dbms

Database Applications Laboratory Manual

Exercise 1:

Consider the following relations:

Student (snum: integer, sname: string, major: string, level: string, age: integer)

Class (name: string, meets at: string, room: string, d: integer)

Enrolled (snum: integer, cname: string)

Faculty (fid: integer, fname: string, deptid: integer)

The meaning of these relations is straightforward; for example, Enrolled has one record per student-class pair such that the student is enrolled in the class. Level is a two character code with 4 different values (example: Junior: JR etc)

Write the following queries in SQL. No duplicates should be printed in any of the answers.

i. Find the names of all juniors (level = JR) who are enrolled in a class taught by Prof. Harshith

ii. Find the names of all classes that either meet in room R128 or have five or more Students enrolled.

iii. Find the names of all students who are enrolled in two classes that meet at the same time.

iv. Find the names of faculty members who teach in every room in which some class is taught.

v. Find the names of faculty members for whom the combined enrollment of the courses that they teach is less than five.

Department of Computer Science & Engineering Page 1


SQL>desc student;

Table Column Data Type Length Precision Scale Primary Key Nullable DefaultComment

STUDENT SNUM Number - 4 0 - - -

SNAME Varchar2 10 - - - - -

MAJOR Varchar2 5 - - - - -

SLEVEL Varchar2 7 - - - - -

AGE Number - - - - - -

SQL>insert into student('3','SUHAS','CS','FR','20');

SQL>select * from student;

SNUM SNAME MAJOR SLEVEL AGE

3 SUHAS CS FR 20

14 RAJESH CS JR 20

9 VINUTHA IS JR 22

2 VARSHA CS JR 20

5 SHRUTHI CS SO 20

6 VISHAK CS JR 20

7 VISHAL EC SR 21

10 AMIT EC SR 20

12 ATUL EC JR 20

SQL>desc enrolled;

TableColum

nData Type

Length

Precision

Scale

Primary Key

Nullable

Default

Comment

ENROLLED

SNUM Number - - - - - -

CNAME Varchar2 10 - - - - -

SQL>INSERT INTO ENROLLED VALUES(08,'SE');

SQL>SELECT * FROM ENROLLED;



SNUM CNAME

3 DS LAB

4 DS LAB

5 DS LAB

9 DBMS LAB

8 DS LAB

4 EC

3 DBMS LAB

7 EC

9 SE

2 DBMS LAB

2 DS LAB

8 SE

2 DBMS

2 C SHARP

6 DBMS LAB

9 EC

SQL>desc class;

Table

Column

Data Type

Length

Precision

Scale

Primary Key

Nullable

Default

Comment

CLASS

NAME Varchar2 10 - - - - -

MEETSAT

Varchar2 5 - - - - -

ROOM Varchar2 7 - - - - -

D Number - - - - - -

SQL>INSERT INTO CLASS VALUES('SE','9AM','R138',555);

SQL>SELECT * FROM CLASS;

NAME MEETSAT ROOM D

DBMS 11AM R128 333

SS 9AM R128 111

DS LAB 11AM R138 444

DBMS 1PM R128 555

EC 10AM R138 222

DBMS 11AM R128 333

DBMS 10AM R128 555

SE 9AM R138 555



SQL>desc faculty;

TableColum

nData Type

Length

Precision

Scale

Primary Key

Nullable

Default

Comment

FACULTY

FID Number - - - - - -

FNAME Varchar2 10 - - - - -

DEPTID Number - - - - - -

SQL>select * from faculty;

FID FNAME DEPTID

333 HARSHITH 2

555 HARSH 3

222 SNEHA 1

111 VANDANA 1

444 HARSHA 2

i) SQL>select distinct sname from student s,enrolled e,class c,faculty f

where s.slevel='JR' and s.snum=e.snum and e.cname=c.name and c.d=f.fid and f.fname='HARSHITH';

SNAME

VARSHA

ii) SQL>select name from class cwhere c.room='R128'UNION (select cname from enrolled group by cname having count(*)>=5);

DBMS

DS LAB

SS

iii) SQL>select sname from student s,enrolled e,class c where s.snum=e.snum and e.cname=c.namegroup by s.snum, sname, c.meetsathaving count(*)>1;

SNAME

VARSHA



iv) SQL>select fname from faculty f,class c where f.fid=c.dgroup by fid, fnamehaving count(distinct c.room) = (select count(distinct room) from class);

FNAME

HARSH

v) SQL>select f.fname from faculty f,class c,enrolled ewhere f.fid=c.d and c.name=e.cnamegroup by f.fid, f.fname having count (*)<5;

FNAME

HARSH

SNEHA

HARSHITH

Exercise 2:



The following relations keep track of airline flight information:

Flights (no: integer, from: string, to: string, distance: integer, Departs: time, arrives: time, price: real)

Aircraft (aid: integer, aname: string, cruisingrange: integer)

Certified (eid: integer, aid: integer)

Employees (eid: integer, ename: string, salary: integer)

Note that the Employees relation describes pilots and other kinds of employees as well; Every pilot certified for some aircraft, and only pilots are certified to fly.

Write each of the following queries in SQL.

i) Find the names of aircraft such that all pilots certified to operate them have salaries more than Rs.80, 000.

ii) For each pilot who is certified for more than three aircrafts, find the eid and the maximum cruising range of the aircraft for which she or he is certified.

iii) Find the names of pilots whose salary is less than the price of the cheapest route from Bengaluru to Frankfurt.

iv) For all aircraft with cruisingrange over 1000 Kms, .find the name of the aircraft and the average salary of all pilots certified for this aircraft.

v) Find the names of pilots certified for some Boeing aircraft.

vi) Find the aids of all aircraft that can be used on routes from Bengaluru to New Delhi.

SQL>create table flights(no number, source varchar(10),dest varchar(10),distance number, departs timestamp, arrives timestamp, price decimal(10,2));

Table created.

SQL>create table aircraft(aid number,aname varchar(10),cruisingrange number);

Table created.

SQL>create table certified(eid number,aid number);

Table created.

SQL>create table employees(eid number,ename varchar(10),salary number);



Table created.

SQL>desc flights;

Table Column Data Type Length Precision Scale Primary Key Nullable Default

Commen

t

FLIGHTS NO Number - - - - - -

SOURCE Varchar2 10 - - - - -

DEST Varchar2 10 - - - - -

DISTANCE Number - - - - - -

DEPARTS Timestamp(6) 11 - 6 - - -

ARRIVES Timestamp(6) 11 - 6 - - -

PRICE Number - 10 2 - - -

SQL>insert into flights values(100,'BANGALURU','FRANKFURT',90000,TIMESTAMP'2012-09-09 01:00:00',TIMESTAMP'2012 09 10 18:00:00',50000);

SQL>select* from flights;

NO SOURCE DEST DISTANCE DEPARTS ARRIVES PRICE

100 BANGALURU FRANKFURT 9000009-SEP-12 01.00.00.000000 AM

10-SEP-12 06.00.00.000000 PM

50000

112 BANGALURU NEW DELHI 100011-SEP-12 09.00.00.000000 AM

11-SEP-12 12.00.00.000000 PM

6000

102 BANGALURU DUBAI 3000013-SEP-12 05.00.00.000000 AM

13-SEP-12 11.00.00.000000 AM

15000

111 BANGALURU MUMBAI 150011-SEP-12 10.00.00.000000 AM

11-SEP-12 01.00.00.000000 PM

6200

101 BANGALURU NEW YORK 9000012-SEP-12 02.00.00.000000 PM

13-SEP-12 05.00.00.000000 PM

60000

110 BANGALURU FRANKFURT 9000014-SEP-12 02.00.00.000000 AM

15-SEP-12 07.00.00.000000 PM

40000

SQL>desc aircraft;

AIRCRAFT AID Number - -

- -

- -



ANAME Varchar2 10-

- -

- -

CRUISINGRANGE Number - -

- -

- -

SQL>insert into AIRCRAFT values(812,'BOEING',800);

SQL>select* from aircraft;

AID ANAME CRUISINGRANGE

999 FLIGHT 11000

998 FLIGHT 21000

888 BOEING 900

812 BOEING 800

SQL>desc certified;

TableColum

nData Type

Length

Precision

Scale

Primary Key

Nullable

Default

Comment

CERTIFIED

EID Number - - - - - -

AID Number - - - - - -

SQL>insert into CERTIFIED values(113,812);

SQL>select* from certified;

EID AID

111 888

112 998

111 812

112 999

113 812

111 999

111 998

113 888

SQL>desc employees;



EMPLOYEES EID Number - - -

- -

-

ENAME Varchar210

- -

- -

-

SALARY Number - - -

- -

-

SQL>insert into employees values (223,'KIRAN',15000);

SQL>select* from employees;

EID ENAME SALARY

113 ANJU 35000

222 CHITRA 10000

333 RAVI 25000

112 RAJ 85000

111 RAJESH 90000

223 KIRAN 15000

334 RAMESH 30000

i) SQL> select a.aid, a.aname from aircraft a, certified c, employees e

where a.aid=c.aid and c.eid=e.eid and e.salary>80000

group by a.aid,a.aname

having count(*)=(select count(*) from employees e where e.salary>80000)

AID ANAME

999 FLIGHT

998 FLIGHT

ii) SQL>Select c.eid, MAX(a.cruisingrange)from certified c, aircraft a

where c.aid=a.aid

group by c.eid

having count(*)>3;

EID MAX(A.CRUISINGRANGE)

111 21000

iii) SQL> SELECT distinct E.ENAME FROM EMPLOYEES E,certified c



WHERE e.eid=c.eid and E.SALARY < ( SELECT MIN (F.PRICE)FROM FLIGHTS F

WHERE F.SOURCE = 'BANGALURU' AND F.DEST = 'FRANKFURT');

ENAME

ANJU

iv) SQL> create table temp(aname varchar(20),salary number);

SQL>insert into temp(aname,salary) select a.aid, e.salary

From AIRCRAFT A,CERTIFIED C,EMPLOYEES E

WHERE A.CRUISINGRANGE>1000 AND A.AID=C.AID AND C.EID=E.EID ;

SQL>select aname,avg(salary) from temp group by aname;

ANAME AVG(SALARY)

998 87500

999 87500

v) SQL>SELECT DISTINCT E.ENAME FROM EMPLOYEES E, CERTIFIED C, AIRCRAFT A

WHERE E.EID = C.EID AND C.AID = A.AID AND A.ANAME = 'BOEING';

ENAMERAJESH

ANJU

vi) SQL>SELECT A.AID FROM AIRCRAFT A

WHERE A.CRUISINGRANGE > (SELECT MIN (F.DISTANCE)

FROM FLIGHTS F

WHERE F.SOURCE = 'BANGALURU'AND F.DEST = 'NEW DELHI');

AID

999

998



Exercise 3:

Consider the following database of student enrollment in courses & books adopted for each course. STUDENT (regno: string, name: string, major: string, bdate:date)

COURSE (course #:int, cname:string, dept:string)

ENROLL ( regno:string, course#:int, sem:int, marks:int)

BOOK _ ADOPTION (course# :int, sem:int, book-ISBN:int)

TEXT (book-ISBN:int, book-title:string, publisher:string, author:string) i. Create the above tables by properly specifying the primary keys and the foreign keys.

ii. Enter at least five tuples for each relation.

iii. Demonstrate how you add a new text book to the database and make this book be adopted by

some department.

iv. Produce a list of text books (include Course #, Book-ISBN, Book-title) in the alphabetical

order for courses offered by the ‘CS’ department that use more than two books.

v. List any department that has all its adopted books published by a specific publisher.

vi. Generate suitable reports.

vii. Create suitable front end for querying and displaying the results.

1. Create the above tables by properly specifying the primary keys and the foreign keys.

SQL> create table students (regno varchar(10)primary key,name varchar(10),major varchar(10),bdate date);

Table created.

SQL>create table courses(course_id int primary key,cname varchar(10),dept varchar(20));

Table created.

SQL>create table enrolls(regno varchar(10),course_id int,se mint,marks int,primary key(regno)references students(regno),foreign key(course_id)references courses(course_id));



Table created.

SQL>create table texts(book_isbn int primary key,book_title varchar(15),publisher varchar(10),author varchar(15));

Table created.

SQL>create table book_adoptions(course_id int,sem int,book_isbn int,primary key(course_id,sem),foreign key(course_id)references courses(course_id),foreign key(book_isbn)references texts(book_isbn));

Table created.

2. Enter at least five tuples for each relation

SQL>insert into students values (‘&regno”,’&name’,’&major’,’&date’);

SQL>insert into COURSES values (‘&course_id”,’&cname’,’&dept’);

SQL>insert into text values (‘&book_isbn’,’&book_title’,’&publisher’,’&author’);

SQL>insert into book_adoption values (‘&course_id’,’&sem’,’&book_isbn’);

SQL>select * from STUDENTS;

REGNO NAME MAJOR BDATE

123 akhilaa data stru 01-DEC-09

567 sinchana ada 01-DEC-09

345 anusha maths 01-DEC-09

234 arnitha data base 01-DEC-09

456 sneha microproce 01-DEC-09



Select * from courses;

COURSE# CNAME DEPT

5 Be tce

1 Be ise

2 Be CS

3 Be ece

4 Be CS

Select * from enroll;


REGNO COURSE# SEM MARKS

234 2 3 21

123 1 2 24

345 3 7 25

456 4 1 19

567 5 5 23


Select * from text;

BOOK_ISBN BOOK_TITLE PUBLISHER AUTHOR

765 Ada nandi padhma red

543 microproc mcgrachill douglashal

876 engee math prism dsc

123 let us c pearson yashwant

987 data struc nandi padhma red

654 data base pearson navathe

Select * from book_adoption;



3. Demonstrate how you add a new text book to the database and make this book be

adopted by some department.

SQL>insert into text values ('444','analog','pearson','rajesh');

Select * from text;

BOOK_ISBN BOOK_TITLE PUBLISHER AUTHOR

765 Ada nandi padhma red




987 data struc nandi padhma red


COURSE# SEM BOOK_ISBN

4 2 123

2 2 123

4 2 123

5 5 543

3 7 765

1 2 987

2 3 876

4 1 654


654 data base pearson navathe

444 Analog Pearson rajesh

SQL>insert into book_adoption values(3,4,444);

4. Produce a list of text books (include Course #, Book-ISBN, Book-title) in the alphabetical order for courses offered by the ‘CS’ department that use more than two books.

select distinct c.course#,t.book_isbn,t.book_title

from text t,book_adoption b,course c

where c.dept='CS' and c.course#=b.course# and b.book_isbn=t.book_isbn


COURSE# SEM BOOK_ISBN

4 2 123

2 2 123

4 2 123

5 5 543

3 7 765

1 2 987

2 3 876

4 1 654

3 4 444


and c.couse# in (select course# from book_adoption b group by course# having count(*)>2)

order by

t.book_title;

5. List any department that has all its adopted books published by a specific publisher.

COURSE# BOOK_ISBN BOOK_TITLE CNAME

765 ada nandi padhma red




Exercise 4:

The following tables are maintained by a book dealer.

AUTHOR (author-id:int, name:string, city:string, country:string)

PUBLISHER (publisher-id:int, name:string, city:string, country:string)

CATALOG (book-id:int, title:string, author-id:int, publisher-id:int, category-id:int, year:int,

price:int)

CATEGORY (category-id:int, description:string)

ORDER-DETAILS (order-no:int, book-id:int, quantity:int)

i. Create the above tables by properly specifying the primary keys and the foreign keys.

ii. Enter at least five tuples for each relation.



iii. Give the details of the authors who have 2 or more books in the catalog and the price of the

books is greater than the average price of the books in the catalog and the year of publication is

after 2000.

iv. Find the author of the book which has maximum sales.

v. Demonstrate how you increase the price of books published by a specific publisher by 10%.

vi. Generate suitable reports.

vii. Create suitable front end for querying and displaying the results.

1. Create the above tables by properly specifying the primary keys and the foreign keys.

Create table authors(aid int primary key,name varchar(15),city varchar(15),country varchar(15));

Table created.

Create table publishers(pi dint primary key,name varchar(15),city varchar(15),country varchar(15));

Table created.

Create table categorys(cid int primary key,description varchar(15));

Table created.

Create table catalogs(bid int primary key,title varchar(15),aid int,pi dint,cid int,year int,price int

Foreign key(aid)references authors(aid),

Foreign key(pid)referencespublishers(pid),

Foreign key(cid)references categorys(cid));

Table created.

Create table order_details(orderno int,bid int,qty int,primary key(orderno,bid),foreign key(bid) references catalogs(bid));

Table created.



2. Enter at least five tuples for each relation.

SQL>desc authors;

Table Column Data Type Length

AUTHOR1 AID Number -

NAME Varchar2 15

CITY Varchar2 15

COUNTRY Varchar2 15

SQL>insert into authors values(‘&aid’,’&name’,’&city’,’&country’);

SQL>select * from authors;

AID NAME CITY COUNTRY

5 SHIVAAM KENGARI INDIA

2 ASHOKA HOSUR INDIA

1 THILAK BANGALORE INDIA

3 RAVI CHENNAI INDIA

SQL>desc publishers;

Table Column Data Type Length



PUBLISHER PID Number -

NAME Varchar2 15

CITY Varchar2 15

COUNTRY Varchar2 15

SQL>insert into publishers values(‘&pid’,’&name’,’&city’,’&country’);

SQL>select * from publishers;

SQL>insert into categorys values(‘&cid’,’&description’);

SQL>select * from categorys;

SQL>insert into catalogs values(‘&bid’, ‘&title’, ‘&aid’, ‘&pid’, ‘&cid’, ‘&year’,’&price’);

SQL>select * from catalogs;

SQL>insert into order_details values(‘&orderno’,’&bid’,’&qty’);

SQL>select * from order_details;

3.Give the details of the authors who have 2 or more books in the catalog and the price of the

books is greater than the average price of the books in the catalog and the year of publication

is after 2000.

SQL>select distinct a.aid,a.name,a.city,a.country from authors a, catalogs c where a.aid=c.aid

and c.price>(select avg(price) from catalogs) and c.year >2000 and a.aid=(select aid from

catalogs group by aid having count(*)>=2);

AID CITY COUNTRY



5 KENGARI INDIA

2 HOSUR INDIA

4. Find the author of the book which has maximum sales.

SQL>select name, qty from order_details o,authors a,catalogs c where o.bid=c.bid and

c.aid=a.aid and qty in(select max(qty) from order_details);

NAME QTY

ASHOKA 5

5. Demonstrate how you increase the price of books published by a specific publisher by 10%.

SQL>Update catalogs set price=price*0.1+price where pid in(select pid from publisher where name=’pearson’);

Two rows updated



Exercise 5:

Consider the following database for a banking enterprise.

BRANCH (branch-name: string, branch-city: string, assets: real)

ACCOUNT (accno: int, branch-name: string, balance: real)

DEPOSITOR (customer-name: string, accno: int)

CUSTOMER (customer-name: string, customer-street: string, customer-city: string)

LOAN (loan-number: int, branch-name: string, amount: real)

BORROWER (customer-name: string, loan-number: int)

(i) Create the above tables by properly specifying the primary keys and foreign keys.

(ii) Enter atleast five tuples for each relation.

(iii) Find all the customers who have atleast two accounts at the main branch.

(iv) Find all the customers who have an account at all the branches located in a specific city.

(v) Demonstrate how you delete all account tuples at every branch located in a specific city.



(vi) Generation of suitable reports.

(vii) Create suitable front end for querying and displaying the results.

Create table branch(bname varchar(20) primary key, bcity varchar(20),assets real);

Table created.

Create table account(accno int primary key, bname varchar(20),balance real,foreign key (bname) references branch(bname));

Table created.

Create table customer(cname varchar(20) primary key, cstreet varchar(20),city varchar(20));

Table created.

create table depositor(cname varchar(20), accno int,primary key(cname,accno),foreign key (cname) references customer(cname),foreign key (accno) references account(accno));

Table created.

Create table loan(lnumber int primary key, bname varchar(20),amount real,foreign key (bname) references branch(bname));

Table created.

Create table borrower(cname varchar(20),lnumber int,foreign key (cname) references customer(cname),foreign key(lnumber) references loan(lnumber));

Table created.

insert into branch values('jayanagar','bangalore',1500)

select * from branch

BNAME BCITYASSET

S

jayanagar bangalore 1500



majestic bangalore 2600

jubleehillshyderabad

4000

panjagutta

hyderabad

3500

krpuram chennai 1020

insert into account values(5600,'krpuram',25000)

Select * from account

ACCNO BNAME BALANCE

5467 jubleehills 16000

5060 majestic 10000

5065 jayanagar 20000


5600 krpuram 25000

5490 panjagutta 15000

5111 jayanagar 15000

insert into customer values('naveen','market road','bangalore')

select * from customer

CNAME CSTREET CITY

sudhakar udayanagar road bangalore

naveen market road bangalore



naveen j bagalur road chennai

pradeep pujagutta road hyderabad

dilip band box road bomanahalli

insert into depositor values('naveen j',5600)

select * from depositor

CNAME ACCNO

dilip 5467

dilip 5468

naveen 5111

naveen j 5600

pradeep 5490

sudhakar 5060

sudhakar 5065

insert into loan values(10,'majestic',150000)

select * from loan

ACCNO BNAME BALANCE


10 majestic 150000

18 jayanagar 150000

16 krpuram 200000



insert into borrower values('sudhakar',10)

select * from borrower

CNAME LNUMBER

dilip 15

naveen 15

naveen j 16

sudhakar 10

3. SELECT d.CNAME FROM DEPOSITOR D,ACCOUNT A,CUSTOMER C, branch b

WHERE D.ACCNO=A.ACCNO AND C.CNAME=D.CNAME AND A.BNAME=B.BNAME

AND A.BNAME IN (SELECT BNAME FROM BRANCH WHERE BCITY='bangalore')

GROUP BY d.CNAME

HAVING COUNT(DISTINCT B.BNAME)>=2;

CNAME

sudhakar

4. Select cname from customer c where not exists(select bname from branch where

city=’bangalore’ minus select bname from depositor d,account a where d.accno=a.accno and

d.cname=c.cname) and exists (select bname from branch where city=’bangalore’);

bNAME

mejestic



5. Delete from account where bname in (select bname from branch where city=’chennai’);

Two rows deleted


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