the table Customer.SEELCT * FROM Customer;

2) To list the first name, last name.SELECT first_name, last_nameFROM Customer;

3) To list the first name and last name of persons in Karnataka.SELECT first_name, last_nameFROM CustomerWHERE state = ‘KARNATAKA’;

4) To list all the columns for invalid persons. SELECT *FROM CustomerWHERE status = ‘I’;

5) To list the names of active customers. SELECT first_name, last_nameFROM CustomerWHERE status = ‘A’;

6) To list the name and address using concatenation. SELECT first_name || ‘ ‘ || last_name, address1 || ‘,’ || address2 || ‘,’ || city || ‘,’ || state || ‘-‘ || pinFROM Customer;

7) To select records where the pin code has not been entered. SELECT *FROM CustomerWHERE pin IS NULL;

8) To select the single occurrence of any value from the table. SELECT DISTINCT stateFROM Customer;

9) To select rows of valid customers from Karnataka. SELECT *FROM Customer

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WHERE state = ‘KARNATAKA’AND status = ‘V’;

10) To select rows of customers from Karnataka or Kerala. SELECT *FROM CustomerWHERE state = KARNATAKA’OR state = ‘KERALA’;

11) To sort the customer data in the alphabetic order of state. SELECT state, first_name, last_name, pinFROM CustomerORDER BY state;

12) To sort in the descending order. SELECT state, first_name, last_name, pinFROM CustomerORDER BY state DESC;

13) To sort the customer data, state wise and within state by the last name. SELECT state, first_name, last_name, pinFROM CustomerORDER BY state, last_name;

14) To retrieve records of Karnataka customers who are valid. SELECT *FROM CustomerWHERE UPPER(state) = ‘KARNATAKA’AND UPPER(status) = ‘V’;

15) To retrieve records of Karnataka/Kerala customers. SELECT *FROM CustomerWHERE UPPER(state) = ‘KARNATAKA’OR UPPER(state) = ‘KERALA’;

16) To retrieve records of Karnataka/Kerala customers who are active. SELECT *FROM CustomerWHERE (UPPER(state) = ‘KARNATAKA’ OR UPPER(state) = ‘KERALA’)AND UPPER(status) = ‘A’;

17) To retrieve records of Karnataka customers with pin code 576101. SELECT *

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FROM CustomerWHERE LOWER(state) = ‘karnataka’AND pin = ‘576101’;

18) To retrieve rows where the state name begins with K and followed by any other char-acter.

SELECT first_name, last_name, stateFROM CustomerWHERE state LIKE ‘K%’;

19) To retrieve rows where the first name contains the word RAJ embedded in it. SELECT first_name, last_name, stateFROM CustomerWHERE first_name LIKE ‘%RAJ%’;

20) To retrieve rows where the address2 contains the word UDUPI or UDIPI in which the 3rd character may be anything.

SELECT first_name, last_name, stateFROM CustomerWHERE address2 LIKE ‘UD_PI’;

21) To retrieve rows where the cust_no has data representing any value between 1003 and 1005, both numbers included.

SELECT *FROM CustomerWHERE cust_no BETWEEN 1003 AND 1005;

22) To retrieve rows of persons born after 9-JAN-70 and before 1-AUG-96. SELECT *FROM CustomerWHERE birth_date BETWEEN ’10-JAN-70’ AND ’31-JUL-96’;

23) To retrieve rows where the city has data which is equal to UDP or MNG or BNG or PJM or MAR.

SELECT *FROM CustomerWHERE city IN (‘UDP’, ‘MNG’, ‘BNG’, ‘PJM’, ‘MAR’);

TABLE DEFINITIONS

SQL> CREATE TABLE Emp (emp_no NUMBER,emp_name VARCHAR(20),

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join_date DATE,join_basic NUMBER(7, 2),PRIMARY KEY (emp_no)

);Table created.

Insert the following data:

EMP NO EMP NAME JOIN DATE JOIN BASIC1001 Subhas bose 01-JUN-96 30001002 Nadeem shah 01-JUN-96 25001003 Charles babbage 01-JUN-96 30001004 Shreyas kumar 01-JUL-96 25001005 George boole 01-JUL-96 2800

SQL> CREATE TABLE Salary (emp_no NUMBER,basic NUMBER(7, 2), commission NUMBER(7, 2),deduction NUMBER(7, 2),salary_date DATE,FOREIGN KEY (emp_no) REFERENCES Emp

);

Table created.

Insert the following data:

EMP NO BASIC COMMISSION DEDUCTION SALARY DATE

1001 3000 200 250 30-JUN-961002 2500 120 200 30-JUN-961003 3000 500 290 30-JUN-961004 2500 200 300 30-JUN-961005 2800 100 250 30-JUN-961001 3000 200 250 31-JUL-961002 2500 120 200 31-JUL-961003 3000 500 290 31-JUL-961004 2500 200 300 31-JUL-961005 2800 100 150 31-JUL-96

QUERIES

1) To sum the salary of each employee.SELECT emp_no, SUM(basic)FROM salaryGROUP BY emp_no;

2) To sum the salary of each employee and sort it on the sum of basic.

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SELECT emp_no, SUM(basic)FROM salaryGROUP BY emp_noORDER BY SUM(basic);

3) To sum the salary of each employee and sort it in descending order on the sum of ba-sic.

SELECT emp_no, SUM(basic)FROM salaryGROUP BY emp_noORDER BY SUM(basic) DESC;

4) To sum the salary of each employee and sort it in descending order on the sum of ba-sic. Display name also

SELECT s.emp_no, e.emp_name, SUM(s.basic)FROM salary s, emp eWHERE s.emp_no = e.emp_noGROUP BY s.emp_no, e.emp_noORDER BY SUM(s.basic) DESC;

5) To group the data by average salary of each employee.SELECT s.emp_no, INITCAP(e.emp_name), AVG(s.basic)FROM salary s, emp eWHERE s.emp_no = e.emp_noGROUP BY s.emp_no, e.emp_noORDER BY AVG(s.basic);

6) To group the basic by month.SELECT TO_CHAR(salary_date, ‘MONTH’) “MONTH”, SUM(basic) “TOTAL BASIC”FROM salaryGROUP BY TO_CHAR(salary_date, ‘MONTH’);

7) To group the data by average salary of each employee and display where average ba-sic is more than 2000..

SELECT s.emp_no, INITCAP(e.emp_name), AVG(s.basic)FROM salary s, emp eWHERE s.emp_no = e.emp_noGROUP BY s.emp_no, e.emp_noHAVING AVG(s.basic) >= 2000ORDER BY AVG(s.basic);

SUBQUERIES

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SELECT * FROM salaryWHERE basic < (SELECT AVG(basic)

FROM salary);

9) To list the employees whose deduction is 150.SELECT * FROM salaryWHERE emp_no IN (SELECT emp_no

FROM salaryWHERE deduction = 150);

10) To list the names of employees and salary details, whose basic is less than the average salary.

SELECT s.*, e.emp_name FROM salary s, emp eWHERE s.emp_no = e.emp_noAND s.basic < (SELECT AVG(basic)

FROM salary);

WEEK 5

2) Queries (along with subqueries) using ANY, ALL, IN, EXISTS, NOT EXISTS, UNIQUE, INTERSECT, Constraints. Example: select the rollno and name of the student who secured 4 th rank in the class.

3) Queries using Aggregate functions (COUNT, SUM, AVG, MAX and MIN), GROUP BY, HAVING and Creation and Dropping of Views.

4) Queries using Conversions, functions (to_char, to_num, and to_date), string function (Conactenation, lpad, rpad, ltrim, rtrim, lower, upper, initcap, length, substr, and instr), date functions (sys-date, next_day, add_months, last_day, months_between, least, great-est, trunk, round, to_char, to_date).

TABLE DEFINITIONS

SQL> CREATE TABLE Emp (emp_no NUMBER,emp_name VARCHAR(20),

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join_date DATE,join_basic NUMBER(7, 2),PRIMARY KEY (emp_no)

);

Table created.

Insert the following data:

EMP NO EMP NAME JOIN DATE JOIN BASIC1001 Subhas bose 01-JUN-96 30001002 Nadeem shah 01-JUN-96 25001003 Charles babbage 01-JUN-96 30001004 Shreyas kumar 01-JUL-96 25001005 George boole 01-JUL-96 2800

SQL> CREATE TABLE Salary (emp_no NUMBER,basic NUMBER(7, 2), commission NUMBER(7, 2),deduction NUMBER(7, 2),salary_date DATE,FOREIGN KEY (emp_no) REFERENCES Emp

);

Table created.

Insert the following data:

EMP NO BASIC COMMISSION DEDUCTION SALARY DATE

1001 3000 200 250 30-JUN-961002 2500 120 200 30-JUN-961003 3000 500 290 30-JUN-961004 2500 200 300 30-JUN-961005 2800 100 250 30-JUN-961001 3000 200 250 31-JUL-961002 2500 120 200 31-JUL-961003 3000 500 290 31-JUL-961004 2500 200 300 31-JUL-961005 2800 100 150 31-JUL-96

QUERIES

11) To sum the salary of each employee.SELECT emp_no, SUM(basic)FROM salaryGROUP BY emp_no;

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12) To sum the salary of each employee and sort it on the sum of basic. SELECT emp_no, SUM(basic)FROM salaryGROUP BY emp_noORDER BY SUM(basic);

13) To sum the salary of each employee and sort it in descending order on the sum of ba-sic.

SELECT emp_no, SUM(basic)FROM salaryGROUP BY emp_noORDER BY SUM(basic) DESC;

14) To sum the salary of each employee and sort it in descending order on the sum of ba-sic. Display name also

SELECT s.emp_no, e.emp_name, SUM(s.basic)FROM salary s, emp eWHERE s.emp_no = e.emp_noGROUP BY s.emp_no, e.emp_nameORDER BY SUM(s.basic) DESC;

15) To group the data by average salary of each employee.SELECT s.emp_no, INITCAP(e.emp_name), AVG(s.basic)FROM salary s, emp eWHERE s.emp_no = e.emp_noGROUP BY s.emp_no, e.emp_noORDER BY AVG(s.basic);

16) To group the basic by month.SELECT TO_CHAR(salary_date, ‘MONTH’) “MONTH”, SUM(basic) “TOTAL BASIC”FROM salaryGROUP BY TO_CHAR(salary_date, ‘MONTH’);

17) To group the data by average salary of each employee and display where average ba-sic is more than 2000..

SELECT s.emp_no, INITCAP(e.emp_name), AVG(s.basic)FROM salary s, emp eWHERE s.emp_no = e.emp_noGROUP BY s.emp_no, e.emp_noHAVING AVG(s.basic) >= 2000ORDER BY AVG(s.basic);

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18) To list the employees who earn less than the average salary.SELECT * FROM salaryWHERE basic < (SELECT AVG(basic)

FROM salary);

19) To list the employees whose deduction is 150.SELECT * FROM salaryWHERE emp_no IN (SELECT emp_no

FROM salaryWHERE deduction = 150);

20) To list the names of employees and salary details, whose basic is less than the average salary.

SELECT s.*, e.emp_name FROM salary s, emp eWHERE s.emp_no = e.emp_noAND s.basic < (SELECT AVG(basic)

FROM salary);

WEEK 6

2) Queries (along with subqueries) using ANY, ALL, IN, EXISTS, NOT EXISTS, UNIQUE, INTERSECT, Constraints. Example: select the rollno and name of the student who secured 4 th rank in the class.

3) Queries using Aggregate functions (COUNT, SUM, AVG, MAX and MIN), GROUP BY, HAVING and Creation and Dropping of Views.

4) Queries using Conversions, functions (to_char, to_num, and to_date), string function (Conactenation, lpad, rpad, ltrim, rtrim, lower, upper, initcap, length, substr, and instr), date functions (sys-date, next_day, add_months, last_day, months_between, least, great-est, trunk, round, to_char, to_date).

TABLE DEFINITIONS

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Branch Schema <branch-name, branch-city, assets>Customer Schema <customer-name, customer-street, customer-city>Loan Schema <loan-number, branch-name, amount> Borrower Schema <customer-name, loan-number>Account Scheme <account-number, branch-name, balance>Depositor Scheme <customer-name, account-number>

BRANCH TABLE

Branch Name Branch City AssetsBrighton Brooklyn 7100000Downtown Brooklyn 9000000Mianus Horseneck 400000North Town Rye 3700000Perryridge Horseneck 1700000Pownal Bennington 300000Redwood Palo Alto 2100000Round Hill Horseneck 800000

CUSTOMER TABLE

Customer Name Customer Street Customer CityAdams Spring PittsfieldBrooks Senator BrooklynCurry North RyeGlenn Sand Hill WoodsideGreen Walnut StamfordHayes Main HarrisonJohnson Alma Palo AltoJones Main HarrisonLindsay Park PittsfieldSmith North RyeTurner Putnam StamfordWilliams Nassau Princeton

LOAN TABLE

Loan Number Branch Name AmountL-11 Round Hill 900L-14 Downtown 1500L-15 Perryridge 1500L-16 Perryridge 1300

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L-17 Downtown 1000L-23 Redwood 2000L-93 Mianus 500

BORROWER TABLE

Customer Name Loan NumberAdams l-16Curry L-93Hayes L-15Jackson L-14Jones L-17Smith L-11Smith L-23Williams L-17

ACCOUNT TABLE

Account Number Branch Name BalanceA-101 Downtown 500A-102 Perryridge 400A-201 Brighton 900A-215 Mianus 700A-217 Brighton 750A-222 Redwood 700A-305 Round Hill 350

DEPOSITOR TABLE

Customer Name Account NumberHayes A102Johnson A-101Johnson A-201Jones A-217Lindsay A-222Smith A-215Turner A-305

QUERIES

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1) To list all the fields from the table Customer. SELECT branch_nameFROM Loan;

2) To list rows after eliminating duplicates.

SELECT distinct branch_nameFROM Loan;

3) To explicitly list rows, including duplicates.

SELECT all branch_nameFROM Loan;

4) To list fields after applying arithmetic operations.

SELECT loan_number, branch_name, amount *100FROM Loan;

5) Find all loan numbers for loans made at the Perryridge branch with loan amountsgreater than Rs1200.

SELECT loan_numberFROM LoanWHERE branch_name = ‘Perryridge’AND amount > 1200;

6) Find all loan numbers for loans with loan amounts between Rs90,000 and Rs100,000.

SELECT loan_numberFROM LoanWHERE amount BETWEEN 90000 AND 100000;

Or

SELECT loan_numberFROM LoanWHERE amount <= 100000AND amount >= 90000;

7) Find all loan numbers for loans with loan amounts not between Rs90,000 andRs100,000.

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SELECT loan_numberFROM LoanWHERE amount NOT BETWEEN 90000 AND 100000;

8) For all customers who have a loan from the bank, find their names, loan numbers and loan amounts.

SELECT customer_name, Borrower.loan_number, amountFROM Borrower, LoanWHERE Borrower.loan_number = Loan.loan_number;

Or

SELECT customer_name, Borrower.loan_number AS loan_id, amountFROM Borrower, LoanWHERE Borrower.loan_number = Loan.loan_number;

9) Find the customer names, loan numbers and loan amounts for all loans at the Perryridge branch.

SELECT customer_name, Borrower.loan_number, amountFROM Borrower, LoanWHERE Borrower.loan_number = Loan.loan_numberAND branch_name = ‘Perryridge’;

Or

SELECT customer_name, T.loan_number, S.amountFROM Borrower AS T, Loan AS SWHERE T.loan_number = S.loan_numberAND branch_name = ‘Perryridge’;

10) Find the names of all branches that have assets greater than atleast one branch located in Brooklyn.

SELECT DISTINCT T.branch_nameFROM Branch as T, Branch as SWHERE T.assets > S.assetsAND S.branch_city = ‘Brooklyn’;

11) Find the names of all customers whose street address includes the sub-string ‘Main’.

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FROM CustomerWHERE customer_street LIKE ‘%Main%’;

12) To list in alphabetic order all customers who have a loan at the Per-ryridge branch.

SELECT DISTINCT customer_nameFROM Borrower B, Loan LWHERE B.loan_number = L.loan_numberAND branch_name = ‘Perryridge’ORDER BY customer_name;

13) To list the entire loan info in descending order of amont.

SELECT *FROM LoanORDER BY amount DESC, loan_number ASC;

14) To find all customers having a loan, an account or both at the bank, without duplicates.

(SELECT customer_nameFROM Depositor)UNION(SELECT customer_nameFROM Borrower);

15) To find all customers having a loan, an account or both at the bank, with duplicates.

(SELECT customer_nameFROM Depositor)UNION ALL(SELECT customer_nameFROM Borrower);

16) To find all customers having both a loan and an account at the bank, without duplicates.

(SELECT customer_nameFROM Depositor)INTERSECT(SELECT customer_name

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FROM Borrower);

17) To find all customers having a loan, an account or both at the bank, with duplicates.

(SELECT customer_nameFROM Depositor)INTERSECT ALL(SELECT customer_nameFROM Borrower);

18) To find all customers who have an account but no loan at the bank, without duplicates.

(SELECT DISTINCT customer_nameFROM Depositor)EXCEPT(SELECT customer_nameFROM Borrower);

19) To find all customers who have an account but no loan at the bank, with duplicates.

(SELECT DISTINCT customer_nameFROM Depositor)EXCEPT ALL(SELECT customer_nameFROM Borrower);

20) Find the average account balance at the Perryridge branch

SELECT branch_name, AVG(balance)FROM AccountWHERE branch_name = ‘Perryridge’;

21) Find the average account balance at the each branch

SELECT AVG(balance)FROM AccountGROUP BY branch_name;

22) Find the number of depositors for each branch .

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SELECT branch_name, COUNT(DISTINCT customer_name)FROM Depositor D, Account AWHERE D.account_number = A.account_numberGROUP BY branch_name;

23) Find the number of depositors for each branch where average account balance is more than Rs 1200.

SELECT branch_name, COUNT(DISTINCT customer_name)FROM Depositor D, Account AWHERE D.account_number = A.account_numberGROUP BY branch_nameHAVING AVG(balance) > 1200;

24) Find the average balance for all accounts.

SELECT AVG(balance)FROM Account;

25) Find the number of tuples in the customer relation.

SELECT COUNT(*)FROM Customer;

26) Find the average balance for each customer who lives in Harrision and has at least three accounts.

SELECT D.customer_name, AVG(balance)FROM Depositor D, Account A, Customer CWHERE D.account_number = A.account_numberAND D.customer_name = C.customer_nameAND C.customer_city = ‘Harrison’GROUP BY D.customer_nameHAVING COUNT(DISTINCT D.account_number) >= 3;

27) Find all the loan number that appear in loan relation with null amount values.

SELECT loan_numberFROM LoanWHERE amount IS NULL;

28) Find all customers who have both a loan and an account at the bank.

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SELECT customer_nameFROM BorrowerWHERE customer_street IN (SELECT customer_nameFROM Depositor);

29) Find all customers who have both an account and a loan at the Per-ryridge branch

SELECT DISTINCT B.customer_nameFROM Borrower B, Loan LWHERE B.loan_number L.loan_numberAND branch_name = ‘Perryridge’AND (branch_name, customer_name) IN(SELECT branch_name, customer_nameFROM Depositor D, Account AWHERE D.account_number = A.account_number);

or

SELECT customer_nameFROM Borrower BWHERE EXISTS (SELECT *FROM Depositor DWHERE D.customer_name = B.customer_name);

30) Find all customers who do not have a loan at the bank, but do not have an account the bank.

SELECT DISTINCT customer_nameFROM BorrowerWHERE customer_name NOT IN(SELECT customer_nameFROM Depositor);

31) Find the names of customers who do have a loan at the bank, and whose names are neither Smith nor Jones.

SELECT DISTINCT customer_nameFROM BorrowerWHERE customer_name NOT IN (‘Smith’, ‘Jones’);

32) Find the names of all branches that have assets greater than those of at least one branch located in Brooklyn.

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SELECT DISTINCT T.branch_nameFROM Branch AS T, Branch AS SWHERE T.assets > S.assetsAND S.branch_city = ‘Brooklyn’;

33) Find the names of all branches that have assets greater than that of each branch located in Brooklyn.

SELECT branch_nameFROM AccountGROUP BY branch_nameHAVING AVG(balance) >= ALL (SELECT AVG(balance)FROM AccountGROUP BY branch_name);

34) Find all customers who have an account at all the branches located in Brooklyn.

SELECT DISTINCT S.customer_nameFROM Depositor AS DWHERE NOT EXISTS ((SELECT branch_nameFROM BranchWHERE branch_city = ‘Brroklyn)EXCEPT(SELECT R.branch_nameFROM Depositor AS T, Account AS RWHERE T.account_number =R.account_numberAND D.customer_name = t.customer_name));

35) Find all customers who have at most one account at the Perryridge branch.

SELECT T.customer_nameFROM Depositor AS TWHERE UNIQUE (SELECT R.customer_nameFROM Depositor AS R, Account AS AWHERE T.customer_name = R.customer_nameAND R.account_number = A.account_numberAND A.branch_name = ‘Perryridge’);

36) Find all customers who have at least two accounts at the Perryridge branch.

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SELECT DISTINCT T.customer_nameFROM Depositor AS TWHERE NOT UNIQUE (SELECT R.customer_nameFROM Depositor AS R, Account AS AWHERE T.customer_name = R.customer_nameAND R.account_number = A.account_numberAND A.branch_name = ‘Perryridge’);

37) Find the average account balance of those branches where the average account balance is greater than 1200.

SELECT branch_name, avg_balanceFROM (SELECT branch_name, AVG(balance)FROM AccountGROUP BY branch_name)AS Branch_avg(branch_name, avg_balance)WHERE avg_balance > 1200;

38) Find the maximum across all branches of the total balance at each branch.

SELECT MAX(tot_balance)FROM (SELECT branch_name, SUM(balance)FROM AccountGROUP BY branch_name)AS Branch_total(branch_name, tot_balance);

39) Find the all customers who have an account but no loan at the bank.

SELECT d-CNFROM (Depositor LEFT OUTER JOIN BorrowerON Depositor.customer_name = Borrower.customer_name)AS db1(d-CN, account_number, b-CN, loan_number)WHERE b-CN is null;

40) Find the all customers who have either an account or a loan (but not both) at the bank.

SELECT customer_nameFROM (Depositor NATURAL FULL OUTER JOIN Borrower)WHERE account_number IS NULLOR loan_number IS NULL;

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WEEK 7

5) Queries (along with subqueries) using ANY, ALL, IN, EXISTS, NOT EXISTS, UNIQUE, INTERSECT, Constraints. Example: select the rollno and name of the student who secured 4 th rank in the class.

6) Queries using Aggregate functions (COUNT, SUM, AVG, MAX and MIN), GROUP BY, HAVING and Creation and Dropping of Views.

7) Queries using Conversions, functions (to_char, to_num, and to_date), string function (Conactenation, lpad, rpad, ltrim, rtrim, lower, upper, initcap, length, substr, and instr), date functions (sys-date, next_day, add_months, last_day, months_between, least, great-est, trunk, round, to_char, to_date).

DUAL (ORACLE WORK TABLE):

1) To display system date.SELECT SYSDATE FROM DUAL;

2) To display arithmetic calculations.

SELECT 2*2 FROM DUAL;

3) To display the logged user.SELECT USER FROM DUAL;

4) To display system time.SELECT TO_CHAR(SYSDATE, ‘HH:MI:SS’) FROM DUAL;

5) To display current month.SELECT TO_CHAR(SYSDATE, ‘MONTH’) FROM DUAL;

6) To display system date in specified format.SELECT TO_CHAR(SYSDATE, ‘DD/MM/YY’) FROM DUAL;

7) To display system date in specified format.

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SELECT TO_CHAR(SYSDATE, ‘MM’) FROM DUAL;

8) To display date arithmetic.SELECT ADD_MONTHS(SYSDATE, 5) FROM DUAL;

9) To display date arithmetic.SELECT LAST_DAY(SYSDATE) FROM DUAL;

10) To display date arithmetic.SELECT MONTHS_BETWEEN(SYSDATE, ’01-APR-09’) FROM DUAL;

11) To display date arithmetic.SELECT NEXT_DAY(SYSDATE, ‘MON’) FROM DUAL;

GROUP FUNCTIONS:

12) To display average basic salary of the employees.SELECT SUM(basic) FROM salary;

13) To display minimum basic salary of the employees.SELECT MIN(basic) FROM salary;

14) To display maximum basic salary of the employees.SELECT MAX(basic) FROM salary;

15) To display sum of basic salaries of all the employees.SELECT SUM(basic) FROM salary;

16) To display the number of records in salary table.SELECT COUNT(*) FROM salary;

STRING FUNCTIONS:

17) To display a field value after left padding.SELECT LPAD('PAGE-1', 10, '*') FROM DUAL;

18) To display a field value after left padding.SELECT RPAD('PAGE-1', 10, '*') FROM DUAL;

19) To display a field value after converting to lower case.SELECT LOWER(‘A’) FROM DUAL;

20) To display a field value after converting to upper case.SELECT LOWER(‘a’) FROM DUAL;

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21) To display a field value after converting to initial capital case.SELECT INITCAP(‘HOW ARE YOU?’) FROM DUAL;

22) To display a substring of a field value.SELECT SUBSTR(‘CSE2A’, 4, 2) FROM DUAL;

23) To display the length of a field value.SELECT LENGTH(’HOW LONG AM I?’) FROM DUAL;

24) To display a field value after trimming the right side.SELECT RTRIM(‘CSE2A’, ‘2A’) FROM DUAL;

25) To display a field value after trimming the left side.SELECT LTRIM(‘CSE2A’, ‘CSE’) FROM DUAL;

WEEK 8 (PL/SQL)

6) (i) Creation of simple PL/SQL program which includes declaration section, executable section and exception handling section ( ex: Stu-dent marks can be selected from the table and printed for those who secured first class and an exception can be raised if no records were found).(ii) Insert data into student table and use COMMIT, ROLLBACK and SAVEPOINT in SQL block.

7) Develop a program that includes the features NESTED IF, CASE and CASE expression. The program can be extended using the NUL-LIF and COALESCE functions.

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8) Program development using WHILE LOOPS, numeric FOR LOOPS, nested loops using ERROR handling, BUILT IN exceptions, USER defined exceptions, RAISE APPLICATION ERROR.

9) Program development using creation of procedure, passing parame-ters IN and OUT procedures.

10) Program development using creation of stored function, invoke functions in SQL statements and write complex functions.

11) Program development using creation of package specification, package bodies, private objects, package variables and cursors and calling stored packages.

12) Develop programs using features of parameters in a CURSOR, FOR UPDATE CURSOR, WHERE CURRENT of clause and CUR-SOR variables.

Syntax to write a sql program

Declare<declaration stmts>Begin<executable stmts>[exception <exceptional stmts>]----- optionalEnd;/---end of buffer

Example: 1

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Create a file DBFOR.SQL, to execute the FOR loop and display the variable.

At SQL Prompt type, ed dbfor to open notepad and type the below program:Program

declarecnt number;begindbms_output.put_line('This is a demo of FOR loop ');for cnt in 1..5 loop

dbms_output.put_line('loop number ' || cnt);end loop;end;/set serveroutput off

Save the file and at SQL prompt run as:ExecutionSQL>set serveroutput onSQL> start dbfor (press enter) ORSQL> @dbfor

OUTPUT:-

This ia a demo of FOR loop

loop number 1

loop number 2

loop number 3

loop number 4

loop number 5

PS:For syntax:For <var> in <start_num> .. <endnum> loop

<statement(s);>End loop;

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Example: 2Create a file DBREVFOR.SQL, to execute the REVERSE FOR loop and display the variable.

Programbegin

dbms_ouput.put_line(‘This is a demo of REVERSE FOR loop’);for cnt in reverse 1..10 loop

if mod(cnt, 2) = 0 thendbms_output.put_line(‘loop counter ‘ || cnt);

end if;end loop;

end;/

OUTPUT:-

This is a demo of REVERSE FOR loop

loop counter 10loop counter 8loop counter 6loop counter 4loop counter 2

PS:

Reverse For syntax:For <var> in reverse <start_num> .. <endnum> loop

<statement(s);>End loop;

Other forms of if syntax are:If <condition> then

<action(s);>End if;

If <condition> then<action(s);>

Else<action(s);>

End if;

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If <condition> then<action(s);>

Elsif <condition> then<action(s);>

else<action(s);>

End if;

Example: 3Create a file DBLOOP.SQL, to execute the LOOP loop and display the variable.

Programset serveroutput ondeclare

cnt number(2) := 0;begin

dbms_ouput.put_line(‘This is a demo of LOOP loop’);loop

cnt := cnt + 1;exit when cnt > 10;dbms_output.put_line(‘loop counter ‘ || cnt);

end loop;end;/set serveroutput off

OUTPUT:-

This is the demo of LOOP loop

loop counter 1

loop counter 2

loop counter 3loop counter 4loop counter 5loop counter 6

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loop counter 7loop counter 8loop counter 9loop counter 10

PS:Loop syntax:loop

<statement(s);>Exit when <condition>;

End loop;

Example: 4Create a file DBWHILE.SQL, to execute the WHILE loop and display the variable.

Programset serveroutput ondeclare

cnt number(2) := 1;begin

dbms_ouput.put_line(‘This is a demo of WHILE loop’);while cnt <= 10 loop dbms_output.put_line(‘loop counter: ‘ || to_char(cnt, ‘999’));

cnt := cnt + 1;end loop;

end;/set serveroutput off

OUTPUT:-

This is a demo of WHILE loop

loop counter : 1

loop counter : 2

loop counter : 3loop counter : 4loop counter : 5loop counter : 6

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loop counter : 7loop counter : 8loop counter : 9loop counter : 10

PS:while syntax:while <condition> loop

<statement(s);>End loop;

Example: 4Write a program EMPDATA.SQL, to retrieve the employee details of an employee whose number is input by the user .

Program-- PROGRAM TO RETRIEVE EMP DETAILSset serveroutput on

prompt Enter Employee Number: accept ndeclare

dname emp.emp_name%type;dbasic emp.emp_basic%type;ddesig emp.desig%type;

beginselect emp_name, basic, designinto dname, dbasic, ddesigfrom empwhere emp_no = &n;dbms_ouput.put_line(‘Employee Details:);dbms_output.put_line(‘Name: ‘ || dname); dbms_output.put_line(‘Basic: ‘ || dbasic);dbms_output.put_line(‘Designation: ‘ || ddesig);

end;/

OUTPUT:-

enter employee number:

13

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old 9:where eno =&n;

new 9:where eno=13;

employee details

Name:allen

basic:9500

desig:mech

set serveroutput off

PS:Similarly you can use other SQL statements in the PL/SQL block

Exercises:

1) Write a PL/SQL code, EX_INVNO.SQL, block for in-verting a number using all forms of loops.

ANSWER:-

declare

n number(20):=123;

s number(13):=0;

d number(3):=1;

r number(3):=10;

begin

dbms_output.put_line('the number is :' || n);

while n>0 loop

d:=mod(n,10);

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s:=(s*r)+d;

n:=n/r;

end loop;

dbms_output.put_line('inverted values' || s);

end;

/

OUTPUT:-

the number is:123

inverted value is:321

2) Write a PL/SQL code, EX_SUMNO.SQL that prints the sum of ‘n’ natural numbers.

ANSWER:-

prompt enter number:

accept number n

declare

isum number(2):=0;

i number;

n number:=&n;

begin

for i in 1..n loop

isum:=isum+i;

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end loop;

dbms_output.put_line('sum is ' || isum);

end;

/

OUTPUT:-

enter the number:7

sum is 28

3) Write a PL/SQL code, EX_AREA.SQL, of block to cal-culate the area of the circle for the values of radius varying from 3 to 7. Store the radius and the corre-sponding values of calculated area in the table AREA_VALUES.

ANSWER:-

set serveroutput on

declare

area number(5);

rad number(3);

pi number(4):=3.14;

begin

for rad in 3..7 loop

area:=pi*rad*rad;

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dbms_output.put_line('area is' || area);

insert into area_values values(area,rad);

end loop;

end;

/

OUTPUT:-

area is :27area is :48area is :75area is :108area is :147

SQL>select * from area_values;area rad____ ____27 348 475 5108 6147 7

WEEK 9 (PL/SQL)

13) (i) Creation of simple PL/SQL program which includes decla-ration section, executable section and exception handling section ( ex: Student marks can be selected from the table and printed for those who secured first class and an exception can be raised if no records were found).(ii) Insert data into student table and use COMMIT, ROLLBACK and SAVEPOINT in SQL block.

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14) Develop a program that includes the features NESTED IF, CASE and CASE expression. The program can be extended using the NULLIF and COALESCE functions.

15) Program development using WHILE LOOPS, numeric FOR LOOPS, nested loops using ERROR handling, BUILT IN exceptions, USER defined exceptions, RAISE APPLICATION ERROR.

16) Program development using creation of procedure, passing pa-rameters IN and OUT procedures.

17) Program development using creation of stored function, invoke functions in SQL statements and write complex functions.

18) Program development using creation of package specification, package bodies, private objects, package variables and cursors and calling stored packages.

19) Develop programs using features of parameters in a CURSOR, FOR UPDATE CURSOR, WHERE CURRENT of clause and CUR-SOR variables.

Example: 1Create a file (NEWINS.SQL), to insert into a new table, NEWEMP, the record of any employee whose number is input by the user.

1. Create the table NEWEMP <emp_no, emp_name, join_date, basic).2. Open an editor and type the following program.

Programprompt Enter Employee Number: accept userno number

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declaredno number(4);dname varchar2(30);ddate date;dbasic number(10);

beginselect emp_no, emp_name, join_date, basicinto dno, dname, ddate, dbasicfrom empwhere emp_no = &userno;

if sql%rowcount > 0then

insert into newempvalues (dno, dname, ddate, dbaisc);

end if;end;/

3. Save the file as NEWINS4. Execute the program as

SQL> start newins

Example: 2Create a file (NEWINS2.SQL), to insert into a new table, NEWEMP, the record of any employee whose number is input by the user. Also display on the screen the employee details and to handle errors like user entering a number which does not exist in the table.

Programprompt Enter Employee Number: accept userno numberdeclare

dno number(4);dname varchar2(30);ddate date; dbasic number(10);

beginselect emp_no, emp_name, join_date, basicinto dno, dname, ddate, dbasicfrom empwhere emp_no = &userno;

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if sql%rowcount > 0then

insert into newempvalues (dno, dname, ddate, dbasic);

dbms_output.put_line(‘Record inserted into NEWEMP’);dbms_output.put_line(DNO || ‘ ‘ || DNAME || ‘ ‘ || DDATE || ‘ ‘ ||

DBASIC);end if;

exceptionwhen no_data_found then

dbms_output.put_line (‘Record ‘ || &userno || ‘ does not exist’);end;/

Example: 3Create a file (CALCTAX.SQL), to calculate tax for a specific employee and display name and tax.

Program prompt Enter Employee Number: accept userno number

declaretot_basic number(10, 2);tax number(10, 2);name varchar2(30);

beginselect emp_name, basicinto name, tot_basicfrom empwhere emp_no = &userno;

if tot_basic = 0 or tot_basic is null then

dbms_output.put_line(‘NO BASIC’);elsif tot_basic <= 2000then

tax := tot_basic * .02; dbms_output.put_line(NAME || ‘ TOTAL BASIC: ‘ || TOT_BASIC); dbms_output.put_line(NAME || ‘ TOTAL TAX: ‘ || TAX);

elsetax := tot_basic * .04;

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dbms_output.put_line(NAME || ‘ TOTAL BASIC: ‘ || TOT_BASIC); dbms_output.put_line(NAME || ‘ TOTAL TAX: ‘ || TAX);

end if;

exceptionwhen no_data_found then

dbms_output.put_line (‘Record ‘ || &userno || ‘ does not exist’);end;/

PS:EXECPTIONSWhen a program is executed certain errors are automatically recognized and certain error situations must be recognized by the program itself. Errors in general are referred to as Exceptions.Exceptions can be either System defined or User defined.Certain system exceptions raise the following flags:CURSOR_ALREADY_OPEN – Displayed when the user tries to open a cursor that is already openDUP_VAL_ON_INDEX – when user tries to insert a duplicate value into a unique columnINVALID_CURSOR – when user references an invalid cursor or attempts an illegal cursor operationINVALID_NUMBER – when user tries to use something other than a number where one is called forLOGIN_DENIED – when connect request for user has been deniedNO_DATA_FOUND – this flag becomes TRUE when SQL select statement failed to retrieve any rowsNOT_LOGGED_ON – user is not connected to ORACLEPROGRAM_ERROR – user hits a PL/SQL internal error STORAGE_ERROR – user hits a PL/SQL memory errorTIMEOUT_ON_RESOURCE – user has reached timeout while waiting for an Oracle resourceTRANSACTION_BACKED_OUT – a remote server has rolled back the transactionTOO_MANY_ROWS – the flag becomes TRUE when SQL select statement retrieves more than one row and it was supposed to retrieve only 1 rowVALUE_ERROR – user encounters an arithmetic, conversion, truncation or constraint error

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ZERO_DIVIDE – flag becomes TRUE if SQL select statement tries to divide a number by 0OTHERS – this flag is used to catch any error situations not coded by the programmerIn the exception section and must appear last in the exception section

User defined exceptions must be declared in the declare section with the reserved word, EXCEPTION.

Syntax for user defined exception:<exception-name> EXCEPTION;

This exception can be brought into action by the command,RAISE <exception-name>

When the exception is raised, processing control is passed to the EXCEPTION section of the PL/SQL block.The code for the exception must be defined in the EXCEPTION section of the PL/SQL block.

WHEN <exception-name> THEN<action>;

Exercises:

1) Write a PL/SQL code block that will accept an account number from the user and debit an amount of RS2000 from the account. If the ac-count has a minimum balance of 500 after amount is debited the process should set a freeze on the account by setting the status to F.(use table schema Accounts (acno, balance, status)

2) Write a PL/SQL block of code to achieve the following:If the price of the product is >4000 then change the price to 4000. The price change is to be recorded in the old price table along with product number and date on which the price was last changed. (use table schemas Product(pno, price) and Old_Price(pno, date_of_change, oldprice)

WEEK 10 (PL/SQL)

20) (i) Creation of simple PL/SQL program which includes decla-ration section, executable section and exception handling section ( ex: Student marks can be selected from the table and printed for those

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who secured first class and an exception can be raised if no records were found).(ii) Insert data into student table and use COMMIT, ROLLBACK and SAVEPOINT in SQL block.

21) Develop a program that includes the features NESTED IF, CASE and CASE expression. The program can be extended using the NULLIF and COALESCE functions.

22) Program development using WHILE LOOPS, numeric FOR LOOPS, nested loops using ERROR handling, BUILT IN exceptions, USER defined exceptions, RAISE APPLICATION ERROR.

23) Program development using creation of procedure, passing pa-rameters IN and OUT procedures.

24) Program development using creation of stored function, invoke functions in SQL statements and write complex functions.

25) Program development using creation of package specification, package bodies, private objects, package variables and cursors and calling stored packages.

26) Develop programs using features of parameters in a CURSOR, FOR UPDATE CURSOR, WHERE CURRENT of clause and CUR-SOR variables.

Example: 1Create a PL/SQL program using cursors, to retrieve first tuple from the department relation.(use table dept(dno, dname, loc))

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Program declare

vdno dept.deptno%type;vdname dept.dname%type;vloc dept.loc%type;cursor c1 is select * from dept;or // cursor c1 is select * from dept where rowno = 1;

beginopen c1;fetch c1 into vdno,vdname,vloc;dbms_output.put_line('vdno = ' ||vdno|| ' vdname = '||vdname||' vloc

= '||vloc);close c1;

end;/

PS:Cursors are used when the SQL select statement is expected to return more than 1 row. A cursor must be declared and its definition contains a query and is defined in the DECLARE section of the program.A cursor must be opened before processing and closed after processing. (Similar to how files are opened and closed in a C program).

Syntax to define a cursor:CURSOR <CURSOR-NAME> IS <SELECT STATEMENT>

Syntax to open the cursor:OPEN <CURSOR-NAME>

Syntax to store data in the cursor:FETCH <CURSOR-NAME> INTO <VAR1>, <VAR2>, <VAR3>….

ORFETCH <CURSOR-NAME> INTO <RECORD-NAME>

Syntax to close the cursor:CLOSE <CURSOR-NAME>

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Example: 2Create a PL/SQL program using cursors, to retrieve each tuple from the department relation.(use table dept(dno, dname, loc))

Program declare

vdept dept%rowtype;cursor c1 is select * from dept;

beginfor vdept in c1 loop

dbms_output.put_line('vdno = ' ||vdept.deptno|| ' vdname = '||vdept.dname||' vloc = '||vdept.loc);

end loop;

end;/

PS:The cursor for loop can be used to process multiple records. The advantage of cursor for loop is that the loop itself will open the cursor, read the records into the cursor from the table until end of file and close the cursor.

Syntax for cursor FOR LOOP:FOR <VARIABLE> IN <CURSOR-NAME> LOOP

<STATEMENTs>END LOOP;

Example: 3

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Create a PL/SQL program using cursors, to display the number, name, salary of the three highest paid employees.(use table emp(empno, ename,sal))

Program declare

no emp.empno%type;name emp.ename%type;salary emp.sal%type;cursor c1 is select empno, ename, sal from emp order by sal desc;

beginopen c1;loop

fetch c1 into no,name,salary;exit when c1 %notfound;exit when c1 %rowcount >3;dbms_output.put_line(no||name||salary);

end loop;close c1;

end;/

PS:Cursors Attributes:There are 4 cursor attributes used to provide information on the status of a cursor.%NOTFOUND – To determine if a row was retrieved

Used after FETCHNOTFOUND is TRUE if row is not retrievedNOTFOUND is FALSE if row is retrieved

%FOUND – To determine if a row was retrieved.Used after FETCHFOUND is TRUE if row is retrievedFOUND is FALSE if row is not retrieved

%ROWCOUNT – To determine the number of rows retrievedROWCOUNT is 0 when cursor is openedROWCOUNT returns the number of rows retrieved

%ISOPEN – To determine the cursor is open

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ISOPEN is TRUE if a cursor is openISOPEN is FALSE if a cursor is not open

Example: 4Create a PL/SQL program using cursors, to delete the employees whose salary is more than 3000.

Program declare

vrec emp%rowtype;cursor c1 is select * from emp where sal>3000 for update;begin

open c1;loop

fetch c1 into vrec;exit when c1 %notfound;delete from emp where current of c1;dbms_output.put_line('Record deleted');

end loop;close c1;

end;/

PS:In order to DELETE or UPDATE rows, the cursor must be defined with the FOR UPDATE clause.

Example: 5 Create a PL/SQL program using cursors, to update the salary of each employee by the avg salary if their salary is less than avg salary.

Program

declare vrec emp%rowtype;avgsal number(10,2);

cursor c1 is select * from emp for update;

begin

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select avg(sal) into avgsal from emp;for vrec in c1 loop

if vrec.sal < avgsal thenvrec.sal := avgsal;update emp set sal = vrec.sal where current of c1;dbms_output.put_line('Record updated');

end if;end loop;

end;/

PS:Variable Attributes:%TYPE - is used in PL/SQL to declare a variable to be of the same type as a previously declared variable or to be of the same type as a column in a table.

TOTBASIC SALARY.BASIC%TYPE;will declare TOTBASIC of the same type as BASIC column from the table SALARY.

%ROWTYPE – declares a variable which is actually a record which has the same structure as a row from a table.

SALREC SALARY%ROWTYPE;will declare SALREC as a record variable equivalent to the row from the table SALARY.

Example: 6 Create a PL/SQL program using cursors, to insert into a table, NEWEMP, the record of ALL MANAGERS. Also DISPLAY on the screen the NO, NAME, JOIN_DATE. Handle any user defined exceptions.(use table emp(emp_no, emp_name, join_date, desig))

Program

set serveroutput ondeclare

ctr number(2) := 2;dno number(4);dname varchar2(30);ddate date;

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cursor cur_mgr is select emp_no, emp_name, join_datefrom emp where upper(desig) = ‘MGR’;

no_manager_found exception;

beginopen cur_mgr;loop

fetch cur_mgrinto dno, dname, ddate;

exit when cur_mgr%notfound;ctr := ctr + 1;

dbms_output.put_line(ctr || ‘Record inserted into NEWEMP’);

dbms_output.put_line(dno || ‘ ‘ || dname || ‘ ‘ ddate);

insert into new empvalues (dno, dname, ddate);

end loop;

if cur_mgr%rowcount = 0then

close cur_mgr;raise no_manager_found;

end if;

dbms_output.put_line(‘TOTAL number of records’ || ctr);close cur_mgr;

exceptionwhen no_manager_found then

dbms_output.put_line(‘NO RECORS FOUND’);

end;

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/

Exercises:1) Create a PL/SQL program using cursors, to insert into a table,

NEWEMP, for any designation input by the user from the keyboard. Handle any user defined exceptions.

2) Code a program to calculate Tax for any employee whose number is in-put from the keyboard. Display appropriate error message if data does not exist in the table.

WEEK 11 (PL/SQL)

27) (i) Creation of simple PL/SQL program which includes decla-ration section, executable section and exception handling section ( ex: Student marks can be selected from the table and printed for those who secured first class and an exception can be raised if no records were found).

(ii) Insert data into student table and use COMMIT, ROLLBACK and SAVEPOINT in SQL block.

28) Develop a program that includes the features NESTED IF, CASE and CASE expression. The program can be extended using the NULLIF and COALESCE functions.

29) Program development using WHILE LOOPS, numeric FOR LOOPS, nested loops using ERROR handling, BUILT IN exceptions, USER defined exceptions, RAISE APPLICATION ERROR.

30) Program development using creation of procedure, passing pa-rameters IN and OUT procedures.

31) Program development using creation of stored function, invoke functions in SQL statements and write complex functions.

32) Program development using creation of package specification, package bodies, private objects, package variables and cursors and calling stored packages.

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33) Develop programs using features of parameters in a CURSOR, FOR UPDATE CURSOR, WHERE CURRENT of clause and CUR-SOR variables.

Example: 1Code a procedure to calculate the sales made to a particular customer.{ create table trn (itmid number(10),

cstid number(10), trnqty number(10));

create table itmmast (itmid number(10), itmprice number(10,2));

create table cstmast ( cstid number(10), name varchar2(30));}

Step 1: Open the editorStep 2: Type the code below in a file named, TOTSALES.

ProgramCREATE OR REPLACE PROCEDURE TOTSALES

(CID IN CSTMAST.CSTID%TYPE, SAL OUT NUMBER)IS

id TRN.ITMID%TYPE;qty TRN.TRNQTY%TYPE;price ITMMAST.ITMPRICE%TYPE;sales NUMBER(10, 2) := 0;

cursor cur_tr isselect trn.itmid, trnqty, itmpricefrom trn, itmmastwhere trn.cstid = cidand trn.itmid = itmmast.itmid;

beginopen cur_tr;loop

fetch cur_tr into id, qty, price;if cur_tr%rowcount = 0then

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raise_application_error(-20020, ‘ERREOR!!!THERE IS NO DATA’);

end if;exit when cur_tr%notfound;sales := sales + qty * price;

end loop;

close cur_tr;sal := sales;

end;/

Step 3: Save the TOTSALES.SQL file.Step 4: Return to SQL Prompt and compile asSQL> start TOTSALES; (press enter) Step 5: On the screen you will get the message Procedure created. If you have errors type SQL> show errors Step 6: To execute the procedure at SQL prompt type SQL> variable sl number SQL> execute totsales(2001, :sl)SQL> print sl

PS:Procedural Objects

Groups of SQL and PL/SQL statements can be stored in the database. The code stored once in the database can be used by multiple applications. Since the code is in the database, which is in the server, processing is faster.

Procedures and functions are also referred to as sub-programs as they can take parameters and be invoked.

Various types of procedural objects are: Procedures, Functions, Packages.

Procedures:Procedures are sub-programs, which will perform an action and functions are subprograms that are generally coded to compute some value.

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The clients execute the procedure or function and the processing is done in the server. Procedures can receive and return values from and to the caller.Communication is passed to a procedure through a parameter and communication is passed out of a procedure through a parameter.When calling a procedure, the parameters passed can be declared to be IN, OUT or IN OUT.The IN parameter is used to pass values to the procedure being called. It behaves like a constant inside the procedure, i.e., cannot be assigned values inside the procedure.The OUT parameter is used to pass values out of a procedure to the caller of the procedure. It behaves like a uninitialized variable inside the procedure.The IN OUT parameter is used to pass values to the procedure being called and it is used to pass values to the caller of the procedure. The IN OUT variable behaves like a regular variable inside the procedure.

Functions:Functions are also a collection of SQL and PL/SQL code which can return a value to the caller.Unlike procedures, functions can return a value to the caller. This value is returned through the use of the RETURN keyword within the function. A function can return a single value to the caller. Functions do not allow the OUT and IN OUT arguments.

Packages:Packages are groups of procedures, functions, variables and SQL statements in a single unit. It consists of the package definition/specification and package body.A package specification consists of the list of functions, procedures, variables, constants, cursors and exceptions that will be available to users of the package.A package body consists of the PL/SQL blocks and specifications for all of the public objects listed in the package specification. It may also include code that is run every time the package is invoked, regardless of the part of the package that is executed.The name of the package body should be the same as the name of the package specification.

To delete procedural objects:SQL> drop procedure <procedure-name>

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SQL> drop function <function-name>SQL> drop package<package-name>

Example: 2Code a function to return the square of a given number.

Step 1: Open the editorStep 2: Type the code below in a file named, SQR.

ProgramCREATE OR REPLACE FUNCTION SQR

(NO NUMBER)RETURN NUMBERISBEGIN

return no*no;END;/

Step 3: To test the function:a. At SQL prompt type:

SQL> select sqr(10) from dual; b. At SQL prompt, type the following

SQL> variable sq numberSQL> execute :sq := sqr(10)

c. In the editor, type the followingset serveroutput onbegin

dbms_output.put_line(‘Square of 10 is ‘ || sqr(10));end;/set serveroutput off

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Example: 3Code a function to return the net salary of a given employee.

ProgramCREATE OR REPLACE FUNCTION NETSAL

(id in salary.emp_no%type)RETURN NUMBERIS

netsal salary.basic%type;BEGIN

select sum(basic) + sum(commission) – sum(deduction)into netsalfrom salarywhere emp_no = id;return (netsal);

end;/

To test the function:At SQL prompt, type

SQL> variable sal numberSQL> execute :sal := netsal(1001)SQL> print sal

Example: 4Code a package

Step 1: Open the editor and create the packageStep 2: Type the code below in a file named, MAHEPACK.

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ProgramCREATE OR REPLACE PACKAGE MAHEPACKAS

function netsal(id in salary.emp_no%type)

return number;procedure tax

(id in salary.emp_no%type, tax out number); procedure totsales

(cid in cstmast.cstid%type, sal out number);END;/

Step 3: Save the above file and open the editor to create the package body

Program

CREATE OR REPLACE PACKAGE BODY MAHEPACKAS

function netsal(id in salary.emp_no%type)

return numberis

netsal salary.basic%type;begin

select sum(basic) + sum(commission) – sum(deduction)into netsalfrom salarywhere emp_no = id;return (netsal);

end;

procedure tax(id in salary.emp_no%type, tax out number)

isnetsalary number(10, 2);

beginnetsalary := netsal(id);if netsalary < 2000

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thentax := netsalary * 0.02;

elsif netsalary < 4000then

tax := netsal * 0.04;else

tax := netsalary * 0.01;end if;

end;

procedure totsales(cid in cstmast.cstid%type, sal out number)

isid TRN.ITMID%TYPE;qty TRN.TRNQTY%TYPE;price ITMMAST.ITMPRICE%TYPE;sales NUMBER(10, 2) := 0;

cursor cur_tr isselect trn,itmid, trnqty, itmpricefrom trn, itmmastwhere trn.cstid = cidand trn.itmid = itmmast.itmid;

beginsales := 0;open cur_tr;loop

fetch cur_tr into id, qty, price;if cur_tr%rowcount = 0then

raise_application_error(-20020, ‘ERREOR!!!THERE IS NO DATA’);

end if;exit when cur_tr%notfound;sales := sales + qty * price;

end loop;

close cur_tr;sal := sales;

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end;END;/

Step 4: Save the above file and to create the package, at SQL prompt typeSQL> start mpackSQL> start mpackb

Step 5: To execute, at SQL prompt typeSQL> variable tx numberSQL> execute mahepack.tax(1001, :tx)SQL> print txOrSQL> variable nsal numberSQL> execute :nsal := mahepack.netsal(1001)SQL> print nsal

PS:Show Errors: SHOW ERRORS is used to display the line number and error of the most recent compilation errors. SQL> SHOW ERRORS

Raise Application Error: RAISE_APPLICATION_ERROR procedure is one of Oracles utilities which help the user to manage the error conditions in the applications by specifying user-defined error numbers and messages.It takes 2 input parameters – the error number (which must be between -20000 and -20999) and the error message to display.It terminates the procedure execution, rolls back any effects of the procedure, returns any user-specified error number and error message.

WEEK 12 (PL/SQL)

34) (i) Creation of simple PL/SQL program which includes decla-ration section, executable section and exception handling section ( ex: Student marks can be selected from the table and printed for those

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who secured first class and an exception can be raised if no records were found).

(ii) Insert data into student table and use COMMIT, ROLLBACK and SAVEPOINT in SQL block.

35) Develop a program that includes the features NESTED IF, CASE and CASE expression. The program can be extended using the NULLIF and COALESCE functions.

36) Program development using WHILE LOOPS, numeric FOR LOOPS, nested loops using ERROR handling, BUILT IN exceptions, USER defined exceptions, RAISE APPLICATION ERROR.

37) Program development using creation of procedure, passing pa-rameters IN and OUT procedures.

38) Program development using creation of stored function, invoke functions in SQL statements and write complex functions.

39) Program development using creation of package specification, package bodies, private objects, package variables and cursors and calling stored packages.

40) Develop programs using features of parameters in a CURSOR, FOR UPDATE CURSOR, WHERE CURRENT of clause and CUR-SOR variables.

Example: 1Write a row trigger to insert the existing values of the salary table into a new table when the salary table is updated.(Salary < emp_no, basic, commission, deduction, salary_date, department>Salaryaud < emp_no, basic, commission, deduction, salary_date, department>)

Step 1: Open the editorStep 2: Type the code below in a file named, TRSAL.

ProgramCREATE TRIGGER UPDSALBEFORE UPDATE ON SALARY

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FOR EACH ROWBEGIN

insert intosalaryaudvalues

(:old.emp_no, :old.basic, :old.commission, :old.deduction, :old.salary_date, :old.department);END/

Step 3: Save the above file

Step 4: To create the trigger, at SQL prompt typeSQL> start trsal

Step 5: To test the trigger, update values in salary table and see if data is inserted in salaryaud table.

Example: 2Write a trigger to restrict the user from using the emp table on Tuesday.

Programcreate or replace trigger tr2 before insert or update or delete on empbegin

if (rtrim(to_char(sysdate, 'day')) = 'tuesday') then

raise_application_error(-20121, 'Cannot delete on Tuesday');end if;

end;/

Example: 3Write a PL/SQL block of code that first inserts a record in an Emp table. Update salaries of emp 1001 and emp 1002 by Rs 2000 and Rs 1500. Then check to see that the total salary does not exceed Rs 20000. If total salary is greater than Rs 20000 then undo the updates made to emp 1001 and emp 1002.

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ProgramDECLARE

total_sal number(9);BEGIN

insert into empvalues (‘1009’, ‘Ram’, 1000);

SAVEPOINT no_update;update empset sal = sal + 2000where emp_id = 1001;

update empset sal = sal + 1500where emp_id = 1002;

select sum(sal)into total_salfrom emp;

if total_sal > 20000then

ROLLBACK TO SQVEPOINT no_update;end if;

COMMIT;END;/

PS:The above program first inserts a record into emp table. It then marks and saves the current position in the transaction by using the SAVEPOINT. It updates the salaries, if the salaries are exceeding 20000 it rollsback to the save point, ie ignores the 2 updates and only commits the insert. If the salaries do not exceed 20000 then the insert and 2 updates are committed. Example: 4

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Write a PL/SQL code of block, to calculate the area of the circle for the values of radius varying from 1 to 10. Store the odd radius values and the corresponding areas in a table.

Program declare pi constant number(4,2) := 3.14; radius number(5); areaa number(14,2); begin radius := 1; while radius <= 10 loop areaa := pi*power(radius,2);

casewhen radius = 1then

insert into area values (radius, areaa); when radius = 3 then

insert into area values (radius, areaa); when radius = 53 then

insert into area values(radius,areaa); when radius = 7then

insert into area values(radius,areaa); when radius = 9then

insert into area values(radius,areaa);else

dbms_output.put_line(‘EVEN RADIUS, NOT INSERTING!!!’);end case;

radius:=radius+1;end loop;

end;/

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Overview of SQL DDL, DML and DCL Commands.

DDL is Data Definition Language statements. Some examples:CREATE - to create objects in the database

ALTER - alters the structure of the database

DROP - delete objects from the database

TRUNCATE - remove all records from a table, including all

spaces allocated for the records ar

removed

COMMENT - add comments to the data dictionary

GRANT - gives user's access privileges to database

REVOKE - withdraw access privileges given with the GRANT

command

DML is Data Manipulation Language statements. Some

examples:

SELECT - retrieve data from the a database82

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INSERT - insert data into a table

UPDATE - updates existing data within a table

DELETE - deletes all records from a table, the space for the

records remain

CALL - call a PL/SQL or Java subprogram

EXPLAIN PLAN - explain access path to data

LOCK TABLE - control concurrency

DCL is Data Control Language statements. Some examples:

COMMIT - save work done

SAVEPOINT - identify a point in a transaction to which you can

later roll back

ROLLBACK - restore database to original since the last COMMIT

SET TRANSACTION - Change transaction options like what

rollback segment to use

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Basic SQL DDL Commands.

To practice basic SQL DDL Commands such as CREATE, DROP, etc.

1. SQL - CREATE TABLE

Syntax: CREATE TABLE tablename (column_name data_ type constraints, …)

Example:

INPUT:

SQL> CREATE TABLE Emp ( EmpNo short CONSTRAINT PKey PRIMARY KEY,

EName VarChar(15), Job Char(10) CONSTRAINT Unik1 UNIQUE,

Mgr short CONSTRAINT FKey1 REFERENCES EMP (EmpNo),

Hiredate Date, DeptNo short CONSTRAINT FKey2 REFERENCES DEPT(DeptNo));

RESULT: Table created.

SQL>Create table prog20 (pname varchar2(20) not null), doj date not null,dob date not null, sex varchar(1) not null, prof1 varchar(20),prof2 varchar(20),salary number(7,2) not null);

RESULT:

Table created.

SQL>desc prog20;

Name Null? Type

--------------------------------- -------- ----------------------------

PNAME NOT NULL VARCHAR2(20)

DOJ NOT NULL DATE

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DOB NOT NULL DATE

SEX NOT NULL VARCHAR2(1)

PROF1 VARCHAR2(20)

PROF2 VARCHAR2(20)

SALARY NOT NULL NUMBER(7,2)

2. SQL - ALTER TABLE

INPUT:

SQL>ALTER TABLE EMP ADD CONSTRAINT Pkey1 PRIMARY KEY (EmpNo);

RESULT: Table Altered.

Similarly, ALTER TABLE EMP DROP CONSTRAINT Pkey1;

3. SQL - DROP TABLE

– Deletes table structure – Cannot be recovered – Use with caution

INPUT:

SQL> DROP TABLE EMP; Here EMP is table name

RESULT: Table Dropped.

4. TRUNCATE TRUNCATE TABLE <TABLE NAME>;

Basic SQL DML Commands.

To practice basic SQL DML Commands such as INSERT, DELETE, etc.

1. SQL - INSERT INTO

Syntax: INSERT INTO tablename VALUES (value list)

Single-row insert

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INSERT INTO S VALUES(‘S3’,’SUP3’,’BLORE’,10)

Inserting one row, many columns at a time

INSERT INTO S (SNO, SNAME) VALUES (‘S1’, ‘Smith’);S1’ Smith’

Inserting many rows, all/some columns at a time.

INSERT INTO NEW_SUPPLIER (SNO, SNAME)

SELECT SNO, SNAME FROM S

WHERE CITY IN (‘BLORE’,’MADRAS’)

Other Examples:

INPUT:

SQL>Insert into prog values (‘kkk’,’05-may-56’);

RESULT: 1 row created.

INPUT:

SQL>Insert into prog20 values(‘Hema’,’25-sept-01’28-jan-85’,’f’,’c’,’c++’,’25000’);

RESULT: 1 row created.

INPUT:

SQL>Insert into prog values(‘&pname’,’&doj’);

SQL> Insert into prog values('&pname','&doj');

Enter value for pname: ravi

Enter value for doj: 15-june-81

RESULT:

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old 1: Insert into prog values('&pname','&doj')

new 1: Insert into prog values('ravi','15-june-81')

1 row created.

2. SQL - UPDATE

Syntax: UPDATE tablename SET column_name =value [ WHERE condition]

Examples:

UPDATE S SET CITY = ‘KANPUR’ WHERE SNO=‘S1’

UPDATE EMP SET SAL = 1.10 * SAL

SQL> update emp set sal=20000 where empno=7369;

1 row updated.

3. SQL - DELETE FROM Syntax: DELETE FROM tablename WHERE conditionExamples: DELETE FROM SP WHERE PNO= ‘P1’ DELETE FROM SP

INPUT:

SQL>Delete from emp where empno=7369;

RESULT: 1 row deleted.

Basic SQL DCL Commands.

To practice basic SQL DCL Commands such as COMMIT, ROLLBACK etc.

1. COMMIT

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Save changes (transactional).

Syntax:

COMMIT [WORK] [COMMENT 'comment_text'] COMMIT [WORK] [FORCE 'force_text' [,int] ]

FORCE - will manually commit an in-doubt distributed transactionforce_text - transaction identifier (see the DBA_2PC_PENDING view)int - sets a specific SCN.If a network or machine failure prevents a distributed transaction from committing properly, Oracle will store any commit comment in the data dictionary along with the transaction ID.

INPUT:

SQL>commit;

RESULT: Commit complete.

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2. ROLLBACK

Undo work done (transactional).

Syntax:

ROLLBACK [WORK] [TO [SAVEPOINT]'savepoint_text_identifier'];

ROLLBACK [WORK] [FORCE 'force_text'];

FORCE - will manually rollback an in-doubt distributed transaction

INPUT:

SQL>rollback;

RESULT:Rollback complete.

3. SAVEPOINTSave changes to a point (transactional).

Syntax:

SAVEPOINT text_identifier

Example: UPDATE employees SET salary = 95000 WHERE last_name = 'Smith';

SAVEPOINT justsmith;

UPDATE employees SET salary = 1000000;

SAVEPOINT everyone;

SELECT SUM(salary) FROM employees;

ROLLBACK TO SAVEPOINT justsmith; 89

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COMMIT;

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Writing and Practice of Simple Queries.

To write simple queries and practice them.

1. Get the description of EMP table.

SQL>desc emp;

RESULT:Name Null? Type -------------------------------- ----------------------- ------------------------- EMPNO NOT NULL NUMBER(4) ENAME VARCHAR2(10) JOB VARCHAR2(9) MGR NUMBER(4) HIREDATE DATE SAL NUMBER(7,2) COMM NUMBER(7,2) DEPTNO NUMBER(3) AGE NUMBER(3) ESAL NUMBER(10)

2. Get the description DEPT table.

SQL>desc dept;

RESULT:Name Null? Type--------------------------------- --------------------- ---------------------------DEPTNO NOT NULL NUMBER(2)DNAME VARCHAR2(14)LOC VARCHAR2(13)

3.List all employee details.

SQL>select * from emp;

RESULT:

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EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO AGE ESAL-------- ---------- --------- ---------- --------- ---------- ---------- ---------- ---------- -----------------7369 SMITH CLERK 7902 17-DEC-80 800 0 20 25 07499 ALLEN SALESMAN 7698 20-FEB-81 1600 300 30 25 07521 WARD SALESMAN 7698 22-FEB-81 1250 500 30 25 07566 JONES MANAGER 7839 02-APR-81 2975 500 20 25 07698 BLAKE MANAGER 7839 01-MAY-81 2850 1400 30 25 0

4.List all employee names and their salaries, whose salary lies between 1500/- and 3500/- both inclusive.

INPUTSQL>select ename from emp where sal between 1500 and 3500;RESULT

ENAME----------ALLENJONESBLAKECLARKSCOTTTURNERFORDrusselgreg

9 rows selected.

5. List all employee names and their and their manager whose manager is 7902 or 7566 0r 7789.

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INPUT SQL>select ename from emp where mgr in(7602,7566,7789);RESULT

ENAME-------SCOTTFORD

6. List all employees which starts with either J or T.

INPUT SQL>select ename from emp where ename like ‘J%’ or ename like ‘T%’;

RESULT:

ENAME---------JONESTURNERJAMES

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7. List all employee names and jobs, whose job title includes M or P.

INPUT SQL>select ename,job from emp where job like ‘M%’ or job like ‘P%’;

RESULT:ENAME JOB---------- ---------JONES MANAGERBLAKE MANAGERCLARK MANAGERKING PRESIDENT

8. List all jobs available in employee table.

INPUT SQL>select distinct job from emp;

RESULT:JOB---------ANALYSTCLERKMANAGERPRESIDENTSALESMANassistantclerk

7 rows selected.

9. List all employees who belongs to the department 10 or 20.

INPUT SQL>select ename from emp where deptno in (10,20);

RESULT:ENAME----------SMITHJONESCLARKSCOTT

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KINGADAMSFORDMILLER

8 rows selected.

10. List all employee names , salary and 15% rise in salary.

INPUT SQL>select ename , sal , sal+0.15* sal from emp;

RESULT:ENAME SAL SAL+0.15*SAL---------- ---------- ------------SMITH 800 920ALLEN 1600 1840WARD 1250 1437.5JONES 2975 3421.25MARTIN 1250 1437.5BLAKE 2850 3277.5CLARK 2450 2817.5

7 rows selected.

11. List minimum , maximum , average salaries of employee.

INPUT SQL>select min(sal),max(sal),avg(sal) from emp;

RESULT:

MIN(SAL) MAX(SAL) AVG(SAL)--------- ---------- ---------- 3 5000 1936.94118

12. Find how many job titles are available in employee table.

INPUT SQL>select count (distinct job) from emp;

RESULT:COUNT(DISTINCTJOB)------------------

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7

13. What is the difference between maximum and minimum salaries of employees in the organization?

INPUT SQL>select max(sal)-min(sal) from emp;

RESULT:MAX(SAL)-MIN(SAL)----------------- 4997

14. Display all employee names and salary whose salary is greater than minimum salary of the company and job title starts with ‘M’.

INPUT SQL>select ename,sal from emp where job like ‘M%’ and sal > (select min (sal) from emp);

RESULTENAME SAL---------- ----------JONES 2975BLAKE 2850CLARK 2450

15. Find how much amount the company is spending towards salaries.

INPUT SQL>select sum (sal) from emp;

RESULT SUM(SAL)--------- 32928

16. Display name of the dept. with deptno 20.

INPUT SQL>select ename from emp where deptno = 20;

RESULTENAME----------SMITH

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JONESSCOTTADAMS

17. List ename whose commission is NULL.

INPUT SQL>select ename from emp where comm is null;ENAME

RESULT ----------CLARKSCOTTKINGADAMSJAMESFORD

6 rows selected.

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18. Find no.of dept in employee table.

INPUT SQL>select count (distinct ename) from emp;

RESULT

COUNT(DISTINCTENAME

-------------------- 17

19. List ename whose manager is not NULL.

INPUT SQL>select ename from emp where mgr is not null;

RESULTENAME----------SMITHALLENWARDJONESMARTIN

5 rows selected.Writing Queries using GROUP BY and other clauses.

To write queries using clauses such as GROUP BY, ORDER BY, etc. and retrieving information by joining tables.

Source tables: emp, dept, programmer, software, study.

Order by : The order by clause is used to display the results in sorted order.Group by : The attribute or attributes given in the clauses are used to form groups. Tuples with the same value on all attributes in the group by clause are placed in one group.Having: SQL applies predicates (conditions) in the having clause after groups have been formed, so aggregate function be used.

1. Display total salary spent for each job category.

INPUT SQL>select job,sum (sal) from emp group by job;RESULT

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JOB SUM(SAL)--------- ----------ANALYST 6000CLERK 23050MANAGER 8275PRESIDENT 5000SALESMAN 5600assistant 2200clerk 2003

7 rows selected.

2. Display lowest paid employee details under each manager.

INPUT SQL>select ename, sal from emp where sal in (select min(sal) from emp group by mgr);RESULT

ENAME SAL---------- ----------chai 3JAMES 950MILLER 1000ADAMS 1100russel 2200

5 rows selected.

3. Display number of employees working in each department and their department name.

INPUT SQL> select dname, count (ename) from emp, dept where emp.deptno=dept.deptno group by dname;RESULT

DNAME COUNT(ENAME)-------------- ------------ACCOUNTING 3RESEARCH 5SALES 9

4. Display the sales cost of package developed by each programmer.

INPUT SQL>select pname, sum(scost) from software group by pname;

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RESULTPNAME SUM(SCOST)-------------------- ----------john 12000kamala 12000raju 12333

3 rows selected.

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5. Display the number of packages sold by each programmer.

INPUT SQL>select pname, count(title) from software group by pname;

RESULTPNAME COUNT(TITLE)-------------------- ------------john 1kamala 1raju 1ramana 1rani 15 rows selected.

6. Display the number of packages in each language for which the development cost is less than thousand.

INPUT SQL>select devin, count(title) from software where dcost < 1000 group by devin;RESULT

DEVIN COUNT(TITLE)---------- ------------cobol 1

7. Display each institute name with number of students.

INPUT SQL>select splace, count(pname) from study group by splace;RESULT

SPLACE COUNT(PNAME)-------------------- ------------BDPS 2BITS 1BNRILLIANI 1COIT 1HYD 1

5 rows selected.

8. How many copies of package have the least difference between development and selling cost, were sold?

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INPUT SQL>select sold from software where scost – dcost=(select min(scost – dcost) from software);RESULT

SOLD--------- 11

9. Which is the costliest package developed in Pascal.

INPUT SQL>select title from software where devin = ‘PASCAL’ and dcost = (select max(dcost)from software where devin = ‘PASCAL’);RESULT no rows selected

10. Which language was used to develop most no .of packages.

INPUT SQL>select devin, count (*) from software group by devin having count(*) = (select max(count(*) ) from software group by devin); RESULT

DEVIN COUNT(*)---------- ----------jsp 2

11.Who are the male programmers earning below the average salary of female programmers?

INPUT SQL>select pname from programmer where sal < (select avg(sal) from programmer where sex = ‘F’) and sex = ‘M’;RESULT

PNAME--------------------vijay

12. Display the details of software developed by the male programmers earning more than 3000/-.

INPUT SQL>select programmer.pname, title, devin from programmer, software where sal > 3000 and sex = ‘M’ and programmer.pname = software.pname; RESULT

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no rows selected

13. Display the details of software developed in c language by female programmers of pragathi.

INPUT SQL>select software.pname, title, devin, scost, dcost, sold from programmer, software, study where devin = ‘c’ and sex =’F’ and splace = ‘pragathi’ and programmer.pname = software.pname and software.pname = study.pname;

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14. Which language has been stated by the most of the programmers as proficiency one?

INPUT SQL>select prof1, count(*) from programmer group by prof1 having count (*) = (select max (count (*) ) from programmer group by prof1);

Writing Nested Queries.

To write queries using Set operations and to write nested queries.

Set Operations:

UNION - OR

INTERSECT - AND

EXCEPT - - NOT

NESTED QUERY:- A nested query makes use of another sub-query to compute or retrieve the information.

1. Find the name of the institute in which the person studied and developed the costliest package.

INPUT SQL>select splace, pname from study where pname = (select pname from software where scost = (select max (scost) from software);RESULT

SPLACE PNAME------------ -------------SAHBHARI MARY

2. Find the salary and institute of a person who developed the highest selling package.

INPUT SQL> select study.pname, sal, splace from study, programmer where study.pname = programmer.pname and study.pname = (select pname from software where scost = (select max (scost) from software));

RESULT

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PNAME SAL SPLACE----------- ------ -----------MARY 4500 SABHARI

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3. How many packages were developed by the person who developed the cheapest package.

INPUT SQL>select pname, count (title) from software where dcost = (select min(dcost) from software) group by pname;

RESULT PNAME COUNT(TITLE)

------------- ----------------------VIJAY 1

4. Calculate the amount to be recovered for those packages whose development cost has not yet recovered.

INPUT SQL>select title , (dcost-scost) from software where dcost > scost;

5. Display the title, scost, dcost, difference of scost and dcost in the descending order of difference.

INPUT SQL> select title, scost, dcost, (scost - dcost) from software descending order by (scost-dcost);

6. Display the details of those who draw the same salary.

INPUT SQL> select p.pname, p.sal from programmer p, programmer t where p.pname <> t.pname and p.sal = t.sal;(or)

INPUT SQL>select pname,sal from programmer t where pname<>t.pname and sal= t.sal;

Writing Queries using functions.

AIM: To write queries using single row functions and group functions.

1. Display the names and dob of all programmers who were born in january.

INPUT SQL>select pname , dob from programmer where to_char (dob,’MON’)=’JAN’;

2. Calculate the experience in years of each programmer and display along with programmer name in descending order.

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INPUT SQL> select pname, round (months_between(sysdate, doj)/12, 2) "EXPERIENCE" from programmer order by months_between (sysdate, doj) desc;

3. List out the programmer names who will celebrate their birthdays during current month.

INPUT SQL>select pname from programmer where to_char(dob,’MON’) like to_char (sysdate, ‘MON’);

4. Display the least experienced programmer’s details.

INPUT SQL>select * from programmer where doj = (select max (doj) from programmer);

5. Who is the most experienced programmer knowing pascal.

INPUT SQL>select pname from programmer where doj = (select min (doj) from programmer);

6. Who is the youngest programmer born in 1965.

INPUT SQL> select pname , dob from programmer where dob = (select max (dob) from programmer where to_char (dob,'yy') = 65);

7. In which year, most of the programmers are born.

INPUT SQL>select to_char (dob , ‘YY’) from programmer group by to_char (dob, ‘YY’) having count(*) = (select max (count(*)) from programmer group by to_char(dob,’YY’);

8. In which month most number of programmers are joined.

INPUT SQL>select to_char (doj,’YY’) from programmer group by to_char (doj,’YY’) having count (*) = (select max (count(*)) from programmer group by to_char (doj,’YY’);

9. What is the length of the shortest name in programmer table ?

INPUT SQL>select length (pname) from programmer where length (pname) = select min ( length (pname) from programmer);

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10. Display the names of the programmers whose name contains up to 5 characters.

INPUT SQL>select pname from programmer where length (pname) <=5;

11. Display all packages names in small letters and corresponding programmer names in uppercase letters.

INPUT SQL>select lower (title), upper (pname) from software;

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Writing Queries on views.

AIM: To write queries on views.

1. Create a view from single table containing all columns from the base table.

SQL>create view view1 as (select * from programmer);

2. Create a view from single table with selected columns.

SQL>create a view view2 as (select pname,dob,doj,sex,sal from programmer);

3. Create a view from two tables with all columns.

SQL>create view xyz as select * from programmer full natural join software;

4. Create a view from two tables with selected columns.

SQL> create view lmn as (select programmer, pname, title, devin from programmer, software where sal < 3000 and programmer.pname = software.pname);

5. Check all DML commands with above 4 views.

INPUT SQL> insert into view1 values (‘ramu’,’12-sep-03’,’28-jan-85’,’f’,’dbase’,’oracle’,74000);

RESULT

1 row created;

INPUT SQL>update view1 set salary =50000 where pname like ‘raju’;

RESULT 1 row updated.

Note: update command does not works for all queries on views.

INPUT SQL>delete from view1 where pname like ‘raju’;

RESULT 1 row deleted.

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6. Drop views which you generated.

INPUT SQL>drop view view1;

RESULT View dropped;

INPUT SQL>drop view view2;

RESULT View dropped;

INPUT SQL>drop view xyz;

Writing PL/SQL block for insertion into a table.

To write a PL/SQL block for inserting rows into EMPDET table with the following Calculations:HRA=50% OF BASICDA=20% OF BASICPF=7% OF BASICNETPAY=BASIC+DA+HRA-PF

INPUT

DECLAREENO1 empdet.eno%type;ENAME1 empdet.name%type;DEPTNO1 empdet.deptno%type;BASIC1 empdet.basic%type;HRA1 empdet.HRA%type;DA1 empdet.DA%type;PF1 empdet.pf%type;NETPAY1 empdet.netpay%type;

BEGINENO1:=&ENO1;ENAME1:='&ENAME1';DEPTNO1:=&DEPTNO1;BASIC1:=&BASIC1;HRA1:=(BASIC1*50)/100;DA1:=(BASIC1*20)/100;PF1:=(BASIC1*7)/100;

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NETPAY1:=BASIC1+HRA1+DA1-PF1;

INSERT INTO EMPDET VALUES (ENO1, ENAME1, DEPTNO1, BASIC1, HRA1, DA1, PF1, NETPAY1);END;

RESULT:

SQL> @BASICEnter value for eno1: 104old 11: ENO1:=&ENO1;new 11: ENO1:=104;Enter value for ename1: SRINIVAS REDDYold 12: ENAME1:='&ENAME1';new 12: ENAME1:='SRINIVAS REDDY';Enter value for deptno1: 10old 13: DEPTNO1:=&DEPTNO1;new 13: DEPTNO1:=10;Enter value for basic1: 6000old 14: BASIC1:=&BASIC1;new 14: BASIC1:=6000;

PL/SQL procedure successfully completed.

SQL>/Enter value for eno1: 105old 11: ENO1:=&ENO1;new 11: ENO1:=105;Enter value for ename1: CIRAJold 12: ENAME1:='&ENAME1';new 12: ENAME1:='CIRAJ';Enter value for deptno1: 10old 13: DEPTNO1:=&DEPTNO1;new 13: DEPTNO1:=10;Enter value for basic1: 6000old 14: BASIC1:=&BASIC1;new 14: BASIC1:=6000;

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PL/SQL procedure successfully completed.

SQL> SELECT * FROM EMPDET;RESULT ENO NAME DEPTNO BASIC HRA DA PF NETPAY--------- ------------------------------ --------- --------- --------- --------- --------- ----------------------- 101 SANTOSH 10 5000 2500 1000 350 8150 102 SHANKAR 20 5000 2500 1000 350 8150 103 SURESH 20 5500 2750 1100 385 8965 104 SRINIVASA REDDY 10 6000 3000 1200 420 9780 105 CIRAJ 10 6000 3000 1200 420 9780

Writing PL/SQL block for checking armstrong number

To write a PL/SQL block to check whether given number is Armstrong or not.

INPUT

DECLARE num number(5); rem number(5); s number(5):=0; num1 number(5);BEGIN num:=&num; num1:=num;

while(num>0)loop

rem:=mod(num,10); s:=s+power(rem,3);

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num:=trunc(num/10);End loop;

if (s=num1)then dbms_RESULT.put_line(num1||' IS ARMSTRONG NUMBER '); else dbms_RESULT.put_line(num1||' IS NOT ARMSTRONG NUMBER '); End if;END;/

RESULT:SQL>@armEnter value for num: 153old 7: num:=&num;new 7: num:=153;153 IS ARMSTRONG NUMBER

PL/SQL procedure successfully completed.

SQL> /Enter value for num: 123old 7: num:=&num;new 7: num:=123;123 IS NOT ARMSTRONG NUMBER

PL/SQL procedure successfully completed.Writing a PL/SQL block for checking a number even or odd.

AIM: To write a PL/SQL block to check whether a given number is Even or Odd.

INPUT

DECLARE num number(5);

rem number;BEGIN num:=&num;

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rem:=mod(num,2);if rem=0then

dbms_RESULT.put_line(' Number '||num||' is Even'); else

dbms_RESULT.put_line(' Number '||num||' is Odd'); end if;END;

RESULT:

SQL>start evenEnter value for num: 6old 5: num:=&num;new 5: num:=6;Number 6 is Even

PL/SQL procedure successfully completed.

SQL> /Enter value for num: 3old 5: num:=&num;new 5: num:=3;Number 3 is OddPL/SQL procedure successfully completed.

Writing PL/SQL block to find sum of digits of a given number.

To write a PL/SQL block to find Sum of Digits of a given Number.

INPUT

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DECLARE num number(5);

rem number(5); sm number(5):=0; num1 number(5);BEGIN num:=&num; num1:=num;

while(num>0) looprem:=mod(num,10);

sm:=sm+rem; num:=trunc(num/10);

end loop;dbms_RESULT.put_line('SUM OF DIGITS OF '||num1||' IS: '||sm);

end;/ RESULT:SQL> @sumINPUT truncated to 2 charactersEnter value for num: 123old 7: num:=&num;new 7: num:=123;SUM OF DIGITS OF 123 IS: 6PL/SQL procedure successfully completed.

SQL> @sumINPUT truncated to 2 charactersEnter value for num: 456old 7: num:=&num;new 7: num:=456;SUM OF DIGITS OF 456 IS: 15PL/SQL procedure successfully completed.

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Writing PL/SQL block for generating Fibonacci series.To write a PL/SQL block to Generate Fibonacci SeriesINPUT

DECLARE num number(5);

f1 number(5):=0; f2 number(5):=1; f3 number(5); i number(5):=3;BEGIN num:=&num; dbms_RESULT.put_line('THE FIBONACCI SERIES IS:'); dbms_RESULT.put_line(f1); dbms_RESULT.put_line(f2);

while(i<=num) loop f3:=f1+f2;

dbms_RESULT.put_line(f3); f1:=f2;

f2:=f3; i:=i+1;

end loop;END;/RESULT:SQL> start fibEnter value for num: 10old 8: num:=&num;new 8: num:=10;THE FIBONACCI SERIES IS:01123581321

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34PL/SQL procedure successfully completed.

Writing PL/SQL block for checking palendrome.

To write a PL/SQL block to Check the Given String is Palindrome or Not.

INPUT

DECLAREname1 varchar2(20);name2 varchar2(20);l number(5);

BEGINname1:='&name1';l:=length(name1);while l>0 loop

name2:=name2||substr(name1,l,1);l:=l-1;

end loop;dbms_RESULT.put_line('REVERSE OF STRING IS:'||NAME2);if(name1=name2) then

dbms_RESULT.put_line(name1||' IS PALINDROME ');else

dbms_RESULT.put_line(name1||' IS NOT PALINDROME ');end if;

END;/RESULTEnter value for name1: LIRILold 6: name1:='&name1';new 6: name1:='LIRIL';REVERSE OF STRING IS:LIRILLIRIL IS PALINDROME

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PL/SQL procedure successfully completed.

SQL> /Enter value for name1: MADAMold 6: name1:='&name1';new 6: name1:='MADAM';REVERSE OF STRING IS:MADAMMADAM IS PALINDROME

PL/SQL procedure successfully completed.

Writing PL/SQL block to demonstrate Cursors.

To write a Cursor to display the list of Employees and Total Salary Department wise.

INPUTDECLARE cursor c1 is select * from dept; cursor c2 is select * from emp; s emp.sal%type;

BEGINfor i in c1 loop

s:=0;dbms_RESULT.put_line('----------------------------------------------');

dbms_RESULT.put_line('Department is :' || i.deptno ||' Department name is:' || i.dname); dbms_RESULT.put_line('-------------------------------------------'); for j in c2 loop if ( i.deptno=j.deptno) then s:=s+j.sal; dbms_RESULT.put_line(j.empno|| ' '|| j.ename || ' '|| j.sal ); end if;

end loop;

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dbms_RESULT.put_line('----------------------------------------------');dbms_RESULT.put_line('Total salary is: '|| s);dbms_RESULT.put_line('----------------------------------------------');

end loop;END;

RESULT:

SQL> @abc------------------------------------------------------------------------------Department is :10 Department name is : ACCOUNTING------------------------------------------------------------------------------7782 CLARK 24507839 KING 50007934 MILLER 1300-----------------------------------------------------------------------------Total salary is: 8750----------------------------------------------------------------------------------------------------------------------------------------------------------Department is :20 Department name is:RESEARCH------------------------------------------------------------------------------7369 SMITH 8007566 JONES 29757788 SCOTT 30007876 ADAMS 11007902 FORD 3000-----------------------------------------------------------------------------Total salary is: 10875------------------------------------------------------------------------------------------------------------------------------------------------------------Department is :30 Department name is:SALES------------------------------------------------------------------------------7499 ALLEN 16007521 WARD 12507654 MARTIN 12507698 BLAKE 28507844 TURNER 1500

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7900 JAMES 950------------------------------------------------------------------------------Total salary is: 9400------------------------------------------------------------------------------------------------------------------------------------------------------------Department is :40 Department name is:OPERATIONS------------------------------------------------------------------------------------------------------------------------------------------------------------Total salary is: 0------------------------------------------------------------------------------PL/SQL procedure successfully completed.

Writing PL/SQL CURSORTo write a Cursor to display the list of employees who are Working as a Managers or Analyst.

INPUTDECLARE

cursor c(jb varchar2) is select ename from emp where job=jb; em emp.job%type;

BEGINopen c('MANAGER');dbms_RESULT.put_line(' EMPLOYEES WORKING AS MANAGERS

ARE:');loop

fetch c into em;

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exit when c%notfound;dbms_RESULT.put_line(em);

end loop;close c;

open c('ANALYST');dbms_RESULT.put_line(' EMPLOYEES WORKING AS ANALYST

ARE:');loop

fetch c into em;exit when c%notfound;dbms_RESULT.put_line(em);

end loop;close c;

END;

RESULT:

EMPLOYEES WORKING AS MANAGERS ARE:JONESBLAKECLARKEMPLOYEES WORKING AS ANALYST ARE:SCOTTFORD

PL/SQL procedure successfully completed.

Writing PL/SQL CURSORTo write a Cursor to display List of Employees from Emp Table in PL/SQL block

INPUT

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DECLAREcursor c is select empno, ename, deptno, sal from emp ;i emp.empno%type;j emp.ename%type;k emp.deptno%type;l emp.sal%type;

BEGINopen c;dbms_RESULT.put_line('Empno, name, deptno, salary of employees

are:= ');loop

fetch c into i, j, k, l;exit when c%notfound;dbms_RESULT.put_line(i||' '||j||' '||k||' '||l);

end loop;close c;

END;

RESULT:SQL> @EMPEmpno,name,deptno,salary of employees are:=7369 SMITH 20 8007499 ALLEN 30 16007521 WARD 30 12507566 JONES 20 29757654 MARTIN 30 12507698 BLAKE 30 28507782 CLARK 10 24507788 SCOTT 20 30007839 KING 10 50007844 TURNER 30 15007876 ADAMS 20 11007900 JAMES 30 9507902 FORD 20 30007934 MILLER 10 1300PL/SQL procedure successfully completed.

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Writing PL/SQL CURSORTo write a Cursor to find employee with given job and deptno.INPUTDECLARE

cursor c1(j varchar2, dn number) is select empno, ename from emp where job=j and deptno=dn;

row1 emp%rowtype;jb emp.job%type;d emp.deptno%type;

BEGINjb:='&jb';d:=&d;open c1(jb,d);fetch c1 into row1.empno,row1.ename;if c1%notfound then

dbms_RESULT.put_line('Employee does not exist');else

dbms_RESULT.put_line('empno is:'||row1.empno||' ' ||'employee name is:'||row1.ename);

end if;END;

RESULT:SQL> @CUREnter value for jb: MANAGERold 7: jb:='&jb';new 7: jb:='MANAGER';Enter value for d: 20old 8: d:=&d;new 8: d:=20;empno is:7566 employee name is:JONESPL/SQL procedure successfully completed.

SQL> /Enter value for jb: CLERKold 7: jb:='&jb';new 7: jb:='CLERK';

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Enter value for d: 40old 8: d:=&d;new 8: d:=40;Employee does not existPL/SQL procedure successfully completed.

Writing PL/SQL BLOCK using string functions. To write a PL/SQL block to apply String Functions on a given input String.

INPUTDECLARE a varchar2(20);

l number(5);BEGIN

a:='&a';l:=length(a);dbms_RESULT.put_line('Using Lower Function:' || lower(a));dbms_RESULT.put_line('Using UPPER Function:' || upper(a));dbms_RESULT.put_line('Using Initcap Function:' || initcap(a));dbms_RESULT.put_line('Using Substring Function:' || substr(a,l,1));dbms_RESULT.put_line('Using Substring Function:' || substr(a,1,3));dbms_RESULT.put_line('Using Ltrim function for xxxabcxxxx:' ||

ltrim('xxxabcxxxx','x'));dbms_RESULT.put_line('Using Rtrim function for xxxabcxxxx:'||

rtrim('xxxabcxxxx','x')); dbms_RESULT.put_line('Using Lpad function :'|| lpad(a,l+4,'*'));dbms_RESULT.put_line('Using Rpad function :'|| rpad(a,l+4,'*'));

END;

RESULT:

SQL>@STREnter value for a: santosh reddy

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old 5: a:='&a';new 5: a:='santosh reddy';Using Lower Function:santosh reddyUsing UPPER Function:SANTOSH REDDYUsing Initcap Function:Santosh ReddyUsing Substring Function:yUsing Substring Function:sanUsing Ltrim function for xxxabcxxxx:abcxxxxUsing Rtrim function for xxxabcxxxx:xxxabcUsing Lpad function :****santosh reddyUsing Rpad function :santosh reddy****

PL/SQL procedure successfully completed.SQL> /Enter value for a: UMA SHANKARold 5: a:='&a';new 5: a:='UMA SHANKAR';Using Lower Function:uma shankarUsing UPPER Function:UMA SHANKARUsing Initcap Function:Uma ShankarUsing Substring Function:RUsing Substring Function:UMAUsing Ltrim function for xxxabcxxxx:abcxxxxUsing Rtrim function for xxxabcxxxx:xxxabcUsing Lpad function :****UMA SHANKARUsing Rpad function :UMA SHANKAR****

PL/SQL procedure successfully completed

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Writing PL/SQL triggers

To write a TRIGGER to ensure that DEPT TABLE does not contain duplicate of null values in DEPTNO column.

INPUT

CREATE OR RELPLACE TRIGGER trig1 before insert on dept for each rowDECLARE

a number;BEGIN

if(:new.deptno is Null) thenraise_application_error(-20001,'error::deptno cannot be null');

elseselect count(*) into a from dept where deptno=:new.deptno;if(a=1) then

raise_application_error(-20002,'error:: cannot have duplicate deptno');

end if;end if;

END;

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RESULT:

SQL> @triggerTrigger created.

SQL> select * from dept; DEPTNO DNAME LOC --------- -------------- ------------- 10 ACCOUNTING NEW YORK 20 RESEARCH DALLAS 30 SALES CHICAGO 40 OPERATIONS BOSTON

SQL> insert into dept values(&deptnp,'&dname','&loc');Enter value for deptnp: nullEnter value for dname: marketingEnter value for loc: hydold 1: insert into dept values(&deptnp,'&dname','&loc')new 1: insert into dept values(null,'marketing','hyd')insert into dept values(null,'marketing','hyd') *ERROR at line 1:ORA-20001: error::deptno cannot be nullORA-06512: at "SCOTT.TRIG1", line 5ORA-04088: error during execution of trigger 'SCOTT.TRIG1'

SQL> /Enter value for deptnp: 10Enter value for dname: managerEnter value for loc: hydold 1: insert into dept values(&deptnp,'&dname','&loc')new 1: insert into dept values(10,'manager','hyd')insert into dept values(10,'manager','hyd') *ERROR at line 1:ORA-20002: error:: cannot have duplicate deptnoORA-06512: at "SCOTT.TRIG1", line 9ORA-04088: error during execution of trigger 'SCOTT.TRIG1'

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SQL> /Enter value for deptnp: 50Enter value for dname: MARKETINGEnter value for loc: HYDERABADold 1: insert into dept values(&deptnp,'&dname','&loc')new 1: insert into dept values(50,'MARKETING','HYDERABAD')

1 row created.SQL> select * from dept;

DEPTNO DNAME LOC --------- -------------- ------------- 10 ACCOUNTING NEW YORK 20 RESEARCH DALLAS 30 SALES CHICAGO 40 OPERATIONS BOSTON 50 MARKETING HYDE

Locking Table.AIM: To learn commands related to Table Locking

LOCK TABLE Statement Manually lock one or more tables.

Syntax: LOCK TABLE [schema.] table [options] IN lockmode MODE [NOWAIT]

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LOCK TABLE [schema.] view [options] IN lockmode MODE [NOWAIT]

Options: PARTITION (partition) SUBPARTITION (subpartition) @dblink

lockmodes: EXCLUSIVE SHARE ROW EXCLUSIVE SHARE ROW EXCLUSIVE ROW SHARE* | SHARE UPDATE*

If NOWAIT is omitted Oracle will wait until the table is available.

Several tables can be locked with a single command - separate with commas

e.g. LOCK TABLE table1,table2,table3 IN ROW EXCLUSIVE MODE;

Default Locking Behaviour :

A pure SELECT will not lock any rows.

INSERT, UPDATE or DELETE's - will place a ROW EXCLUSIVE lock.

SELECT...FROM...FOR UPDATE NOWAIT - will place a ROW EXCLUSIVE lock.

Multiple Locks on the same rows with LOCK TABLE

Even when a row is locked you can always perform a SELECT (because SELECT does not lock any rows) in addition to this, each type of lock will allow additional locks to be granted as follows.

ROW SHARE = Allow ROW EXCLUSIVE or ROW SHARE or SHARE locks to be granted to the locked rows.

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ROW EXCLUSIVE = Allow ROW EXCLUSIVE or ROW SHARE locks to be granted to the locked rows.

SHARE ROW EXCLUSIVE = Allow ROW SHARE locks to be granted to the locked rows.

SHARE = Allow ROW SHARE or SHARE locks to be granted to the locked rows.

EXCLUSIVE = Allow SELECT queries only

Although it is valid to place more than one lock on a row, UPDATES and DELETE's may still cause a wait if a conflicting row lock is held by another transaction.

Generation of Forms using ORACLE FORM BUILDER To design a form using Oracle Developer 2000

Introduction

Use Form Builder to simplify for the creation of data-entry screens, also known as Forms. Forms are the applications that connect to a database, retrieve information requested by the user, present it in a layout specified by Form designer, and allow the user to modify or add information. Form Builder allows you to build forms quickly and easily. In this Hands-On, you learn how to: Create a Data block for the “Customer” table, Create a layout, Use “content” canvas, Use “execute query”, Navigate a table, Use next, previous record, Enter query, Manipulate table’s record, Insert, Update, Delete and Save record. Form Builder Tool

Open the "Form Builder" tool.

Welcome window

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You will get the ‘Welcome to the Form Builder’ window. If you don’t want to get this window anymore uncheck the ‘Display at startup’ box. You can start your work with any of the following options:Use the data Block WizardBuild a new form manuallyOpen an existing formBuild a form based on a templateThe default is ‘Use the data Block Wizard.’ If you want to build a new form manually, click on "Cancel” or check ‘Build a new form manually’ and click ‘OK.’

Connect to database

In the ‘Object Navigator’ window, highlight "Database Objects." Go to the Main menu and choose "File," then "Connect." In the ‘Connect’ window, login in as “scott” password “tiger,” then click “CONNECT.”

Notice that the box next to ‘Database Objects’ is not empty anymore and it has a ‘+’ sign in it. That will indicate that this item is expandable and you are able to see its entire objects. Click on the ‘+’ sign next to the ‘Database Objects’ to expand all database schemas. Create a Module

In the ‘Object Navigator’ window, highlight module1. This is a default name. Go to the Main menu and choose “File,” select “Save as” to store the new object in the “iself” folder and save it as customer data entry. "c:_de." In this example the ‘DE’ abbreviation stands for Data Entry. Create a Data Block

In the ‘Object Navigator’ window, highlight "Data Blocks,” and click on the "create” icon. The ‘Create’ icon is in the vertical tool bar in the ‘Object Navigator’ window. It is a green ‘+’ sign. If you drag your cursor on the icon a tooltip will show ‘Create.’

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 New Data Block

In the ‘New Data Block’ window, choose the default option “Data Block Wizard” and click "OK." Welcome Data Block

In the ‘Welcome Data Block Wizard’ window click on the “NEXT” icon.

Type of Data Block

Select the type of data block you would like to create by clicking on a radio button. Select the default option ‘Table or View’ and then click “NEXT” again.

Selecting Tables

Click on “browse.” In the ‘Tables’ window, highlight the "cust11” table; then click "OK." Selecting columns for the Data Block Wizard

To choose all columns, click on the two arrow signs in the ‘Data Block Wizard’ window. To choose selected columns, click on the one arrow sign. And then select all columns, and click “next.” 

Layout Wizard

End of the Data Block Wizard and beginning of the Layout WizardIn the ‘Congratulations’ screen, use the default checkmark radio button (Create the data block, then call the Layout Wizard), and click "Finish." You can also use the Data Block Wizard to modify your existing data block. Simply select the data block in the Object Navigator and click the Data Block Wizard toolbar button, or choose ‘Data Block wizard’ from the ‘Tools’ menu. Welcome screen

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In the ‘Welcome to the Layout Wizard’ window, click ”Next.”

Selecting canvas

In the ‘Layout Wizard’ window, select the "new canvas" option. Canvas is a place that you will have your objects such as columns, titles, pictures, etc. If you have already had your canvas, select the canvas and then click on the next. The following are different types of canvases: Content, Stacked, Vertical Toolbar, Horizontal Toolbar, and Tab.

Think of the ‘Content’ canvas as one flat place to have all your objects. In the stacked canvas, you can have multiple layers of objects and it is the same as the tab canvas. You use the vertical or horizontal toolbar canvases for your push buttons. Check the different types of canvases by clicking on the ‘down arrow’ box next to the ‘Type’ field. Select "content," then click “Next.” Selecting Columns for the Layout Wizard

In the ‘Layout Wizard’ window, select all the columns. These are the columns that you want to be displayed on the canvas. Then click “Next.” Change your objects appearances

Change size or prompt if needed. In this window, you can enter a prompt, width, and height for each item on the canvas. You can change the measurement units. As a default the default units for item width and height are points. You can change it to inch or centimeter. When you change size, click “Next.” Selecting a layout style

Select a layout style for your frame by clicking a radio button. Select "Form," if you want one record at a time to be displayed. Select “Tabular,” if you want more than one record at a time to be displayed. Select "Forms," and then click “next.”

Record layout

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Type the "Frame Title" and click "next." Checkmark the ‘Display Scrollbar’ box when you use multiple records or the ‘Tabular’ option.

Congratulation Screen

In the ‘Congratulations’ window, click "Finish."You will see the output layout screen.Make some window adjustments and then run the form. To run the form, click on the ‘Run’ icon. The ‘Run’ icon is on the horizontal toolbar in the ‘CUSTOMER_DE’ canvas. The object module should be compiled successfully before executing the Form.

Execute Query

Click on the "Execute Query" icon below the main menu. If you drag the cursor on the toolbar in the ‘Forms Runtime’ window, a tooltip will be displayed and you see ‘Execute Query.’So to know all your option, drag your cursor to view all the icon descriptions. Next Record

Click on the "Next Record" icon to navigate to the next record.

Previous Record

Click on the "Previous Record" icon to navigate to the previous record.This is an easy way to navigate through the “Customer” table.

Enter Query

Click on the "Enter Query" icon to query selected records.

Insert Record

Click "Insert Record" to add new customer. All items on the forms will be blanked. You can either type all the customer information or duplicate it from pervious record.

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Duplicate Record

To duplicate the previous record, go to the main menu and select the ‘Record’ sub-menu. A drop down menu will be displayed. Select the ‘Duplicate’ option in the sub-menu.

Apply the changes. Remember in this stage, your record was inserted but not committed yet. Next and Previous Record

Click "next record" and "previous record" to navigate through the records and the one was added.

Save transactions

Click "Save" to commit the insert statement. 

Delete Record

Click "Remove Record" to delete the record. Lock a Record

You can also lock the record.

Exit from Form Runtime

Exit the FORM Runtime. If you have not committed any transaction, you will be prompted to save changes. Click “YES” to save changes.Click “OK” for acknowledgement. Don’t forget to save the Form.

RABAD

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Selecting the type of form to create

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Object wizard

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Selecting the canvas on which data block can be displayed

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Form showing the Employee details

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EXPT#28. Generating REPORTS using Oracle Developer 2000 AIM: To design reports using Oracle Developer 2000

IntroductionTabular report shows data in a table format. It is similar in concept to the idea of an Oracle table. Oracle, by default, returns output from your select statement in tabular format. Hands-onIn this Hands-On, your client is a stock broker that keeps track of its customer stock transactions. You have been assigned to write the reports based on their reports layout requirements.Your client wants you to create a simple listing report to show list of the stock trades by using stocks table for their brokerage company Your tasks are:1- Write a tabular report.2- Apply user layout Format mask.3- Run the report.4- Test the repot. You will learn how to: use report wizard, object navigator, report builder, “date model”, property palette, work on query and group box, see report style, use tabular style, navigating through report’s record, change the format mask for dollar, numeric and date items.

Open Report Builder toolOpen the "Report Builder" tool. Connect to databaseIn the Object Navigator, highlight "Database Objects,” choose "File," then select the "Connect" option.In the ‘Connect’ window, login as “iself” password schooling, then click “CONNECT.” Save a reportIn the Object Navigator, highlight the "untitled" report, choose “File,” and select the “Save as” option.

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In the ‘Save as’ window, make sure to save the report in the ISELF folder and name it "rpt01_stock_history,” report number 1 stock history. Data ModelIn the Object Navigator, double click on the "Data Model" icon. Create SQL boxIn the Data Model window, click on the "SQL Query" icon. Then drag the plus sign cursor and click it anywhere in the “Data Model” screen where you wish your object to be. In the ‘SQL Query Statement’ window, write a query to read all the stocks record sorted by their symbol.(SQL Query Statement)SELECT * FROM stocksORDER BY symbolClick “OK.” Change SQL box’s nameIn the Data Model window, in the “SQL” box, right click on the ‘Q_1’ and open its property palette.In its property palette, change the name to Q_STOCKS. Then close the window. Change GROUP box’s nameIn the Data Model, right click on the group box (G_SYMBOL) and open its property palette.In the Group property palette, change the name to ‘G_STOCKS,’ and close the window. Open Report WizardIn the Data Model, click on the ‘Report Wizard’ icon on the horizontal tool bar.In the Style tab, on the Report Wizard window, type ‘Stock History’ in the Title box and choose the report style as ‘Tabular.’Notice that when you change the report style a layout of that report will be displayed on the screen.Choose a different style to display its layout of its report style. Data, Fields, Totals, Labels and Template tabs

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Click “NEXT” to go to the Data tab. In the ‘SQL Query Statement’ verify your query.Click “NEXT” to navigate to the Fields tab, select the fields that you would like to be display in your report. Select all the columns to be display.Click “NEXT” to navigate to Totals tab, select the fields for which you would like to calculate totals. We have none in this hands-on exercise. Click “NEXT” to open the Labels tab, modify the labels and widths for your fields and totals as desired.Click “NEXT” again to go to the Template tab, and choose a template for your report. Your report will inherit the template’s colors, fonts, line widths, and structure.Use the default template and click “finish.” Running a reportNow, you should have your output report on the screen. Resize an objectMaximize the output report and format the report layout. To resize an object , select it and drag its handler to the preferred size. Move an objectTo move an object, select and drag it while the cursor is on the object. This is a simple report.Navigate through the outputTo navigate through the output report in the Report Editor - Live Pre-viewer, click on the "next page" or "previous page" icon on the horizontal toolbar.Do the same with the "first page" or "last page" icon. Use the “zoom in” and “zoom out” icon to preview the report.

Know report’s functionsTo know each icon functionalities, drag your cursor on it and a tooltip will display its function. Change Format MaskTo change the "format mask" of a column, the column should be selected. Then go to the toolbar and click on the “$” icon, "add decimal place," and the “right

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justify” format to the all currency columns (Todays Low, Todays High, and current price) Select the “traded today” column, and click on the ‘,0’ icon (apply commas), and make it right justify.Also, you can change any attributes of field by opening its property palette. To open an object’s property palette, right click on it and select the Property Palette option.Right click on the "trade date" column and open its "property palette."Change the date "Format Mask" property and make it “year 2000 complaint (MM-DD-RR).”

Selecting type of report

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Creating reports

Selecting the format of reports

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Selecting the Table in database

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Selecting the columns in the report

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Modify the labels in a table

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Choose a template to represent the report

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To specify the completion of report generation

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EXPT#29. Providing Security using GRANT and REVOKE.AIM: To learn GRANT and REVOKE commands to restrict privileges.

(1) GRANT Statement Grant privileges to a user (or to a user role)

Syntax:

Grant System-wide Privs:

GRANT system_priv(s) TO grantee [IDENTIFIED BY password] [WITH ADMIN OPTION]

GRANT role TO grantee [IDENTIFIED BY password] [WITH ADMIN OPTION]

GRANT ALL PRIVILEGES TO grantee [IDENTIFIED BY password] [WITH ADMIN OPTION]

Grant privs on specific objects:

GRANT object_priv [(column, column,...)] ON [schema.]object TO grantee [WITH GRANT OPTION] [WITH HIERARCHY OPTION]

GRANT ALL PRIVILEGES [(column, column,...)] ON [schema.]object TO grantee [WITH GRANT OPTION] [WITH HIERARCHY OPTION]

GRANT object_priv [(column, column,...)] ON DIRECTORY directory_name TO grantee [WITH GRANT OPTION] [WITH HIERARCHY OPTION]

GRANT object_priv [(column, column,...)] ON JAVA [RE]SOURCE [schema.]object

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TO grantee [WITH GRANT OPTION] [WITH HIERARCHY OPTION]

grantee: user role PUBLIC

system_privs: CREATE SESSION - Allows user to connect to the database UNLIMITED TABLESPACE - Use an unlimited amount of any tablespace. SELECT ANY TABLE - Query tables, views, or mviews in any schema UPDATE ANY TABLE - Update rows in tables and views in any schema INSERT ANY TABLE - Insert rows into tables and views in any schema Also System Admin rights to CREATE, ALTER or DROP: cluster, context, database, link, dimension, directory, index, materialized view, operator, outline, procedure, profile, role, rollback segment, sequence, session, synonym, table, tablespace, trigger, type, user, view. (full list of system privs)

object_privs: SELECT, UPDATE, INSERT, DELETE, ALTER, DEBUG, EXECUTE, INDEX, REFERENCES roles: SYSDBA, SYSOPER, OSDBA, OSOPER, EXP_FULL_DATABASE, IMP_FULL_DATABASE SELECT_CATALOG_ROLE, EXECUTE_CATALOG_ROLE, DELETE_CATALOG_ROLE

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AQ_USER_ROLE, AQ_ADMINISTRATOR_ROLE - advanced queuing SNMPAGENT - Enterprise Manager/Intelligent Agent. RECOVERY_CATALOG_OWNER - rman HS_ADMIN_ROLE - heterogeneous services

plus any user defined roles you have available

Notes:

Several Object_Privs can be assigned in a single GRANT statemente.g.GRANT SELECT (empno), UPDATE (sal) ON scott.emp TO emma

WITH HIERARCHY OPTION will grant the object privilege on all subobjects, including any created after the GRANT statement is issued.

WITH GRANT OPTION will enable the grantee to grant those object privileges to other users and roles.

"GRANT ALL PRIVILEGES..." may also be written as "GRANT ALL..."

(ii) REVOKE StatementRevoke privileges from users or roles.

Syntax:

Roles: REVOKE role FROM {user, | role, |PUBLIC}

System Privs: REVOKE system_priv(s) FROM {user, | role, |PUBLIC}

REVOKE ALL FROM {user, | role, |PUBLIC}

Object Privs: REVOKE object_priv [(column1, column2..)] ON [schema.]object

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FROM {user, | role, |PUBLIC} [CASCADE CONSTRAINTS] [FORCE]

REVOKE object_priv [(column1, column2..)] ON [schema.]object FROM {user, | role, |PUBLIC} [CASCADE CONSTRAINTS] [FORCE]

REVOKE object_priv [(column1, column2..)] ON DIRECTORY directory_name FROM {user, | role, |PUBLIC} [CASCADE CONSTRAINTS] [FORCE]

REVOKE object_priv [(column1, column2..)] ON JAVA [RE]SOURCE [schema.]object FROM {user, | role, |PUBLIC} [CASCADE CONSTRAINTS] [FORCE]

key: object_privs ALTER, DELETE, EXECUTE, INDEX, INSERT, REFERENCES, SELECT, UPDATE, ALL PRIVILEGES

system_privs ALTER ANY INDEX, BECOME USER, CREATE TABLE, DROP ANY VIEW RESTRICTED SESSION, UNLIMITED TABLESPACE, UPDATE ANY TABLE plus too many others to list here

roles Standard Oracle roles - SYSDBA, SYSOPER, OSDBA, OSOPER, EXP_FULL_DATABASE, IMP_FULL_DATABASE plus any user defined roles you have available

FORCE, will revoke all privileges from a user-defined-type and mark it's dependent objects INVALID.

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The roles CONNECT, RESOURCE and DBA are now deprecated (supported only for backwards compatibility) unless you are still running Oracle 6.0

Error ORA-01927 "cannot REVOKE privileges you did not grant" - This usually means you tried revoking permission from the table owner, e.g.Oracle will not allow REVOKE select on USER1.Table1 from USER1 Owners of objects ALWAYS have full permissions on those objects. This is one reason it makes sense to place tables in one schema and the packaged prodecures used to access those tables in a separate schema.

REFERENCES:

1. Oracle 9i Release 2 (9.2) SQL Reference, www.cs.ncl.ac.uk/teaching/facilities/swdoc/oracle9i/server.920/a96540/toc.htm.

2. Oracle 9i Release 1 (9.0.1) SQL Reference, http://download-east.oracle.com/docs/cd/A91202_01/901_doc/server.901/a90125/toc.htm.

3. An A-Z Index of Oracle SQL Commands (version 9.2) http://www.ss64.com/ora/.

4. Database Systems Instructor: Prof. Samuel Madden Source: MIT Open Courseware (http://ocw.mit.edu).

5. RDBMS Lab Guide, www.campusconnect.infosys.com userid:demo@infosys and passwork:infosys.

6. Orelly PL/SQL Pocket Reference, http://www.unix.org.ua/orelly/oracle/langpkt/index.htm

7. PL/SQL User's Guide and Reference, Release 2 (9.2)

http://www.lc.leidenuniv.nl/awcourse/oracle/appdev.920/a96624/toc.htm.

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VIVA VOICE QUESTIONS AND ANSWERS1. What is database?

A database is a logically coherent collection of data with some inherent meaning, representing some aspect of real world and which is designed, built and populated with data for a specific purpose.

2. What is DBMS? It is a collection of programs that enables user to create and maintain a database. In other words it is general-purpose software that provides the users with the processes of defining, constructing and manipulating the database for various applications.

3. What is a Database system? The database and DBMS software together is called as Database system.

4. Advantages of DBMS? Ø Redundancy is controlled. Ø Unauthorised access is restricted. Ø Providing multiple user interfaces. Ø Enforcing integrity constraints. Ø Providing backup and recovery.

5. Disadvantage in File Processing System? Ø Data redundancy & inconsistency. Ø Difficult in accessing data. Ø Data isolation. Ø Data integrity. Ø Concurrent access is not possible. Ø Security Problems.

6. Describe the three levels of data abstraction? The are three levels of abstraction: Ø Physical level: The lowest level of abstraction describes how data are stored. Ø Logical level: The next higher level of abstraction, describes what data are stored in database and what relationship among those data. Ø View level: The highest level of abstraction describes only part of entire database.

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7. Define the "integrity rules" There are two Integrity rules. Ø Entity Integrity: States that “Primary key cannot have NULL value” Ø Referential Integrity: States that “Foreign Key can be either a NULL value or should be Primary Key value of other relation.

8. What is extension and intension? Extension - It is the number of tuples present in a table at any instance. This is time dependent. Intension - It is a constant value that gives the name, structure of table and the constraints laid on it.

9. What is System R? What are its two major subsystems? System R was designed and developed over a period of 1974-79 at IBM San Jose Research Center. It is a prototype and its purpose was to demonstrate that it is possible to build a Relational System that can be used in a real life environment to solve real life problems, with performance at least comparable to that of existing system. Its two subsystems are Ø Research Storage Ø System Relational Data System.

10. How is the data structure of System R different from the relational structure? Unlike Relational systems in System R Ø Domains are not supported Ø Enforcement of candidate key uniqueness is optional Ø Enforcement of entity integrity is optional Ø Referential integrity is not enforced

11. What is Data Independence? Data independence means that “the application is independent of the storage structure and access strategy of data”. In other words, The ability to modify the schema definition in one level should not affect the schema definition in the next higher level. Two types of Data Independence: Ø Physical Data Independence: Modification in physical level should not affect

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the logical level. Ø Logical Data Independence: Modification in logical level should affect the view level. NOTE: Logical Data Independence is more difficult to achieve

12. What is a view? How it is related to data independence? A view may be thought of as a virtual table, that is, a table that does not really exist in its own right but is instead derived from one or more underlying base table. In other words, there is no stored file that direct represents the view instead a definition of view is stored in data dictionary. Growth and restructuring of base tables is not reflected in views. Thus the view can insulate users from the effects of restructuring and growth in the database. Hence accounts for logical data independence.

13. What is Data Model? A collection of conceptual tools for describing data, data relationships data semantics and constraints.

14. What is E-R model? This data model is based on real world that consists of basic objects called entities and of relationship among these objects. Entities are described in a database by a set of attributes.

15. What is Object Oriented model? This model is based on collection of objects. An object contains values stored in instance variables with in the object. An object also contains bodies of code that operate on the object. These bodies of code are called methods. Objects that contain same types of values and the same methods are grouped together into classes.

16. What is an Entity? It is a 'thing' in the real world with an independent existence.

17. What is an Entity type? It is a collection (set) of entities that have same attributes.

18. What is an Entity set? It is a collection of all entities of particular entity type in the database.

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The collections of entities of a particular entity type are grouped together into an entity set.

20. What is Weak Entity set? An entity set may not have sufficient attributes to form a primary key, and its primary key compromises of its partial key and primary key of its parent entity, then it is said to be Weak Entity set.

21. What is an attribute? It is a particular property, which describes the entity.

22. What is a Relation Schema and a Relation? A relation Schema denoted by R(A1, A2, …, An) is made up of the relation name R and the list of attributes Ai that it contains. A relation is defined as a set of tuples. Let r be the relation which contains set tuples (t1, t2, t3, ..., tn). Each tuple is an ordered list of n-values t=(v1,v2, ..., vn).

23. What is degree of a Relation? It is the number of attribute of its relation schema.

24. What is Relationship? It is an association among two or more entities.

25. What is Relationship set? The collection (or set) of similar relationships.

26. What is Relationship type? Relationship type defines a set of associations or a relationship set among a given set of entity types.

27. What is degree of Relationship type? It is the number of entity type participating.

28. What is DDL (Data Definition Language)? A data base schema is specifies by a set of definitions expressed by a special language called DDL.

29. What is VDL (View Definition Language)? It specifies user views and their mappings to the conceptual schema.

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30. What is SDL (Storage Definition Language)? This language is to specify the internal schema. This language may specify the mapping between two schemas.

31. What is Data Storage - Definition Language? The storage structures and access methods used by database system are specified by a set of definition in a special type of DDL called data storage-definition language.

32. What is DML (Data Manipulation Language)? This language that enable user to access or manipulate data as organised by appropriate data model. Ø Procedural DML or Low level: DML requires a user to specify what data are needed and how to get those data. Ø Non-Procedural DML or High level: DML requires a user to specify what data are needed without specifying how to get those data.

33. What is DML Compiler? It translates DML statements in a query language into low-level instruction that the query evaluation engine can understand.

34. What is Query evaluation engine? It executes low-level instruction generated by compiler.

35. What is DDL Interpreter? It interprets DDL statements and record them in tables containing metadata.

36. What is Record-at-a-time? The Low level or Procedural DML can specify and retrieve each record from a set of records. This retrieve of a record is said to be Record-at-a-time.

37. What is Set-at-a-time or Set-oriented? The High level or Non-procedural DML can specify and retrieve many records in a single DML statement. This retrieve of a record is said to be Set-at-a-time or Set-oriented.

38. What is Relational Algebra?

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It is procedural query language. It consists of a set of operations that take one or two relations as input and produce a new relation.

39. What is Relational Calculus? It is an applied predicate calculus specifically tailored for relational databases proposed by E.F. Codd. E.g. of languages based on it are DSL ALPHA, QUEL.

40. How does Tuple-oriented relational calculus differ from domain-oriented relational calculus The tuple-oriented calculus uses a tuple variables i.e., variable whose only permitted values are tuples of that relation. E.g. QUEL The domain-oriented calculus has domain variables i.e., variables that range over the underlying domains instead of over relation. E.g. ILL, DEDUCE.

41. What is normalization? It is a process of analysing the given relation schemas based on their Functional Dependencies (FDs) and primary key to achieve the properties Ø Minimizing redundancy Ø Minimizing insertion, deletion and update anomalies.

42. What is Functional Dependency? A Functional dependency is denoted by X Y between two sets of attributes X and Y that are subsets of R specifies a constraint on the possible tuple that can form a relation state r of R. The constraint is for any two tuples t1 and t2 in r if t1[X] = t2[X] then they have t1[Y] = t2[Y]. This means the value of X component of a tuple uniquely determines the value of component Y.

43. When is a functional dependency F said to be minimal? Ø Every dependency in F has a single attribute for its right hand side. Ø We cannot replace any dependency X A in F with a dependency Y A where Y is a proper subset of X and still have a set of dependency that is equivalent to F. Ø We cannot remove any dependency from F and still have set of dependency that is equivalent to F.

44. What is Multivalued dependency? Multivalued dependency denoted by X Y specified on relation schema R, where X and Y are both subsets of R, specifies the following constraint on any relation r of R: if two tuples t1 and t2 exist in r such that t1[X] = t2[X] then t3 and t4 should also exist in r with the following properties

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Ø t3[x] = t4[X] = t1[X] = t2[X] Ø t3[Y] = t1[Y] and t4[Y] = t2[Y] Ø t3[Z] = t2[Z] and t4[Z] = t1[Z] where [Z = (R-(X U Y)) ]

45. What is Lossless join property? It guarantees that the spurious tuple generation does not occur with respect to relation schemas after decomposition.

46. What is 1 NF (Normal Form)? The domain of attribute must include only atomic (simple, indivisible) values.

47. What is Fully Functional dependency? It is based on concept of full functional dependency. A functional dependency X Y is full functional dependency if removal of any attribute A from X means that the dependency does not hold any more.

48. What is 2NF? A relation schema R is in 2NF if it is in 1NF and every non-prime attribute A in R is fully functionally dependent on primary key.

49. What is 3NF? A relation schema R is in 3NF if it is in 2NF and for every FD X A either of the following is true Ø X is a Super-key of R. Ø A is a prime attribute of R. In other words, if every non prime attribute is non-transitively dependent on primary key.

50. What is BCNF (Boyce-Codd Normal Form)? A relation schema R is in BCNF if it is in 3NF and satisfies an additional constraint that for every FD X A, X must be a candidate key.

51. What is 4NF? A relation schema R is said to be in 4NF if for every Multivalued dependency X Y that holds over R, one of following is true Ø X is subset or equal to (or) XY = R. Ø X is a super key.

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52. What is 5NF? A Relation schema R is said to be 5NF if for every join dependency {R1, R2, ..., Rn} that holds R, one the following is true Ø Ri = R for some i. Ø The join dependency is implied by the set of FD, over R in which the left side is key of R.

53. What is Domain-Key Normal Form? A relation is said to be in DKNF if all constraints and dependencies that should hold on the the constraint can be enforced by simply enforcing the domain constraint and key constraint on the relation.

54. What are partial, alternate,, artificial, compound and natural key? Partial Key: It is a set of attributes that can uniquely identify weak entities and that are related to same owner entity. It is sometime called as Discriminator. Alternate Key: All Candidate Keys excluding the Primary Key are known as Alternate Keys. Artificial Key: If no obvious key, either stand alone or compound is available, then the last resort is to simply create a key, by assigning a unique number to each record or occurrence. Then this is known as developing an artificial key. Compound Key: If no single data element uniquely identifies occurrences within a construct, then combining multiple elements to create a unique identifier for the construct is known as creating a compound key. Natural Key: When one of the data elements stored within a construct is utilized as the primary key, then it is called the natural key.

55. What is indexing and what are the different kinds of indexing? Indexing is a technique for determining how quickly specific data can be found.

Types: Ø Binary search style indexing Ø B-Tree indexing Ø Inverted list indexing Ø Memory resident table Ø Table indexing

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56. What is system catalog or catalog relation? How is better known as? A RDBMS maintains a description of all the data that it contains, information about every relation and index that it contains. This information is stored in a collection of relations maintained by the system called metadata. It is also called data dictionary.

57. What is meant by query optimization? The phase that identifies an efficient execution plan for evaluating a query that has the least estimated cost is referred to as query optimization.

58. What is join dependency and inclusion dependency? Join Dependency: A Join dependency is generalization of Multivalued dependency.A JD {R1, R2, ..., Rn} is said to hold over a relation R if R1, R2, R3, ..., Rn is a lossless-join decomposition of R . There is no set of sound and complete inference rules for JD. Inclusion Dependency: An Inclusion Dependency is a statement of the form that some columns of a relation are contained in other columns. A foreign key constraint is an example of inclusion dependency.

59. What is durability in DBMS? Once the DBMS informs the user that a transaction has successfully completed, its effects should persist even if the system crashes before all its changes are reflected on disk. This property is called durability.

60. What do you mean by atomicity and aggregation? Atomicity: Either all actions are carried out or none are. Users should not have to worry about the effect of incomplete transactions. DBMS ensures this by undoing the actions of incomplete transactions. Aggregation: A concept which is used to model a relationship between a collection of entities and relationships. It is used when we need to express a relationship among relationships.

61. What is a Phantom Deadlock? In distributed deadlock detection, the delay in propagating local information might cause the deadlock detection algorithms to identify deadlocks that do not really exist. Such situations are called phantom deadlocks and they lead to

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unnecessary aborts.

62. What is a checkpoint and When does it occur? A Checkpoint is like a snapshot of the DBMS state. By taking checkpoints, the DBMS can reduce the amount of work to be done during restart in the event of subsequent crashes.

63. What are the different phases of transaction? Different phases are Ø Analysis phase Ø Redo Phase Ø Undo phase

64. What do you mean by flat file database? It is a database in which there are no programs or user access languages. It has no cross-file capabilities but is user-friendly and provides user-interface management.

65. What is "transparent DBMS"? It is one, which keeps its Physical Structure hidden from user.

66. Brief theory of Network, Hierarchical schemas and their properties Network schema uses a graph data structure to organize records example for such a database management system is CTCG while a hierarchical schema uses a tree data structure example for such a system is IMS.

67. What is a query? A query with respect to DBMS relates to user commands that are used to interact with a data base. The query language can be classified into data definition language and data manipulation language.

68. What do you mean by Correlated subquery? Subqueries, or nested queries, are used to bring back a set of rows to be used by the parent query. Depending on how the subquery is written, it can be executed once for the parent query or it can be executed once for each row returned by the parent query. If the subquery is executed for each row of the parent, this is called a correlated subquery. A correlated subquery can be easily identified if it contains any references to the parent subquery columns in its WHERE clause. Columns from the subquery cannot be referenced anywhere else in the parent query. The following example

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demonstrates a non-correlated subquery. E.g. Select * From CUST Where '10/03/1990' IN (Select ODATE From ORDER Where CUST.CNUM = ORDER.CNUM)

69. What are the primitive operations common to all record management systems? Addition, deletion and modification.

70. Name the buffer in which all the commands that are typed in are stored

‘Edit’ Buffer

71. What are the unary operations in Relational Algebra? PROJECTION and SELECTION.

72. Are the resulting relations of PRODUCT and JOIN operation the same? No. PRODUCT: Concatenation of every row in one relation with every row in another. JOIN: Concatenation of rows from one relation and related rows from another.

73. What is RDBMS KERNEL? Two important pieces of RDBMS architecture are the kernel, which is the software, and the data dictionary, which consists of the system-level data structures used by the kernel to manage the database You might think of an RDBMS as an operating system (or set of subsystems), designed specifically for controlling data access; its primary functions are storing, retrieving, and securing data. An RDBMS maintains its own list of authorized users and their associated privileges; manages memory caches and paging; controls locking for concurrent resource usage; dispatches and schedules user requests; and manages space usage within its table-space structures.

74. Name the sub-systems of a RDBMS I/O, Security, Language Processing, Process Control, Storage Management, Logging and Recovery, Distribution Control, Transaction Control, Memory Management, Lock Management

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75. Which part of the RDBMS takes care of the data dictionary? How Data dictionary is a set of tables and database objects that is stored in a special area of the database and maintained exclusively by the kernel.

76. What is the job of the information stored in data-dictionary? The information in the data dictionary validates the existence of the objects, provides access to them, and maps the actual physical storage location.

77. Not only RDBMS takes care of locating data it also determines an optimal access path to store or retrieve the data

76. How do you communicate with an RDBMS? You communicate with an RDBMS using Structured Query Language (SQL)

78. Define SQL and state the differences between SQL and other conventional programming Languages SQL is a nonprocedural language that is designed specifically for data access operations on normalized relational database structures. The primary difference between SQL and other conventional programming languages is that SQL statements specify what data operations should be performed rather than how to perform them.

79. Name the three major set of files on disk that compose a database in Oracle There are three major sets of files on disk that compose a database. All the files are binary. These are Ø Database files Ø Control files Ø Redo logs The most important of these are the database files where the actual data resides. The control files and the redo logs support the functioning of the architecture itself. All three sets of files must be present, open, and available to Oracle for any data on the database to be useable. Without these files, you cannot access the database, and the database administrator might have to recover some or all of the database using a backup, if there is one.

80. What is an Oracle Instance? The Oracle system processes, also known as Oracle background processes, provide functions for the user processes—functions that would otherwise be

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done by the user processes themselves Oracle database-wide system memory is known as the SGA, the system global area or shared global area. The data and control structures in the SGA are shareable, and all the Oracle background processes and user processes can use them. The combination of the SGA and the Oracle background processes is known as an Oracle instance

81. What are the four Oracle system processes that must always be up and running for the database to be useable The four Oracle system processes that must always be up and running for the database to be useable include DBWR (Database Writer), LGWR (Log Writer), SMON (System Monitor), and PMON (Process Monitor).

82. What are database files, control files and log files. How many of these files should a database have at least? Why? Database Files The database files hold the actual data and are typically the largest in size. Depending on their sizes, the tables (and other objects) for all the user accounts can go in one database file—but that's not an ideal situation because it does not make the database structure very flexible for controlling access to storage for different users, putting the database on different disk drives, or backing up and restoring just part of the database. You must have at least one database file but usually, more than one files are used. In terms of accessing and using the data in the tables and other objects, the number (or location) of the files is immaterial. The database files are fixed in size and never grow bigger than the size at which they were created Control Files The control files and redo logs support the rest of the architecture. Any database must have at least one control file, although you typically have more than one to guard against loss. The control file records the name of the database, the date and time it was created, the location of the database and redo logs, and the synchronization information to ensure that all three sets of files are always in step. Every time you add a new database or redo log file to the database, the information is recorded in the control files. Redo Logs Any database must have at least two redo logs. These are the journals for the database; the redo logs record all changes to the user objects or system objects. If any type of failure occurs, the changes recorded in the redo logs can be used to bring the database to a consistent state without losing any committed

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transactions. In the case of non-data loss failure, Oracle can apply the information in the redo logs automatically without intervention from the DBA. The redo log files are fixed in size and never grow dynamically from the size at which they were created.

83. What is ROWID? The ROWID is a unique database-wide physical address for every row on every table. Once assigned (when the row is first inserted into the database), it never changes until the row is deleted or the table is dropped. The ROWID consists of the following three components, the combination of which uniquely identifies the physical storage location of the row. Ø Oracle database file number, which contains the block with the rows Ø Oracle block address, which contains the row Ø The row within the block (because each block can hold many rows) The ROWID is used internally in indexes as a quick means of retrieving rows with a particular key value. Application developers also use it in SQL statements as a quick way to access a row once they know the ROWID

84. What is Oracle Block? Can two Oracle Blocks have the same address? Oracle "formats" the database files into a number of Oracle blocks when they are first created—making it easier for the RDBMS software to manage the files and easier to read data into the memory areas. The block size should be a multiple of the operating system block size. Regardless of the block size, the entire block is not available for holding data; Oracle takes up some space to manage the contents of the block. This block header has a minimum size, but it can grow. These Oracle blocks are the smallest unit of storage. Increasing the Oracle block size can improve performance, but it should be done only when the database is first created. Each Oracle block is numbered sequentially for each database file starting at 1. Two blocks can have the same block address if they are in different database files.

85. What is database Trigger? A database trigger is a PL/SQL block that can defined to automatically execute for insert, update, and delete statements against a table. The trigger can e

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defined to execute once for the entire statement or once for every row that is inserted, updated, or deleted. For any one table, there are twelve events for which you can define database triggers. A database trigger can call database procedures that are also written in PL/SQL.

86. Name two utilities that Oracle provides, which are use for backup and recovery. Along with the RDBMS software, Oracle provides two utilities that you can use to back up and restore the database. These utilities are Export and Import. The Export utility dumps the definitions and data for the specified part of the database to an operating system binary file. The Import utility reads the file produced by an export, recreates the definitions of objects, and inserts the data If Export and Import are used as a means of backing up and recovering the database, all the changes made to the database cannot be recovered since the export was performed. The best you can do is recover the database to the time when the export was last performed.

87. What are stored-procedures? And what are the advantages of using them. Stored procedures are database objects that perform a user defined operation. A stored procedure can have a set of compound SQL statements. A stored procedure executes the SQL commands and returns the result to the client. Stored procedures are used to reduce network traffic.

88. How are exceptions handled in PL/SQL? Give some of the internal exceptions' name PL/SQL exception handling is a mechanism for dealing with run-time errors encountered during procedure execution. Use of this mechanism enables execution to continue if the error is not severe enough to cause procedure termination. The exception handler must be defined within a subprogram specification. Errors cause the program to raise an exception with a transfer of control to the exception-handler block. After the exception handler executes, control returns to the block in which the handler was defined. If there are no more executable statements in the block, control returns to the caller. User-Defined Exceptions PL/SQL enables the user to define exception handlers in the declarations area of subprogram specifications. User accomplishes this by naming an exception as in the following example:

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ot_failure EXCEPTION; In this case, the exception name is ot_failure. Code associated with this handler is written in the EXCEPTION specification area as follows: EXCEPTION when OT_FAILURE then out_status_code := g_out_status_code; out_msg := g_out_msg; The following is an example of a subprogram exception: EXCEPTION when NO_DATA_FOUND then g_out_status_code := 'FAIL'; RAISE ot_failure; Within this exception is the RAISE statement that transfers control back to the ot_failure exception handler. This technique of raising the exception is used to invoke all user-defined exceptions. System-Defined Exceptions Exceptions internal to PL/SQL are raised automatically upon error. NO_DATA_FOUND is a system-defined exception. Table below gives a complete list of internal exceptions.

PL/SQL internal exceptions.

PL/SQL internal exceptions.

Exception Name Oracle Error CURSOR_ALREADY_OPEN ORA-06511 DUP_VAL_ON_INDEX ORA-00001 INVALID_CURSOR ORA-01001 INVALID_NUMBER ORA-01722 LOGIN_DENIED ORA-01017 NO_DATA_FOUND ORA-01403 NOT_LOGGED_ON ORA-01012 PROGRAM_ERROR ORA-06501 STORAGE_ERROR ORA-06500 TIMEOUT_ON_RESOURCE ORA-00051 TOO_MANY_ROWS ORA-01422 TRANSACTION_BACKED_OUT ORA-00061 VALUE_ERROR ORA-06502 ZERO_DIVIDE ORA-01476

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In addition to this list of exceptions, there is a catch-all exception named OTHERS that traps all errors for which specific error handling has not been established.

89. Does PL/SQL support "overloading"? Explain The concept of overloading in PL/SQL relates to the idea that you can define procedures and functions with the same name. PL/SQL does not look only at the referenced name, however, to resolve a procedure or function call. The count and data types of formal parameters are also considered. PL/SQL also attempts to resolve any procedure or function calls in locally defined packages before looking at globally defined packages or internal functions. To further ensure calling the proper procedure, you can use the dot notation. Prefacing a procedure or function name with the package name fully qualifies any procedure or function reference.

90. Tables derived from the ERD a) Are totally unnormalised b) Are always in 1NF c) Can be further denormalised d) May have multi-valued attributes

(b) Are always in 1NF

91. Spurious tuples may occur due to i. Bad normalization ii. Theta joins iii. Updating tables from join a) i & ii b) ii & iii c) i & iii d) ii & iii

(a) i & iii because theta joins are joins made on keys that are not primary keys.

92. A B C is a set of attributes. The functional dependency is as follows AB -> B AC -> C C -> B a) is in 1NF b) is in 2NF

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c) is in 3NF d) is in BCNF

(a) is in 1NF since (AC)+ = { A, B, C} hence AC is the primary key. Since C B is a FD given, where neither C is a Key nor B is a prime attribute, this it is not in 3NF. Further B is not functionally dependent on key AC thus it is not in 2NF. Thus the given FDs is in 1NF.

93. In mapping of ERD to DFD a) entities in ERD should correspond to an existing entity/store in DFD b) entity in DFD is converted to attributes of an entity in ERD c) relations in ERD has 1 to 1 correspondence to processes in DFD d) relationships in ERD has 1 to 1 correspondence to flows in DFD

(a) entities in ERD should correspond to an existing entity/store in DFD

94. A dominant entity is the entity a) on the N side in a 1 : N relationship b) on the 1 side in a 1 : N relationship c) on either side in a 1 : 1 relationship d) nothing to do with 1 : 1 or 1 : N relationship

(b) on the 1 side in a 1 : N relationship

95. Select 'NORTH', CUSTOMER From CUST_DTLS Where REGION = 'N' Order By CUSTOMER Union Select 'EAST', CUSTOMER From CUST_DTLS Where REGION = 'E' Order By CUSTOMER The above is a) Not an error b) Error - the string in single quotes 'NORTH' and 'SOUTH' c) Error - the string should be in double quotes d) Error - ORDER BY clause

(d) Error - the ORDER BY clause. Since ORDER BY clause cannot be used in UNIONS

96. What is Storage Manager? It is a program module that provides the interface between the low-level data

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stored in database, application programs and queries submitted to the system.

97. What is Buffer Manager? It is a program module, which is responsible for fetching data from disk storage into main memory and deciding what data to be cache in memory.

98. What is Transaction Manager? It is a program module, which ensures that database, remains in a consistent state despite system failures and concurrent transaction execution proceeds without conflicting.

99. What is File Manager? It is a program module, which manages the allocation of space on disk storage and data structure used to represent information stored on a disk.

100. What is Authorization and Integrity manager? It is the program module, which tests for the satisfaction of integrity constraint and checks the authority of user to access data.

101. What are stand-alone procedures? Procedures that are not part of a package are known as stand-alone because they independently defined. A good example of a stand-alone procedure is one written in a SQL*Forms application. These types of procedures are not available for reference from other Oracle tools. Another limitation of stand-alone procedures is that they are compiled at run time, which slows execution.

102. What are cursors give different types of cursors. PL/SQL uses cursors for all database information accesses statements. The language supports the use two types of cursors Ø Implicit Ø Explicit

103. What is cold backup and hot backup (in case of Oracle)? Ø Cold Backup: It is copying the three sets of files (database files, redo logs, and control file) when the instance is shut down. This is a straight file copy, usually from the disk directly to tape. You must shut down the instance to guarantee a consistent copy. If a cold backup is performed, the only option available in the event of data file

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loss is restoring all the files from the latest backup. All work performed on the database since the last backup is lost. Ø Hot Backup: Some sites (such as worldwide airline reservations systems) cannot shut down the database while making a backup copy of the files. The cold backup is not an available option. So different means of backing up database must be used — the hot backup. Issue a SQL command to indicate to Oracle, on a tablespace-by-tablespace basis, that the files of the tablespace are to backed up. The users can continue to make full use of the files, including making changes to the data. Once the user has indicated that he/she wants to back up the tablespace files, he/she can use the operating system to copy those files to the desired backup destination. The database must be running in ARCHIVELOG mode for the hot backup option. If a data loss failure does occur, the lost database files can be restored using the hot backup and the online and offline redo logs created since the backup was done. The database is restored to the most consistent state without any loss of committed transactions.

104. What are Armstrong rules? How do we say that they are complete and/or sound The well-known inference rules for FDs Ø Reflexive rule : If Y is subset or equal to X then X Y. Ø Augmentation rule: If X Y then XZ YZ. Ø Transitive rule: If {X Y, Y Z} then X Z. Ø Decomposition rule :

If X YZ then X Y. Ø Union or Additive rule: If {X Y, X Z} then X YZ. Ø Pseudo Transitive rule : If {X Y, WY Z} then WX Z. Of these the first three are known as Amstrong Rules. They are sound because it is enough if a set of FDs satisfy these three. They are called complete because using these three rules we can generate the rest all inference rules.

105. How can you find the minimal key of relational schema? 176

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Minimal key is one which can identify each tuple of the given relation schema uniquely. For finding the minimal key it is required to find the closure that is the set of all attributes that are dependent on any given set of attributes under the given set of functional dependency. Algo. I Determining X+, closure for X, given set of FDs F 1. Set X+ = X 2. Set Old X+ = X+ 3. For each FD Y Z in F and if Y belongs to X+ then add Z to X+ 4. Repeat steps 2 and 3 until Old X+ = X+

Algo.II Determining minimal K for relation schema R, given set of FDs F 1. Set K to R that is make K a set of all attributes in R 2. For each attribute A in K a. Compute (K – A)+ with respect to F b. If (K – A)+ = R then set K = (K – A)+

106. What do you understand by dependency preservation? Given a relation R and a set of FDs F, dependency preservation states that the closure of the union of the projection of F on each decomposed relation Ri is equal to the closure of F. i.e., ((PR1(F)) U … U (PRn(F)))+ = F+ if decomposition is not dependency preserving, then some dependency is lost in the decomposition.

107. What is meant by Proactive, Retroactive and Simultaneous Update. Proactive Update: The updates that are applied to database before it becomes effective in real world . Retroactive Update: The updates that are applied to database after it becomes effective in real world . Simulatneous Update: The updates that are applied to database at the same time when it becomes effective in real world .

108. What are the different types of JOIN operations? Equi Join: This is the most common type of join which involves only equality comparisions. The disadvantage in this type of join is that there

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