9 Gtu Query Algebra Solution

8/18/2019 9 Gtu Query Algebra Solution

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Darshan Institute of Engineering & Technology

130703‐Database Management System (DBMS)

GTU Q. Papers - Query

Firoz Sherasiya Page 1

We have following relations:

Supplier(S#,sname,status,city)

Parts(P#,pname,color,weight,city)

SP(S#,P#,quantity)

Answer the following queries in SQL.

1) Find name of supplier for city = ‘Delhi’.

Select sname from supplier

Where city=’delhi’;

2)

Find suppliers whose name start with ‘AB’

Select * from supplier

Where sname like ‘AB%’;

3)

Find all suppliers whose status is 10, 20 or 30.


Where status IN (10, 20, 30);

4)

Find total number of city of all suppliers.

Select count(city) from supplier;

5)

Find s# of supplier who supplies ‘red’ part.

Select sp.s# from sp

Inner join supplier s on s.s#=sp.s#

Inner join parts p on p.p#=sp.p#

Where p.color=’red’;

6) Count number of supplier who supplies ‘red’ part.

Select count(*) from sp

Inner join supplier s on s.s#=sp.s#

Inner join parts on p.p#=sp.p#

Where p.color=’red’;

7) Sort the supplier table by sname.


Order by sname;

8)

Delete records in supplier table whose status is 40.

Delete from supplier where status=40;

9) Add one field in supplier table.

Alter table supplier

Add new_column varchar2(20);

10) Find name of parts whose color is ‘red’

Select pname from parts

Where color=’red’;

11)

Find parts name whose weight less than 10 kg.

Select pname from parts where weight


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15) Find name of supplier who supply maximum parts.

Select sname, max(count(sp.p#)) from supplier

Inner join sp on sp.s#=supplier.s#;

16)

Sort the parts table by pname.

Select * from parts

Order by pname; 17)

Delete records in parts table whose color is ‘blue’.

Delete from parts

Where color=’blue’

18)

Drop one field in parts table.

Alter table parts

Drop column city;

Consider following schema and write SQL for given statements.

student( rollno, name, branch)

exam(rollno, subject_code, obtained_marks , paper_code) papers(paper_code, paper_satter_name, university)

1)

Display name of student who got first class in subject ‘130703’.

Select s.name from student s

Inner join exam e on e.rollno=s.rollno

Where e.obtained_marks>=60 and

e.paper_code=’130703’;

2) Display name of all student with their total mark.

Select sum(e.obtained_marks), s.name from student s


Group by s.name;

3)

Display list number of student in each university.

Select count(s.rollno), p.university from student s


Inner join paper p on p.papercode=e.papercode

Group by p.university;

4)

Display list of student who has not given any exam.

Select s.name from student s

Where s.rollno NOT IN

(select distinct e.rollno from exam e);

Consider following schema and represent given statements in relation algebra form.

Branch(branch_name,branch_city)

Account(branch_name, acc_no, balance)

Depositor(Customer_name, acc_no)

1)

Find out list of customer who have account at ‘abc’ branch.

∏ customer_name (σbranch_name=”abc” (depositor account) )


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2) Find out all customers who have account in ‘Ahmedabad’ city and balance is greater than 10,000.

∏ customer_name (σbranch_city=”ahmedabad” ^ balance>10000 (depositor account)

3) find out list of all branch name with their maximum balance.

∏branch_name G max (balance) (branch account)

Consider following schema and write SQL for given statements.

Student(Rollno, Name, Age,Sex,City) Student_marks(Rollno,Sub1,Sub2,Sub3,Total,Average)

Write query to

1)

Calculate and store total and average marks from Sub1, Sub2 & Sub3.

Update student_marks

set total=sub1+sub2+sub3,

average=(sub1+sub2+sub3)/3;

2)

Display name of students who got more than 60 marks in subject Sub1.

Select name from student where rollno IN (select rollno from student_marks where sub1>60);

3)

Display name of students with their total and average marks.

Select a.name, b.total, b.average from student a, student_marks b

where a.rollno=b.rollno;

4) Display name of students who got equal marks in subject Sub2

Select a.name from student a where a.rollno IN

(select b.rollno from student_marks b where b.rollno IN

(select c.rollno from student_marks c where b.sub2=c.sub2));

Write down the query for the following table where primary keys are underlined.

Person(ss#, name, address) Car(license, year, model) Accident(date, driver, damage‐amount) Owns(ss#, license) Log(license, date, driver)

1) Find the total number of people whose cars were involved in accidents in 2009.

Select count(ss#) from person Where ss# IN

(select ss# from owns where licence IN

(select licence from log where driver IN

(select driver from accident

Where date > ’01‐ jan‐09’ and ’31‐dec‐09’)));

OR

Select count(ss#) from person

Inner join owns on person.ss#=owns.ss#

Inner join log on owns.licence=log.licence

Inner join accident on accident.

Where accident.date > ’01‐ jan‐09’ and accident.date


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OR

Select count(accident.date) from accident

Inner join log on log.date=accident.date and log.driver=accident.driver

Inner join owns on own.licence=log.licence

Inner join person on person.ss#=own.ss#

where person.name = ‘s.sudarshan’;

3) Add a new customer to the database.

Insert into person (ss#, name, address) values (11, ‘ramesh’, ‘rajkot’);

Insert into owns (ss#, license) values (11, ‘L11’);

Insert into car (license, year, model) values (‘L11’, 2011, ‘esteem’);

Insert into log (license, date, driver) values (‘L11’, null, ‘mohan’);

4) Add a new accident recorded for the Santro belonging to “KORTH”

Insert into accident (date, driver, damage‐amount) values (’01‐oct‐11’,’magan’,10000);

Insert into log (license, date, driver) values (‘L08’,’01‐oct‐11’,’magan’);

Insert into car (license, year, model) values (’L08’,’2007’,’alto’);

Consider the employee data. Give an expression in SQL for the following query : Employee(employee‐name, street, city) Works(employee‐name, company‐name, salary)

Company(company‐name, city) Manages(employee‐name, manager‐name)

1)

Find the name of all employees who work for State Bank.

Select employee‐name from works

Where company‐name=’state bank’;

2)

Find the names and cities of residence of all employees who work for State Bank.

Select employee‐name, city from employee

Where employee‐name IN

(select employee‐name from works

Where company‐name=’state bank’); OR

Select e.employee‐name, e.city from employee e

Inner join works w on w.employee‐name= e.employee‐name

Where w.company‐name=’state bank’;

3) Find all employee in the database who do not work for State Bank.

Select employee‐name from work

Where company‐name != ‘state bank’;

4) Find all employee in the database who earn more than every employee of UCO Bank.

Select employee‐name from works

Where salary > (select max (salary) from works

Where company‐name =’uco bank’);

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9 Gtu Query Algebra Solution