Dbms lab Manual

SHRI RAWATPURA SARKAR INSTITUTE OF TECHNOLOGY-II NEW RAIPUR

Department

Of

Computer Science & Engineering

Data Base Management Systems Lab Manual

SHRI RAWATPURA SARKAR INSTITUTE OF TECHNOLOGY-II, NEW RAIPUR

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

1). Open your oracle application by the following navigationStart->all programs->oracle orahome.->application

development->sql.

2). You will be asked for user name, pass word and host stringYou have to enter user name, pass word and host string as given by the administrator. It will be different from one user to another user.

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

If you type your programs at sql prompt then screen will look like follow:SQL> SELECT ename,empno,

2 sal from 3 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 operations directly by selecting menu options provided by it.

To execute the program which saved; do the followingSQL> @ programname.sqlOrSQL> Run programname.sql Then press ‘\’ key and enter.

This how we can write, edit and execute the sql command and programs.Always you have to save your programs in your own logins.

2


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…

DDL commands:

1. The Create Table Command: - it defines each column of the table uniquely. Each column has minimum of three attributes, a name , data type 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),<new col>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));

5. Renaming the tables

Syntax:Rename <oldtable> to <new table>;

3


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 most natural thing to do is load this table with data to be manipulated later.

Syntax:insert into <tablename> (<col1>,<col2>) values(<exp>,<exp>);

9. Delete operations.

a) remove all rowsSyntax: delete from <tablename>;

b) removal of a specified row/sSyntax: delete from <tablename> where <condition>;

10. Updating the contents of a table.

a) updating all rowsSyntax:Update <tablename> set <col>=<exp>,<col>=<exp>;

b) updating seleted records.Syntax:Update <tablename> set <col>=<exp>,<col>=<exp>

where <condition>;4


11. Types of data constrains.a) not null constraint at column level.Syntax:<col><datatype>(size)not null

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

c) unique constraint at table level:Syntax:Create table tablename(col=format,col=format,unique(<col1>,<col2>);

d) primary key constraint at column levelSyntax:<col><datatype>(size)primary key;

e) primary key constraint at table level.Syntax:Create table tablename(col=format,col=format primary key(col1>,<col2>);

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 constraintcheck constraint constraint at column level.Syntax: <col><datatype>(size) check(<logical expression>)

i) check constraint constraint at table level.Syntax: check(<logical expression>)

5


DQL Commands:

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

a) all rows and all columnsSyntax:

Select <col> to <col n> from tablename;

Select * from tablename;

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

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

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 information stored in its tables from unauthoraised viewing and damage.The rights that allow the user of some or all oracle resources on the server are called privileges.

a) Grant privileges using the GRANT statement

The grant statement provides various types of access to database objects 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:

6


The REVOKE statement is used to deny the Grant given on an object.

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

CSE/5th/DBMS Lab/Prepared by Vivek Kumar Sinha

7


Experiment 1

Database Schema for a customer-sale scenario

Customer(Cust id : integer, cust_name: string)Item(item_id: integer, item_name: string, price: integer)Sale(bill_no: integer, bill_data: date, cust_id: integer, item_id: integer, qty_sold: integer)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List all the bills for the current date with the customer names and item numbersd) List the total Bill details with the quantity sold, price of the item and the final amounte) List the details of the customer who have bought a product which has a price>200f) Give a count of how many products have been bought by each customerg) Give a list of products bought by a customer having cust_id as 5h) List the item details which are sold as of todayi) Create a view which lists out the bill_no, bill_date, cust_id, item_id, price, qty_sold, amountj) Create a view which lists the daily sales date wise for the last one week

Aim: Create the tables with the appropriate integrity constraints and Insert around 10 records in each of the tables

SQL> create table customer1 (cust_id number(5) primary key, cust_name varchar2(15));Output: Table created.

SQL> desc customer1;

Output: Name Null? Type----------------------------------------- -------- ----------------CUST_ID NOT NULL NUMBER(5)CUST_NAME VARCHAR2(15)

Valid Test Data

b) SQL> insert into customer1 values(&custid,'&custname');SQL> select * from customer1;Output:CUST_ID CUST_NAME---------- ---------------

100 ramu 101 kamal 102 raju 103 raju sundaram 104 lawrence

8


SQL> create table item(item_id number(4) primary key, item_name varchar2(15),price number(6,2));SQL> dsec item

Output:Name Null? Type……………………………………………………………………………………………………Cust_id NOT NULL NUMBER(4)Item_name VARCHAR2(15)PRICE NUMBER(6,2)SQL>insert into item values(&item_id,’&item_name’,&price);

SQL> select * from item;Output:ITEM_ID ITEM_NAME PRICE……………………………………………………………………………………..2334 geera 6.254532 corn soup 34.652124 lays chips 204531 setwet 99.992319 duracell 45.5

SQL>create table sale(bill_no number(5) primary key,bill_date date, cust_id number(5) references customer(cust_id), item_id number(4) references item(item_id),qty_sold number(4));

Out put: Table Created.

SQL>dsec saleOutput: Name Null? Type………………………………………………………………………………………..BILL_NO NOT NULL NUMBER(4)BILL_DATE DATECUST_ID NUMBER(5)ITEM_ID NUMBER(4)QTY_SOLD NUMBER(4)

SQL>insert into Sale values(&bill_no, ’&bill_date’, &cust_id, &item_id, &qty_sold);

SQL>select * from sale;Output:BILL_NO BILL_DATE CUST_ID ITEM_ID QTY_SOLD………………………………………………………………………………………………………...1450 04-JAN-06 100 2124 21451 04-JAN-06 101 2319 1

9


1452 04-JAN-06 103 4531 21453 04-JAN-06 102 2334 31454 04-JAN-06 104 4532 3

c) List all the bills for the current date with the customer names and item numbersSQL> select c.custname, i.itemid, s.billno from customer c, item I, sale s

where c.custid=s.custid ands.billdate=to_char(sysdate);

CUSTNAME ITEMID BILLNO------------- --------- ---------John 5001 332

d) List the total Bill details with the quantity sold, price of the item and the final amountSQL> select i.price, s.qty,(i.price*s.qty) total from item I, sale s where i.itemid=s.itemid;

PRICE QTY TOTAL------- ----- --------120 2 24020 3 605 2 1010 1 10350 4 1400

e) List the details of the customer who have bought a product which has a price>200SQL> select c.custid, c.custname from customer c, sale s, item i where i.price>200 and

c.custid=s.custid and i.itemid=s.itemid;

CUSTID CUSTNAME--------- --------------

4 duffy

f) Give a count of how many products have been bought by each customerSQL> select custid, count(itemid) from sale group by custid;

CUSTID COUNT(ITEMID)---------- ---------------------1 23 14 15 1

g) Give a list of products bought by a customer having cust_id as 5SQL> select i.itemname from item i, sale s where s.custid=5 and i.itemid-s.itemid;

ITEMNAME--------------Pens

10


h) List the item details which are sold as of todaySQL> select i.itemid, i.itemname from item I, sale s where i.itemid=s.itemid

and s.billdate=to_char(sysdate);

ITEMID ITEMNAME--------- -------------

1234 pencil

i) Create a view which lists out the bill_no, bill_date, cust_id, item_id, price, qty_sold, amountSQL>create view cust as (select s.billno, s.billdate, c.custid, i. iitemid, i.price, s.qty from

customer c,sale s item I where c.custid=s.custid and i.iemid=s.itemid);

view created.

SQL>select * from cust;

BILLNO BILLDATE CUSTID ITEMID PRICE QTY……………………………………………………………………………………………3432 12-JAN-06 3 3244 120 24424 20-FEB-06 1 3456 20 3332 13-MAR-06 1 1234 5 22343 10-MAR 5 5001 10 11331 11-MAR-06 4 76776 350 4

11


Experiment 2

Database Schema for a Student Library scenario

Student(Stud_no : integer, Stud_name: string)Membership(Mem_no: integer, Stud_no: integer)Book(book_no: integer, book_name:string, author: string)Iss_rec(iss_no:integer, iss_date: date, Mem_no: integer, book_no: integer)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List all the student names with their membership numbersd) List all the issues for the current date with student and Book namese) List the details of students who borrowed book whose author is CJDATEf) Give a count of how many books have been bought by each studentg) Give a list of books taken by student with stud_no as 5h) List the book details which are issued as of todayi) Create a view which lists out the iss_no, iss _date, stud_name, book namej) Create a view which lists the daily issues-date wise for the last one week

AIM: Create the tables with the appropriate integrity constraints Insert around 10 records in each of the tables

SQL>create table student(stud_no number(5) primary key,stud_name varchar2(15));

SQL>desc student; Name Null? Type………………………………………………………………………………………..STUD_NO NOT NULL NUMBER(5)STUD_NAME VARCAHR2(15)Valid Test Data:

SQL>insert into student values(&stud_no,’&stud_name’);

SQL>select * from student;

STUD_NO STUD_NAME....................................................................

508 HARISH513 BALAJI518 RAKESH524 PAVAN

12


534 JOYCE

SQL>create table membership(mem_no number(5) primary key,stud_no number(5) references student(stud)no));SQL>dsec membership;

Name Null? Type…………………………………………………………………………………………………….MEM_NO NOT NULL NUMBER(5)STUD_NO NUMBER(5)

SQL>insert into membership values(&mem_no,&stud_no);Enter value for mem_no:5440Enter value for stud_no:510old 1:insert into membership values(&mem_no,&stud_no)new 1:insert into membership values(5440,510)insert into membership values(5440,510)*Errors Observed:

ERROR at line 1:ORA-02291:integrity constraint(HARISH.SYS_C002724)violated-primary key not found

SQL>select * from membership; MEM_NO STUD_NO………………………………………………………………………..

5440 5135441 5085442 5185443 5345444 524

SQL>create table book(book_no number(5) primary key,book_name varchar2(20),author varchar2(2));

SQL>desc book;

Name Null? Type………………………………………………………………………………………..BOOK_NO NOT NULL NUMBER(5)BOOK_NAME VARCHAR2(20)AUTHOR VARCHAR2(20)

SQL>insert into book values(&book_no,’&book_name’,’&author’);SQL>select * from book;

BOOK_NO BOOK_NAME AUTHOR………………………………………………………………………………………………..9123 DBMS Rama Krishna2342 JAVA Robett wilkins

13


4523 Fearless tales Alfred8723 my ambition Harish7821 Harry Potter JK Rowling

SQL>create table lss_rec(iss_no number primary key,iss_date date,mem_no number(5) references membership(mem_no),book_no number(5) references book(book_no));

SQL>desc iss_rec;Name Null? Type………………………………………………………………………………………………………ISS_NO NOT NULL NUMBERISS_DATE DATEMEM_NO NUMBER(5)BOOK_NO NUMBER(5)

SQL>select * from iss_rec;ISS_NO ISS_DATE MEM_NO BOOK_NO…………………………………………………………………………………………………43 05-JAN-06 5443 452381 28-DEC-05 5441 872322 08-DEC-05 5440 782153 07-JAN-06 5442 912335 06-JAN-06 5444 2342

c) List all the student names with their membership numbers

SQL> select s.studname, m.memno from student s, membership m where m.studno=s.studno;

STUDNAME MEMNO------------- --------abhijeet 1001arun 1002arvind 1003ashish 1004ashwin 1005

d) List all the issues for the current date with student and Book names

SQL> select i.issno, s.studname, b.bookname from iss_rec I, membership m, student s, book b2 where i.memno=m.memno and m.studno=s.studno and i.issdate=to_char(sysdate);

ISSNOSTUDNAME BOOKNAME------- ------------ ---------------13 arvind P&S

e) List the details of students who borrowed book whose author is CJDATE

14


SQL> select * from student where studno in(select studno from membership where memno in 2 (select memno from iss_rec where bookno in(select bookno from book where author=’CJDATE’)));

STUDNO STUDNAME---------- -------------

505 ashwin

f) Give a count of how many books have been bought by each student

SQL> select s.studno, count(i.bookno) from student s.membership m, book b, 2 iss_rec I where s.studno=m.studno and b.bookno=i.bookno group by s.studno;

STUDNO COUNT(I.BOOKNO)---------- -----------------------501 5502 5503 5504 5505 5

g) Give a list of books taken by student with stud_no as 5

SQL> select bookname from book where bookno in (select bookno from iss_rec where2 memno in(select memno from membership where3 studno in(select studno from student where studno=5)));

BOOKNAME-------------NT

h) List the book details which are issued as of today

SQL> delete from book where bookno in(select bookno from iss_rec where issdate=to_char(sysdate));delete from book where bookno in (select bookno from iss_rec where issdate=to_char(sysdate))

Errors Observed:

ERROR at line 1:ORA-02292: integrity constraint (SCOTT.SYS_C00840) violated – child record found

15


Experiment 3

Database Schema for a Employee-pay scenario

employee(emp_id : integer , emp_name: string)department(dept_id: integer, dept_name:string)paydetails(emp_id : integer, dept_id: integer, basic: integer, deductions: integer, additions: integer, DOJ: date)payroll(emp_id : integer, pay_date: date)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List the employee details department wised) List all the employee names who joined after particular date e) List the details of employees whose basic salary is between 10,000 and 20,000f) Give a count of how many employees are working in each departmentg) Give a names of the employees whose netsalary>10,000 h) List the details for an employee_id=5i) Create a view which lists out the emp_name, department, basic, dedeuctions, netsalaryj) Create a view which lists the emp_name and his netsalary

16



Create table employee(emp_id number(5) primary key,emp_name varchar2(25));

SQL>desc employee;

Name Null? Type………………………………………………………………………………………..EMP_ID NOT NULL NUMBER(5)EMP_NAME VARCHAR2(25)

Valid Test Data:

SQL>insert into employee values(&emp_id,’&emp_name’);SQL>select * from employee;

EMP_ID EMP_NAME………………………………………………………….

10 Robert21 Coulthard

30 Fernando Alonso 39 Kartikeyan 87 Kimmi

SQL>create table department(dept_id number(5) primary key,dept_name varchar2(20));

SQL>desc department;Name Null? Type………………………………………………………………………………………..DEPT_ID NOT NULL NUMBER(5)DEPT_NAME VARCHAR2(20)

SQL>insert into department values(&dept_id,’&dept_name’);SQL>select * from department;

DEPT_ID DEPT_NAME……………………………………………………………………………..

100 sales101 accounts102 administration103 production104 supervisor

SQL>create table paydetails(emp_id number(5) references employee(emp_id),dept_id number(5) reerences department(dept_id),basic number(7,2),deductions number(5,2),additions number(5,2),doj date);

17


SQL>desc paydetails;

Name Null? Type………………………………………………………………………………………..EMP_ID NUMBER(5)DEPT_ID NUMBER(5)BASIC NUMBER(7,2)DEDUCTIONS NUMBER(5,2)ADDITIONS NUMBER(5,2)DOJ DATE

Different Data Sets:SQL>insert into paydeatils values(&emp_id,&dept_id,&basic,&deductions,&additions,&doj);

SQL>select * from paydeatils;

EMP_ID DEPT_ID BASIC DEDUCTIONS ADDITIONS DOJ…………………………………………………………………………………………………………………..10 101 25023.12 43.09 71.23 08-JAN-9321 100 10500.29 23.98 40.9 01-JAN-06 30 102 6500.5 30.54 15 06-JUL-9739 103 9700.45 32.78 65.09 08-AUG-0387 104 15000 97.66 154.8 24-SEP-04

SQL>create table payroll(emp_id number(5)references employee(emp_id),pay_date date);SQL>desc payroll;

Name Null? Type………………………………………………………………………………………..EMP_ID NUMBER(5)PAY_DATE DATE

SQL>insert into payroll values(&emp_id,’&date’);

SQL>select * from payroll;EMP_ID PAY_DATE………………………………………………………….10 31-JAN-0621 03-FEB-0630 15-JAN-0639 27-JAN-0687 04-FEB-06c) List the employee details department wise

SQL>select empid,deptid from paydet;

EMPID DEPTID18


…………………………401 500402 200403 600404 400405 1200

d) List all the employee names who joined after particular date

SQL>select e,empname from employee e,paydet p where e.empid=p.empid and p.doj>=’05-mar-06’;

EMPNAME…………………AVINASHNITINPHALGUN

e) List the details of employees whose basic salary is between 10,000 and 20,000

sqL> Select empid,empname from employee where salary between 10000 and 20000;

EMPID EMPNAME…………………………….

402 AKHILA403 aaaaaaaa

EMPID EMPNAME…………………………….

AKHILA

f) Give a count of how many employees are working in each departmentSQL>select count(empid),deptid from paydet group by deptid;

COUNT (EMPID) DEPTID………………………………………………………

1 200 1 4001 5001 6001 1200

g) Give a names of the employees whose netsalary>10,000

SQL> select empname from employee where empid in(select empid from paydet where basic-deduction>10000);

EMPNAME………………

19


AVINASHAKHILAHARISHNITINPHALGUN

h) List the details for an employee_id=5

SQL> select * from employee where empid=5;

EMPID EMPNAME------------------------------------------5 Coulthard

20


Experiment 4

Database Schema for a Video Library scenario

Customer(cust_no: integer,cust_name: string)Membership(Mem_no: integer, cust_no: integer)Cassette(cass_no:integer, cass_name:string, Language: String) Iss_rec(iss_no: integer, iss_date: date, mem_no: integer, cass_no: integer)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List all the customer names with their membership numbersd) List all the issues for the current date with the customer names and cassette namese) List the details of the customer who has borrowed the cassette whose title is “ The Legend”f) Give a count of how many cassettes have been borrowed by each customerg) Give a list of book which has been taken by the student with mem_no as 5h) List the cassettes issues for today


SQL>create table customer(cust_no number(5) primary key,cust_name varchar2(20));

SQL>desc customer;

Name Null? Type……………………………………………………………………………………………………………..CUST_NO NOT NULL NUMBER(5)CUST_NAME VARCHAR2(20)

Valid Test Data:

SQL>insert into customer values(&cust_no,’&cust_name’);SQL>select * from customer;

CUST_NO CUST_NAME……………………………………………………………….

50 scott51 pandey52 varshney53 naidu54 bhimbra

21


SQL>create table membership(mem_no number(5) primary key,cust_no number(5) references customer(cust_no));

SQL>dsec membership;Name Null? Type………………………………………………………………………………………………………...MEM_NO NOT NULL NUMBER(5)CUST_NO NUMBER(5)

SQL>insert into memship values(&mem_no,&cust_no);SQL>select * from memship;

MEM_NO CUST_NO…………………………………………………920 50981 51897 52820 53928 54 SQL>create table cassette(cass_no number(5) primary key,Cass_name varchar2(15),language varchar2(15));

SQL>desc cassette;

Name Null? Type………………………………………………………………………………………..CASS_NO NOT NULL NUMBER(5)CASS_NAME VARCHAR2(15)LANGUAGE VARCHAR2(15)

SQL>insert into cassette values(&cass_no,’&cass_name’,’&language’);

SQL>select * from cassette;

CASS_NO CASS_NAME LANGUAGE………………………………………………………………………………………1 tagore telugu2 the lion king English3 anniyan tamil4 indra telugu5 lord of rings English

SQL>create table issu_rec(iss_no number(5) primary key,iss_date date,mem_no number(5)references memship(mem_no),cass_no number(5) references cassette(cass_no));

SQL>desc issu_rec;Name Null? Type………………………………………………………………………………………………………...

22


ISS_NO NOT NULL NUMBER(5)ISS_DATE DATEMEM_NO NUMBER(5)CASS_NO NUMBER(5)

SQL>select * from issu_rec;

ISS_NO ISS_DATE MEM_NO CASS_NO……………………………………………………………………………………22 07-JAN-06 920 123 10-JAN-00 981 226 10-JAN-06 897 53 01-JAN-06 820 434 31-DEC-05 928 3

c) List all the customer names with their membership numbersSQL>select c.custname,m.memno from customer1 c,membership1 m where c.custno=m.custno;

CUSTNAME MEMNO……………….. ………………..NIKHIL 51VIVEK 52SHRAVAN 58VAMSI 57SHIVA 56

d) List all the issues for the current date with the customer names and cassette names

SQL>select i.issno,c.custname,cc.cassettename from customer1 c,membership1 m,cassette cc,issrec1 I where i.issdate=to_char(sysdate) and c.custno=m.custno and i.cassno=cc.cassno and i.memno=m.memno;

OutPut:

no rows selected.

23


Experiment 5

Database Schema for a student-Lab scenario

Student(stud_no: integer, stud_name: string, class: string)Class(class: string, descrip: string)Lab(mach_no: integer, Lab_no: integer, description: String)Allotment(Stud_no: Integer, mach_no: integer, dayof week: string)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List all the machine allotments with the student names, lab and machine numbersd) List the total number of lab allotments day wisee) Give a count of how many machines have been allocated to the ‘CSIT’ classf) Give a machine allotment etails of the stud_no 5 with his personal and class detailsg) Count for how many machines have been allocatedin Lab_no 1 for the day of the week as

“Monday”h) How many students class wise have allocated machines in the labsi) Create a view which lists out the stud_no, stud_name, mach_no, lab_no, dayofweekj) Create a view which lists the machine allotment details for “Thursday”


SQL>create table stu(stud_no number(5) primary key,stud_nam varchar2(20),class varchar2(20));

SQL> desc stu;

Name null? TypeSTUD_NO NOT NULL NUMBER(5)STUD_NAM VARCHAR2(20)CLASS VARCHAR2(20)

Valid Data Sets:

SQL> insert into stu values(&stud_no,’&stud_nam’,’&class’);SQL> select * from stu;

STUD_NO STUD_NAM CLASS39 LEON CSE34 VIKAS CSIT18 MATHEW ECE8 HANSEN MECH24 ALEXIS EEE

24


SQL> Create table class (class varchar2(20), descript varchar2(10));SQL> Describe class;

Name null type

CLASS VARCHAR2(10)DESCRIPT VARCHAR2(20)

SQL> create table lab(match_no number(5), lab_no number(5), description varchar2(20));

SQL> desc lab;

Name null type

MACH_NO NOT NULL NUMBER(5)LAB_NO NUMBER(5)DESCRIPTION VARCHAR2(20)

SQL> insert into lab values(&mach_no,&lab_no,’&description’);

SQL> select * from lab;

MATCH_NO LAB_NO DESCRIPTION--------------- --------- --------------------23 7 physics78 2 chemistry87 1 edc12 10 cds8 3 java lab

SQL> create table allotment(stud_no number(5) references stu(stud_no), match_no number(5) references lab(mach_no),Doweek varchar2(20));

SQL> desc allotment;

Name Null? Type-------------- ------- ---------STUD_NO NUMBER(5)MACH_NO NUMBER(5)DOWEEK VARCHAR2(20)

SQL>select * from allotment;

STUD_NO MACH_NO DOWEEK------------- -- ------------ ------------

25


39 23 sat34 87 mon18 78 tue8 12 wed24 12 thu

c) List all the machine allotments with the student names, lab and machine numbers

SQL>select s.studname,l.machno from student1 s,lab l,allotment a where a.machno=l.machno and a.studno=s.studno;

STUDNAME MACHNO………………………………………..ABHIJEET 1KALYAN 22 ASHWIN 3ARKA 4ARVIND 5

d) List the total number of lab allotments day wise

SQL>select l.machno,l.descrip,a.day from lab l,allotment a where a.machno=l.machno;

MACHNO DESCRIP DAY……………………………………………………………………1 UNIX MONDAY22 UNIX TUESDAY3 XP WEDNESDAY4 WINDOWS THRUSDAY5 ME FRIDAY

e) Give a count of how many machines have been allocated to the ‘CSIT’ class

SQL>select count(machno)from allotment where studno in(select studno from student1 where class=’CSIT’);

COUNT (MACHNO)…………………….. 1

f) Give a machine allotment etails of the stud_no 5 with his personal and class details

SQL>select a.studno,a.machno,s.studname,s.class from allotment a,student1 s where a.studno=s.studno and a.studno=503;

STUDNO MACHNO STUDNAME CLASS………………………………………………………………………………………………………

26


503 5 ARVIND CSE

g) Count for how many machines have been allocatedin Lab_no 1 for the day of the week as “Monday”

h) How many students class wise have allocated machines in the labs

SQL>select count(studno) “allocated students in the labs”,class from student1 where studno in(select studno from allotment) group by class;

allocated students in the lab CLASS…………………………………………………………………………… 2 CSE 1 ECE 1 EEE 1 IT

27


Experiment 6

Write a program to find largest number from the given three numbers.

Aim: To find largest number from the given three numbers.

Algorithm:

Step 1: Declare the variable A, B, and C.Step 2: Store the valid data.Step 3: Compare variable A with B and A with CStep 4: If the value stored in variable A is big, it displays “A is Big”. (IF conditional statement should be used)Step 5: Compare variable B with CStep 6: If the value stored in variable B is big, it displays “B is Big”.Step 7: other wise it displays “C is Big”

Declare

A number;

B number;

C number;

Begin

A:=&a;

B:=&b;

C:=&c;

If a > b && a> c then

Dbms_output.put_line(‘ A is big ‘);

Else

If( b>c && b> a ) then

Dbms_output.put_line(‘ B is big ‘);

Else

28


Dbms_output.put_line(‘ C is big ‘);

End if;

End if;

End;

Valid Data Sets:

Enter the value of a:1Enter the value of b:2Enter the value of c:3

OUTPUT:

C is big

Invalid Data sets :

Enter the value of a:yEnter the value of b:xEnter the value of c:a

Output:

Invalid data types.

29


Experiment 7

Simple programs using loop, while and for iterative control statement.

a) To generate first 10 natural numbers using loop, while and for.

AIM: To generate first 10 natural numbers using loop, while and for.

Step 1: Declare the variable I.Step 2: Store the valid data 1 in I.Step 3: Use LOOP statementStep 4: Display the first value.Step 5: Increment the value of I by 1 value.Step 6: check the value up to 10 no. and repeat the loopStep 7: If condition exceeds the given value 10, the loop will be terminated.

/* using loop statement */

Declare

I number;

Begin

I:=1;

Loop

Dbms_output.put_line(I);

I:=I+1;

Exit when I>10;

End loop;

End;

Algorithm: for WHILE loop

Step 1: Declare the variable I.Step 2: Store the valid data 1 in I.Step 3: Use WHILE statement

30


Step 4: Check the value of I with value 10.Step 5: if the value of I reached to 10 the loop will be terminatedStep 6: otherwise display value of IStep 7: increment the next value of I using I=I+1.

/* using while */

Declare

I number;

Begin

I:=1;

While (I<=10)

loop


I:=I+1;

End loop;

End;

31


Algorithm:Step 1: Declare the variable I.Step 2: Store the value 1 in var. I.Step 3: Use For… LOOP statementStep 4: Display the first value of I.Step 5: Increment the value of I by 1 value.Step 6: check the value up to 10 no. and repeat the loopStep 7: if the loop exceeds the value 10 then the loop will be terminated.

/* using for loop*/

Begin

For I in 1..10

loop


End loop;

End;

Valid Test Data:

OUTPUT

12345678910

32


Experiment 8

Program to check whether given number is Armstrong or not.

AIM: to check whether given number is Armstrong or not.

Algorithm:

Step 1: Declare the variable N, S, D and DUP.Step 2: Store the value in var. N and var. DUP..Step 3: check for the value of N, which is not equal to 0.Step 4: divide value stored in N by 10 and store it var. D. (D=n%10).Step 5: the reminder will be multiply 3 times and store it in Var. S. Step 6: The coefficient will be calculated using FLOOR function. And store it in var. N.Step 7: repeat the Steps 3, 4, 5, and 6 till loop will be terminated.Step 8: Check whether the stored value and calculated values are sameStep 9: if both the values are same, then display “The given number is

Armstrong”Step 10: Otherwise display “it is not Armstrong” and terminate the loop.

Declare

N number;S number;D number;

Begin

N:=&n;

S:=0;

While(n!=0)

Loop

D=n%10;S:=s+(D*D*D);N:=floor(n/10);

End loop;

If (DUP=S) then

DBMS_output.put_line(‘number is armstrong’);

Else

33


DBMS_output.put_line(‘number is not armstrong’);

End if;

End;

Test Valid Data Set:

Enter value of n

153

OUTPUT:

number is Armstrong.

34


Experiment 9

Write a program to generate all prime numbers below 100.

AIM: to generate all prime numbers below 100.

Declare

I number;

J number;

C number;

Begin

While(i<=100)

Loop

C:=0;J:=1;

While(j<=i)

Loop

If(floor(i%j)=0) then

C:= C+1;

End if;

J:=j+1;

End loop;

If(c=2) then

Dbms_output.put_line(i);

End if;

Endloop; End;

35


Valid Test Data

OUTPUT:

2

3

5

7

11

.

.

97

36


Experiment 10

Write a program to demonstrate the GOTO statement.

AIM: to demonstrate the GOTO statement

Declare

I number;

Begin

I:=1;

If(i>=0) then

GOTO here;

Else

Dbms_output.put_line( ‘ I is negative’);

End if;

<<here>>

Dbms_output.put_line( ‘ I is positive’);

End;

Valid Test Data

OUTPUT:

I is positive

37


11. Write a program to demonstrate %type and %rowtype attributes

AIM: to demonstrate %type and %rowtype attributes

HW/SW requirements: Processor : AMD Athelon ™ 1.67 GHz

RAM : 256 MBHard Disk : 40 GBSoftware : Oracle, PlSQL

Declare

My_Empno emp.empno%type;My_Ename emp.ename%type;My_Emprow emp%rowtype;No number;

Begin

No:=&no;

Select empno,ename into my_empno,my_ename from emp where empno=no;

If(SQl%rowcount=1) then

Dbms_output.put_line(‘empno is’ || my_empno || ‘ename is ‘ || my_ename);

Else

Dbms_output.put_line( ‘error’);

End if;

Select * into my_emprow from emp where empno=no;

If(SQl%rowcount=1) then

Dbms_output.put_line(‘empno is’ || my_emprow.empno || ‘ename is ‘ || my_emprow.ename);

Else

Dbms_output.put_line( ‘error’);

38


End if;

End;

Valid Test Data

Enter the value for no:

7788

OUTPUT

empno is 7788 ename is vinay s.

empno is 7788 ename is vinay s.

39


12. Write a program to demonstrate predefined exceptions

AIM: to demonstrate predefined exceptions

Declare

A number

B number;

C number;

Begin

A:=&a;

B:=&b;

C:=a/b;

Dbms_output.put_line(‘division is ‘ || C);

Exception

If (ZERO_DIVIDE) then

Dbms_output.put_line(‘b could not be zero’);

End if;

End;

Valid Test Data:

Enter the value for a:

10

Enter the value for b:

0

OUTPUT:

b could not be zero

40


14. Write a program to demonstrate user defined exceptions

AIM: to demonstrate user defined exceptions

Declare

A number

B number;

C number;

Mydivide_zero EXCEPTION;

Begin

A:=&a;

B:=&b;

If(B=0) then

Raise Mydivide_zero;

else

C:=a/b;

Dbms_output.put_line(‘division is ‘ || C);

End if;

Exception

If (mydivide_zero) then

Dbms_output.put_line(‘b could not be zero’);

End if;

End;

Valid Test Data:

Enter the value for a:

10

41


Enter the value for b:

0

OUTPUT:

b could not be zero

42


15. Create a Cursor which update the salaries of an Employee as follows.

1. if sal<1000then update the salary to 1500.

2. if sal>=1000 and <2000 then update the salary to 2500.

3. if sal>=2000 and <=3000 then update the salary to 4000.

And also count the no.of records have been updated.*/

Declare

Cursor my_cur is select empno,sal from emp;

Xno emp.empno%type;Xsal emp.sal%type;C number;

Begin

Open my_cur;

C:=0;

Loop

Fetch my_cur into xno,xsal;

If(xsal<1000) then

Update emp set sal=3000 where empno=xno;

C:=c+1;

Else if(xsal>=2000) and xsa<3000) then


C:=c+1;

End if;

End if;

Exit when my_cur%NOTFOUND ;

43


End loop;

Close my_cur;

Dbma_output.put_line(c||’records have been successfully updated’);

End;

Sql>@a.sql;

records have been successfully updated

pl/sql procedure successfully completed.

Valid Test Data

Before executing the cursor, the records in emp table as follows

Sql>select * from emp;

OUTPUT:

EMPNO ENAME JOB MGR HIREDATE SAL COMMD EPTNO

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

7369 SMITH CLERK 7902 17-DEC-80 2000 20

7499 ALLEN SALESMAN 7698 20-FEB-81 1600 300 30

7521 WARD SALESMAN 7698 22-FEB-81 1250 500 30

EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO

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

7566 JONES MANAGER 7839 02-APR-81 2975 20

7654 MARTIN SALESMAN 7698 28-SEP-81 1250 1400 30

7698 BLAKE MANAGER 7839 01-MAY-81 2850 30

…

….

… 14 rows selected.

44


16. create a procedure which generate all the prime numbers below the given number and count the no.of prime numbers.

Create or replace procedure prime_proc(n IN number,tot OUT number) as

i number;

c number;

j number;

Begin

i:=1;

tot:=0;

while(i<=n)

loop

j:=1;

c:=0;

while(j<=i)

loop

if(mod(I,j)=0) then

c:=c+1;

end if;

j:=j+1;

end loop;

if(c=2) then

dbms_output.put_line(i);

tot:=tot+1;

end if;

45


i:=i+1;

end loop;

end;

/

Sql>procedure created.

declare

t number;

begin

prime_proc(10,t);

dbms_output.put_line(‘the total prime no .are’||t);

end;

Valid Test Data:

sql>set serveroutput on

OUTPUT

sql>/

2

3

5

7

The total prime no.are 4

Pl/sql procedure successfully completed.

46


17. create a procedure which updates the salaries of an employees as follows.

1.if sal<1000 then update the salry to 1500.

2.if sal>=1000 and <=2400 then update the salary to 2500.*/

Create or replace procedure myproc as

Cursor my_cur is select empno,sal from emp;

Xno emp.empno%type;

Xsal emp.sal%type;

C number;

Begin

Open my_cur;

C:=0;

Loop

Fetch my_cur into xno,xsal;

If(xsal<1000) then


C:=c+1;

Else

Is(xsal>=1000 and xsal<=2400) then


C:=c+1;

End if;

End if;

Exit when my_cur%NOTFOUND;

End loop;

47


Close my_cur;

Dbms_output.put_line(c||’records have been successfully updated’);

End;

/

Valid Test Data:

Procedure created.

Sql>exec myproc;

OUTPUT:

Records have been successfully completed.

/* create function which add two given numbers. (Simple programs) */

Create or replace function add_fun(a number,b number) return

Number as

C number;

Begin

C:=a+b;

Return c;

End;

/

Function created.

/*add_fun specification*/

Declare

Result number;

48


Begin

Result:=add_fun(10,20);

Dbms_output.put_line(‘the sum of 10 and 20 is’||result);

End;

Sql>/

The sum of 10 and 20 is 30

Pl/sql procedure successfully completed.

/*create a function which count total no.of employees having salary less than 6000.*/

/*function body*/

Create or replace function count_emp(esal number)return number as

Cursor vin_cur as Select empno,sal from emp;

Xno emp.empno%type;

Xsal emp.sal%type;

C number;

Begin

Open vin_cur;

C:=0;

loop

fetch vin_cur into xno,xsal;

if(xsal<esal) then

c:=c+1;

end if;

exit when vin_cur%notfound;

49


end loop;

close vin_cur;

return c;

end;

/

Function created.

/*function specification*/

Declare

Ne number;

Xsal number;

Begin

Ne:=count_emp(xsal);

Dbms_output.put_line(xsal);

Dbma_output.put_line(‘there are ‘||ne||;employees’);

End;

/

OUTPUT

There are 8 employees.

50


51


52

Dbms lab Manual

Engineering

Transcript of Dbms lab Manual