DBE Assignment#04

System Programming Sunita M. Dol, CSE Dept

Walchand Institute of Technology, Solapur

HANDOUT#04

Aim:

Basic SQL: Write simple queries in SQL on the schema created for a specific application.

Theory:

SQL is based on set and relational operations with certain modifications and enhancements. A

typical SQL query has the form:

select A1, A2, ..., An

from r1, r2, ..., rm

where P

– Ai represents an attribute,

– Ri represents a relation and

– P is a predicate.

The SELECT Clause

The select clause list the attributes desired in the result of a query. SQL allows duplicates in

relations as well as in query results. To force the elimination of duplicates, insert the keyword

distinct after select. The keyword all specifies that duplicates not be removed. An asterisk in the

select clause denotes “all attributes”. The select clause can contain arithmetic expressions

involving the operation, +, –, *, and /, and operating on constants or attributes of tuples.

The WHERE Clause

The where clause specifies conditions that the result must satisfy. Comparison results can be

combined using the logical connectives and, or, and not. Comparisons can be applied to results

of arithmetic expressions. SQL includes a between comparison operator.

The FROM Clause

The from clause lists the relations involved in the query.

The RENAME Operation

The SQL allows renaming relations and attributes using the as clause:

old-name as new-name

Tuple Variables

Tuple variables are defined in the from clause via the use of the as clause. Keyword as is

optional and may be omitted

borrower as T ≡ borrower T



String Operations

SQL includes a string-matching operator for comparisons on character strings. The operator

“like” uses patterns that are described using two special characters:

percent (%). The % character matches any substring.

underscore (_). The _ character matches any character.

SQL supports a variety of string operations such as

concatenation (using “||”)

converting from upper to lower case (and vice versa)

Finding string length, extracting substrings, etc.

Ordering the display of tuples

SQL offers the user some control over the order in which tuples in a relation are displayed. The

order by clause causes the tuples in the result of a query to appear in sorted order. We may

specify desc for descending order or asc for ascending order, for each attribute; ascending order

is the default.

Set Operations:

The SQL operations union, intersect, and except operate on relations and correspond to the

relational-algebra operations ∪ , ∩, and −. Like union, intersection, and set difference in

relational algebra, the relations participating in the operations must be compatible; that is, they

must have the same set of attributes.

In union operation, the number of duplicate tuples in the result is equal to the total

number of duplicates that appear in both d and b. Thus, if Jones has three accounts and

two loans at the bank, then there will be five tuples with the name Jones in the result.

In intersection operation, the number of duplicate tuples that appear in the result is equal

to the minimum number of duplicates in both d and b. Thus, if Jones has three accounts

and two loans at the bank, then there will be two tuples with the name Jones in the result.

In except operation, the number of duplicate copies of a tuple in the result is equal to the number

of duplicate copies of the tuple in d minus the number of duplicate copies of the tuple in b,

provided that the difference is positive. Thus, if Jones has three accounts and one loan at the

bank, then there will be two tuples with the name Jones in the result. If, instead, this customer

has two accounts and three loans at the bank, there will be no tuple with the name Jones in the

result.

Queries and Output: Select-From-Where Clause

Find the names of all branches in the loan relation

SQL> select branch_name from loan;



BRANCH_NAME

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

Round Hill

Downtown

Perryridge

Perryridge

Downtown

Redwood

Mianus

7 rows selected.

List the tuples of loan relation with amount multiplied by 100.

SQL> select * from loan;

LOAN_NUMBE BRANCH_NAME AMOUNT

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

L-11 Round Hill 900

L-14 Downtown 1500

L-15 Perryridge 1500


L-17 Downtown 1000

L-23 Redwood 2000

L-93 Mianus 500

7 rows selected.

SQL> select loan_number,branch_name,amount*100 from loan;

LOAN_NUMBE BRANCH_NAME AMOUNT*100

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

L-11 Round Hill 90000

L-14 Downtown 150000



L-17 Downtown 100000

L-23 Redwood 200000

L-93 Mianus 50000

7 rows selected.



Find all loan numbers for the loans made at Perryridge branch with amount greater than

$1200.



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

L-11 Round Hill 900

L-14 Downtown 1500



L-17 Downtown 1000

L-23 Redwood 2000

L-93 Mianus 500

7 rows selected.

SQL> select loan_number from loan where branch_name='Perryridge' and amount>1200;

LOAN_NUMBE

----------

L-15

L-16

Rename Operation



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

L-11 Round Hill 900

L-14 Downtown 1500



L-17 Downtown 1000

L-23 Redwood 2000

L-93 Mianus 500

7 rows selected.

SQL> select customer_name,borrower.loan_number as loan_id,amount

2 from borrower,loan



3 where borrower.loan_number=loan.loan_number;

CUSTOMER_NAME LOAN_ID AMOUNT

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

Smith L-11 900

Hayes L-15 1500

Adams L-16 1300

Williams L-17 1000

Jones L-17 1000

Smith L-23 2000

Curry L-93 500

7 rows selected.

Tuple Variables

Find the names of all branches that have assets greater than at least one branch located in

Brooklyn.

SQL> select * from branch;

BRANCH_NAME BRANCH_CITY ASSETS

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

Brighton Brooklyn 7100000

Downtown Brooklyn 9000000

Mianus Horseneck 400000

North Town Rye 3700000

Perryridge Horseneck 1700000

Pownal Bennington 300000

Redwood Palo Alto 2100000

Round Hill Horseneck 8000000

8 rows selected.

SQL> select distinct T.branch_name

2 from branch T,branch S

3 where T.assets>S.assets

4 and S.branch_city='Brooklyn';

BRANCH_NAME

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

Round Hill



Downtown

String Operations

Find the names of all customers whose street address includes substring 'Main'.

SQL> select * from customer;

CUSTOMER_NAME CUSTOMER_STREET

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

CUSTOMER_CITY

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

Adams Spring

Pittsfield

Brooks Senator

Brooklyn

Curry North

Rye


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

CUSTOMER_CITY

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

Glenn Sand Hill

Woodside

Green Walnut

Stamford

Hayes Main

Harrison


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

CUSTOMER_CITY

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

Johnson Alma

Palo Alto



Jones Main

Harrison

Lindsay Park

Pittsfield


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

CUSTOMER_CITY

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

Smith North

Rye

Turner Putnam

Stamford

Williams Nassau

Princeton

12 rows selected.

SQL> select customer_name from customer where customer_street like '%Main%';

CUSTOMER_NAME

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

Hayes

Jones


List the entire loan relation in descending order of amount.If several loans have the same

amount then order them in ascending order by loan number.



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

L-11 Round Hill 900



L-14 Downtown 1500



L-17 Downtown 1000

L-23 Redwood 2000

L-93 Mianus 500

7 rows selected.

SQL> select * from loan

2 order by amount desc,loan_number asc;


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

L-23 Redwood 2000

L-14 Downtown 1500



L-17 Downtown 1000

L-11 Round Hill 900

L-93 Mianus 500

7 rows selected.

Set Operations

The set of all customers who have an account at the bank

SQL> select * from depositor;

CUSTOMER_NAME ACCOUNT_NU

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

Hayes A-102

Johnson A-101

Johnson A-201

Jones A-217

Lindsay A-222

Smith A-215

Turner A-305

7 rows selected.



SQL> select customer_name from depositor;

CUSTOMER_NAME

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

Hayes

Johnson

Johnson

Jones

Lindsay

Smith

Turner

7 rows selected.

The set of customers who have loan at the bank.

SQL> select * from borrower;

CUSTOMER_NAME LOAN_NUMBE

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

Adams L-16

Curry L-93

Hayes L-15

Jones L-17

Smith L-11

Smith L-23

Williams L-17

7 rows selected.

SQL> select customer_name from borrower;

CUSTOMER_NAME

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

Adams

Curry

Hayes

Jones

Smith

Smith

Williams



7 rows selected.

Find all the bank customer having loan,an account or both at the bank.

SQL> (select customer_name from depositor)

2 union

3 (select customer_name from borrower);

CUSTOMER_NAME

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

Adams

Curry

Hayes

Johnson

Jones

Lindsay

Smith

Turner

Williams

9 rows selected.


2 union all


CUSTOMER_NAME

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

Hayes

Johnson

Johnson

Jones

Lindsay

Smith

Turner

Adams

Curry

Hayes

Jones



CUSTOMER_NAME

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

Smith

Smith

Williams

14 rows selected.

Find all the bank customer having loan and an account at the bank.


2 intersect


CUSTOMER_NAME

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

Hayes

Jones

Smith

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

SQL> (select customer_name from depositor) minus (select customer_name from borrower);

CUSTOMER_NAME

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

Johnson

Lindsay

Turner

Conclusion: We have written simple queries in SQL using

Select-from-where

Rename operation

Tuple variables

String Operation


Set operations

References:



Database system concepts by Abraham Silberschatz, Henry F. Korth, S. Sudarshan

(McGraw Hill International Edition) sixth edition.

Database system concepts by Abraham Silberschatz, Henry F. Korth, S. Sudarshan

(McGraw Hill International Edition) fifth edition.

http://codex.cs.yale.edu/avi/db-book/db4/slide-dir/






DBE Assignment#04

Engineering

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