chapter 1 · a mini project should be implemented by the students in teams. the maximum size of a...

Database Application Laboratory 1

Dr. A.I.T.,Dept of CSE

CHAPTER 1

Course objectives:

1. Provide a strong formal foundation in database concepts and technology

2. Familiarize the students with the database environments towards an information-oriented data-

processing oriented framework.

3. Understand the relational data model and systematic database design approaches covering

conceptual design, logical design.

4. To present the concepts and techniques relating to query processing by SQL

PART – A

INSTRUCTIONS:

1. The exercises are to be solved in an RDBMS environment like Oracle or DB2.

2. Suitable tuples have to be entered so that queries are executed correctly.

3. Relevant queries other than the ones listed along with the exercises may also be asked in the

examinations.

4. Questions must be asked based on lots.

1. Consider the following relations:

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

CLASS(cname:string, meets at:time, room:string, fid:integer)

ENROLLED(snum:integer, cname:string)

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

The meaning of these relations is straight forward; 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 duplicate 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 tech is less than five.

Sub Title : DATABASE APPLICATIONS LABORATORY

Sub Code: CSL57 No. of Credits: 1.5= 0 :0 : 1.5 (L-T-P) No. of lecture hours / week :

3

Exam Duration : 3 hrs CIE + SEE = 50 + 50 =100



(vi) Generation of suitable reports.

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 is certified for some aircraft, and only pilots are certified to fly.

Write each of the following queries in SQL

vi) Generation of suitable reports

3. Consider the following database of student enrollment in courses and 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) Generation of suitable reports.

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 foreign keys.

(i) Find the names of aircraft such that all pilots certified to operate them have salaries more than

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 the pilots whose salary is less than the price of the cheapest route from

Bengaluru to Frankfurt.

(iv) For all aircraft with cruising range over 1000 Kms, find the name of the aircraft and the

average salary of all pilots certified for this aircraft.



ii) Enter atleast five tuples for each tables.

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) Generation of suitable reports.

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 at least five tuples for each relation.

(iii) Find all the customers who have at least 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.

PART – B

A mini project should be implemented by the students in teams. The maximum size of a team

can be 2 from the same batch. The students have to finalize a project topic by discussing with the

faculty. The mini project must be carried out in the college only.

The tasks when implementing mini project would be:

1. Understand the complete domain knowledge of the application and derive the complete

data requirement specification for the mini project.

2. Design the ER diagram for the application

3. Design Relational Scheme diagram for the application

4. Normalization of the relational design.

5. Implement minimum 5 queries for the application

6. Documentation & submission of report.

General guidelines:

• Database for the project - Oracle / MySQL/ DB2 / SQL Server etc.

Sample Mini Projects.

• Inventory Control System.

• Material Requirement Processing.

• Hospital Management System.

• Railway Reservation System.

• Web Based User Identification System.



• Timetable Management System.

• Hotel Management System

• Placement management system

• Library management system

• Personal Information System.

Note: In the examination, the marks will be evaluated based on Part A and project

demonstration, project report and viva-voce.

Course Outcomes:

1. Understand underlying concepts of database technologies.

2. Design and implement a database schema for a given problem-domain.

3. Understand, analyze, and apply common SQL Statements including DDL, DML and DCL

statements to perform different operations.

4. Design and implement a database for a given problem according to well-known design

principles that balance data retrieval performance with data consistency.

TEXT BOOK:

1. Fundamental of Database Systems by Elmasri and Navathe, 7th Edition, Addison-Wesley,

2015 ISBN-10: 0133970779, ISBN-13: 978-0133970777

REFERENCE BOOKS/WEBLINKS:

1. Database Management Systems by Raghu Ramakrishnan and Johannes Gehrke – 3rd Edition,

McGraw-Hill, 2006.

2. An Introduction to Database Systems by C.J. Date, A. Kannan, S. Swamynathan, 8th Edition,

Pearson Education,2013.

3. Data Base system Concepts by Silberschatz, Korth and Sudharshan, 5th edition McGraw Hill,

2011.

Cos Mapping with POs

CO1 PO2,PO3

CO2 PO2,PO3

CO3 PO1,PO3,PO4,PO5

CO4 PO3,PO4,PO5



CHAPTER 2

INTRODUCTION

OVERVIEW

Databases store data and metadata. Data are the individual facts that are used to derive

information. Metadata describe the content, quality, condition, availability, and characteristics of

data. Database Management Systems (DBMS) is used to modify the data.

There are varieties of database types: Sequential files, Hierarchical databases, Network

databases, and Relational databases. The recent database type is Object-Relational database,

which is essentially a relational database with some Object properties.

Relational databases become popular because it was easy to modify the schema. It is very easy to

add tables and columns to the schema, and doing so does not affect the remainder of the schema,

and more important, does not affect the applications that access the database schema. Older

databases required the databases to be restructured and the applications to be modified. Avoiding

database and application maintenance is an important benefit and the reason for the switch.

Relational databases consist of independent tables. Relational Database Management Software

(RDBMS) does not know how the records in the table are related.

ORACLE

Oracle is the largest database manufacturer and the second-largest software manufacturer in the

world. The company began as a relational database manufacturer. In the beginning, Oracle touted

its software as “being able to run on any platform”. This openness has been most attractive to

companies; an Oracle has tried to maintain its image as an open product. Oracle was at a good

place when industry became extremely interested in moving away from network databases and

the mainframe.

Oracle Database 10g is an entry-level database based on the Oracle Database 10g Release 2 code

base that's free to develop, deploy, and distribute; fast to download; and simple to administer.

Oracle Database is a great starter database for:

• Developers working on PHP, Java, .NET, and Open Source applications

• DBAs who need a starter database for training and deployment

• Independent Software Vendors (ISVs) and hardware vendors who want a starter database

• Educational institutions and students who need database for their curriculum



The main use of the Oracle database system is to store and retrieve data for applications. The

Oracle database includes different languages and interfaces that allow programmers to access

and manipulate the data in the database. The ANSI standard Structured Query Language (SQL)

provides basic functions for data manipulation, transaction control, and record retrieval from the

database. However, most end users interact with Oracle through applications that provide an

interface that hides the underlying SQL and its complexity.

BASICS

Structured Query Language (SQL), which is an ANSI standard language for interacting with

relational databases, is the main tool for extracting the information.

A database is a representation of a real-world thing called an Entity. Examples of entities are

vehicles, employees, customers, fish, buildings, and even things such as baseball teams. The

database stores facts about the entity in an organized framework, model, or schema. These facts

are called attributes.

An Instance is one occurrence of an entity. Each entity must have an identifier, which is one or

more attributes that make each entity instance unique from any other instance. The identifier

should contain a value that does not change. Examples of identifiers are student IDs, payroll

numbers, or social security numbers. If the entity does not have an attribute that can be used as

an identifier, an artificial identifier can be created. The identifier on an entity is often called a

primary key. A foreign key is a set of attributes of the considered table that exists as a primary

key attributes in another table. Database records are matched (joined) through the use of primary

and foreign keys.

Normalization is a process consisting of series of steps, which is used to group the database

attributes. The purpose of this design is to ensure that the tables within the database are space

efficient and performance efficient.

• Zero Normal Form—each of the relations (tables) has a unique identifier (primary key).

• First Normal Form—Separate the repeating groups of attributes or multi-valued attributes

into a relation of their own. Be sure to form composite keys.

• Second Normal Form—Establish full functional dependency by separating out attributes

that are not fully dependent on the full primary keys.

• Third Normal Form—Remove transitive dependencies by separating attributes that are

dependent on a non key attribute.



CHAPTER 3

STUDENT DATABASE

Consider the following relations:

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

CLASS(cname:string, meets at:time, room:string, fid:integer)

ENROLLED(snum:integer, cname:string)

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

Schema-Diagram:

STUDENT

snum sname major level age

CLASS

cname meets_at room fid

ENROLLED

snum cname

FACULTY

fid fname deptid

Creation of Tables:

create table student

(

snum int primary key,

sname varchar2(10),

major varchar2(10),

levels varchar2(2),

age number(2)

);

create table faculty

(

fid number(5) primary key,

fname varchar2(10),

deptid number(5)

);



create table class

(

cname varchar2(10) primary key,

meetsat varchar2(10),

room varchar2(5),

fid number(5) references faculty(fid)

);

create table enrolled

(

snum int references student(snum),

cname varchar2(10) references class(cname)

);

Insertion of records:

insert into student values(1,'pratima','cs','jr',20);

insert into student values(2,'jyoti','is','jr',20);

insert into student values(3,'divya','is','sr',21);

insert into student values(4,'akki','civil','ss',18);

insert into student values(5,'usha','ec','sj',20);

insert into faculty values(11,'harshith',111);

insert into faculty values(22,'harish',111);

insert into faculty values(33,'archana',222);

insert into faculty values(44,'sita',333);

insert into faculty values(55,'anchal',222);

insert into class values('1st sem','9am-10am',125,11);

insert into class values('2nd sem','10am-11am',128,11);

insert into class values('3rd sem','11am-12pm',128,33);

insert into class values('4th sem','11am-12pm',126,11);

insert into class values('5th sem','10am-11am',128,22);

insert into enrolled values(1,'1st sem');

insert into enrolled values(2,'2nd sem');

insert into enrolled values(3,'3rd sem');

insert into enrolled values(4,'4th sem');

insert into enrolled values(5,'5th sem');

select * from enrolled;

SNUM CNAME

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

1 1st sem

2 2nd sem

3 4th sem

4 3rd sem

5 5th sem

3 3rd sem



4 4th sem

select * from student;

SNUM SNAME MAJOR LEVEL AGE

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

1 pratima cs jr 20

2 shruti is jr 20

3 chetan is sr 21

4 akki civil ss 18

5 usha ec sj 20

select * from faculty;

FID FNAME DEPTID

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

11 harshith 111

22 harish 111

33 archana 222

44 sita 333

55 anchal 222

select * from class;

CNAME MEETSAT ROOM FID

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

1st sem 9am-10am 125 11

2nd sem 10am-11am 128 22

3rd sem 11am-12pm 128 33

4th sem 11am-12pm 126 11

5th sem 10am-11am 128 22

6th sem 10am-11am 128 11

Queries:

(i) Find the names of all juniors(level=jr) who are enrolled in a class taught by

Prof.Harshith

select distinct s.sname

from student s,class c,enrolled e, faculty f

where f.fid=c.fid and

c.cname=e.cname and

e.snum=s.snum and

f.fname='harshith' and levels='jr' ;

SNAME

----------

Pratima



(ii)Find the names of all classes that either meet in room R128 or have five or more

students enrolled.

select distinct(c.cname)

from class c,enrolled e

where c.cname=e.cname and

room='128' or e.cname IN (select cname

from enrolled

group by cname

having count(snum)>=5);

CNAME

----------

5th sem

2nd sem

3rd sem

(iii).find the names of all students who are enrolled in two classes that meet at the same

time.

select distinct s.sname

from student s

where s.snum in (select e1.snum

from enrolled e1,enrolled e2,class c1,class c2

where e1.cname=c1.cname and

e2.cname=c2.cname and

e1.snum=e2.snum and

e1.cname<>c2.cname and

c1.meetsat=c2.meetsat);

SNAME

----------

akki

chetan

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

taught.

select f.fname

from faculty f

where not exists ((select c.room

from class c)

minus

(select c1.room

from class c1

where c1.fid=f.fid));

FNAME

----------

Harshith



(v) Find the names of faculty members for whom the combined enrollment of the courses

that they take is less than five.

select distinct f.fname from faculty f

where 5> (select count(e.snum)

from class c,enrolled e

where c.cname=e.cname and

f.fid=c.fid);

FNAME

----------

harish

archana

sita

anchal

harshithy



CHAPTER 4

AIRCRAFT DATABASE

The following relations keep track of airline flight information:

FLIGHTS(no:integer, from:string, to:string, distance:integer, departs:date, arrives:date,

price:real)

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

CERTIFIED(eid:integer, aid:integer)

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

Schema-Diagram:

FLIGHTS

no from to distance departs arrives price

AIRCRAFT

aid aname cruisingrange

CERTIFIED

eid aid

EMPLOYEES

eid ename salary

Creation of Tables:

create table flights

(

no integer primary key,

frm varchar(20),

end varchar(20),

dist int,

dept date,

arr date,

price real

);



create table aircraft

(

aid int primary key,

aname varchar2(15),

crange int

);

create table certified

(

eid int refernces employees(eid),

aid int references aircraft(aid),

primary key(cid,aid)

);

create table employees

(

eid int primary key,

ename varchar2(15),

salary real

);


insert into flights values('255','bangalore','frankfurt','200','01-aug-2011','01-aug-2011','5000’);

insert into flights values('256','bangalore','frankfurt','200','01-aug-2011','01-aug-2011','8000’);

insert into flights values('257','bangalore','delhi','200','01-aug-2011','01-aug-2011','5000');

insert into flights values('258','bangalore','delhi','200','01-aug-2011','01-aug-2011','6000');

insert into flights values('259','bangalore','mangalore','200','01-aug-2011','01-aug-2011','8000’);

insert into aircraft values('685','boeing15','1000');

insert into aircraft values('686','boeing10','2000');

insert into aircraft values('687','skytrain','1000');

insert into aircraft values('688','avenger','100');

insert into certified values('101','685');







insert into employees values(101,'asha',90000);

insert into employees values(102,'arun',85000);

insert into employees values(103,'anand',3000);

insert into employees values(104,'ramya',4000);



select * from flights;

NO FRM END DIST DEPT ARR PRICE

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

255 bangalore frankfurt 200 01-AUG-11 01-AUG-11 5000

256 bangalore frankfurt 200 01-AUG-11 01-AUG-11 8000

257 bangalore delhi 200 01-AUG-11 01-AUG-11 5000

258 bangalore delhi 200 01-AUG-11 01-AUG-11 6000

259 bangalore mangalore 200 01-AUG-11 01-AUG-11 8000

select * from certified;

CID AID

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

101 685

101 686

101 687

101 688

102 686

103 685

103 687

select * from employees;

EID ENAME SALARY

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

101 asha 90000

102 arun 85000

103 anand 3000

104 ramya 4000

select * from aircraft;

AID ANAME CRANGE

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

685 boeing15 1000

686 boeing10 2000

687 skytrain 1000

688 avenger 100

Queries:

(i) Find the names of aircraft such that all pilots ceritified to operate them have salaries

more than 80,000.

select distinct a.aname

from aircraft a,certified c,employees e

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



ANAME

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

avenger

skytrain

boeing15

boeing10

(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.

select distinct c.cid,max(a.crange)

from certified c,aircraft a

where c.aid=a.aid

group by c.cid

having count(c.cid)>3;

CID MAX(A.CRANGE)

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

101 2000

(iii) Find the names of the pilots whose salary is less than the price of the cheapest route

from bengaluru to frankfurt.

select e.ename

from employees e

where e.salary<(select min(price)

from flights

where frm='bangalore' and end='frankfurt');

ENAME

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

anand

ramya

(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.

select name, avgsal

from(select a.aid,a.aname as name,avg(e.salary)

as avgsal from aircraft a,certified c,employees e

where a.aid=c.aid and c.cid=e.eid and crange>1000

group by a.aid,a.aname);

NAME AVGSAL

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

boeing10 87500



CHAPTER 5

STUDENT ENROLLMENT DATABASE Consider the following database of student enrollment in courses & books adopted for each

course:

STUDENT (regno: string, name: string, major: string, bdate: date)

COURSE (courseno: int, cname: string, dept: string)

ENROLL (regno: string, courseno: int, sem: int, marks: int)

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

TEXT (book-ISBN: int, book-title: string, publisher: string, author: string)

Schema-Diagram:

STUDENT

regno name major bdate

COURSE

courseno cname dept

ENROLL

regno courseno sem marks

BOOK _ ADOPTION

courseno sem book-ISBN

TEXT

book-ISBN book-title publisher author



Creation of Tables:

create table student

(

regno varchar(5) primary key,

name varchar(10),

major varchar(10),

bdate date

);

create table course

(

cno int primary key,

cname varchar(10),

dept varchar(10)

);

create table enroll

(

regno references student,

cno references course,

sem int primary key,

marks int

);

create table text

(

isbn int primary key,

title varchar(15),

publisher varchar(10),

author varchar(10)

);

create table bad

(

cno references course,

sem references enroll,

isbn references text

);


insert into student values('cs42','mikhil','cse','17-dec-1986');

insert into student values('cs48','mujeeb','cse','02-sep-1986');

insert into student values('ec26','pradeep','cse','16-aug-1987');

insert into student values('ee37','majid','ise','28-may-1986');



insert into student values('is48','wajid','ise','28-may-1986');

insert into course values(1,'.net','computer');

insert into course values(2,'j2ee','computer');

insert into course values(3,'mis','infosci');

insert into course values(4,'fs','infosci');

insert into course values(5,'oracle','computer');

insert into enroll values('cs42',1,6,98);

insert into enroll values('cs48',2,3,97);

insert into enroll values('ec26',5,5,50);

insert into enroll values('is48',3,7,90);

insert into enroll values('ee37',4,4,80);

insert into text values(101,'let us c','lpe','fahad');

insert into text values(102,'c++','abc','mujeeb');

insert into text values(103,'oracle','def','othman');

insert into text values(104,'.net','lpe','naushad');

insert into text values(105,'j2ee','pearson','mikhil');

insert into bad values(1,6,101);





select * from student;

REGNO NAME MAJOR BDATE

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

cs42 mikhil cse 17-DEC-86

cs48 mujeeb cse 02-SEP-86

ec26 pradeep cse 16-AUG-87

ee37 majid ise 28-MAY-86

is48 wajid ise 28-MAY-86

select * from course;

CNO CNAME DEPT

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

1 .net computer

2 j2ee computer

3 mis infosci

4 fs infosci

5 oracle computer



select * from enroll;

REGNO CNO SEM MARKS

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

cs42 1 6 98

cs48 2 3 97

ec26 5 5 50

is48 3 7 90

ee37 4 4 80

select * from text;

ISBN TITLE PUBLISHER AUTHOR

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

101 let us c lpe fahad

102 c++ abc mujeeb

103 oracle def othman

104 .net lpe naushad

105 j2ee pearson mikhil

select * from bad;

CNO SEM ISBN

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

1 6 101

1 6 103

1 6 102

4 4 104

5 7 105

Queries:

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

adopted by some department.

insert into text values(106,'JAVA','pearson','Avril');


select * from text;

ISBN TITLE PUBLISHER AUTHOR

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

101 let us c lpe fahad

102 c++ abc mujeeb

103 oracle def othman

104 .net lpe naushad

105 j2ee pearson mikhil

106 JAVA pearson Avril



select * from bad;

CNO SEM ISBN

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

1 6 101

1 6 103

1 6 102

4 4 104

5 7 105

2 7 106

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 2 books.

SELECT s.cno,t.title,b.isbn

FROM course s,bad b,text t

where s.cno=b.cno

AND b.isbn=t.isbn

AND s.dept='computer'

AND b.cno in (select b.cno

from bad b,course s

where b.cno=s.cno

group by b.cno

having count(*)>2)

order by t.title;

CNO TITLE ISBN

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

1 c++ 102

1 let us c 101

1 oracle 103

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

select s.dept, t.publisher

from course s, bad b, text t

where s.cno=b.cno

and t.isbn=b.isbn

and publisher='lpe';

DEPT PUBLISHER

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

computer lpe

infosci lpe



CHAPTER 6

BOOK DEALER DATABASE

The following tables are maintained by a book dealer:

AUTHOR (author_id: int, author_name: string, city: string, country: string)

PUBLISHER (publ_id: int, publ_name: string, publ_city: string, publ_country: string)

CATEGORY (category_id: int, description: string)

CATALOG (book-id: int, title: string, author_id: int, publisher_id: int, category_id: int,

cyear: int, price: int)

ORDER-DETAILS (order_no: int, book_id: int, qty: int)

Schema-Diagram:

AUTHOR

author_id author_name city country

PUBLISHER

publ_id publ_name publ_city publ_country

CATEGORY

category_id description

CATALOG

book-id title author_id publ_id category_id cyear price

ORDER-DETAILS

order_no book_id qty



Creation of Tables:

create table author

(

aid int primary key,

name varchar(15),

city varchar(15),

country varchar(15)

);

create table publisher

(

pid int primary key,

name varchar(15) ,

city varchar(15) ,

country varchar(15)

);

create table category

(

cid int primary key,

description varchar(15)

);

create table catalog

(

bid int primary key,

title varchar(15),

aid references author,

pid references publisher,

cid references category,

year int,

price int

);

create table odetails

(

ono int,

bid references catalog,

qty int

);


insert into author values (1,'mik','bang','ind');

insert into author values (2,'muj','bang','ind');

insert into author values (3,'prad','tri','aus');

insert into author values (4,'maj','anan','ame');

insert into author values (5,'waj','anan','euro');



insert into publisher values (101,'pearson','bang','ind');

insert into publisher values (102,'tata','mumbai','aus');

insert into publisher values (103,'sapna','che','euro');

insert into publisher values (104,'abc','tri','ame');

insert into publisher values (105,'xyz','anan','ind');

insert into category values (1001,'computer');

insert into category values (1002,'electronics');

insert into category values (1003,'maths');

insert into category values (1004,'science');

insert into category values (1005,'electrical');

insert into catalog values (111,'lib1',1,101,1001,2002,500);








insert into odetails values (1,111,2);








select * from author;

AID NAME CITY COUNTRY

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

1 mik bang ind

2 muj bang ind

3 prad tri aus

4 maj anan ame

5 waj anan euro

select * from publisher;

PID NAME CITY COUNTRY

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

101 pearson bang ind

102 tata mumbai aus

103 sapna che euro

104 abc tri ame

105 xyz anan ind



select * from category;

CID DESCRIPTION

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

1001 computer

1002 electronics

1003 maths

1004 science

1005 electrical

select * from catalog;

BID TITLE AID PID CID YEAR PRICE

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

111 lib1 1 101 1001 2002 500

112 lib2 2 102 1002 2000 800

113 lib3 3 103 1003 2003 200

114 lib4 4 104 1001 2006 350

115 lib5 5 105 1004 2007 100

116 lib6 2 103 1005 2007 600

117 lib7 2 105 1002 2007 450

118 lib8 1 101 1001 2002 500

select * from odetails;

ONO BID QTY

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

1 111 2

2 112 3

3 111 5

4 113 1

5 114 2

6 115 1

1 114 2

2 113 2

Queries:

iii. Give the detail of the author who have 2 or more books in the catalog and the price of

the book is greater than the average price of the book in the catalog and the year of

publication after 2000

select name, city, country

from author

where aid in (select aid from catalog where year>2000 and

price>(select avg(price) from catalog)

group by aid

having count(*)>1);



NAME CITY COUNTRY

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

mik bang ind

muj bang ind

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

select a.aid, name

from author a, catalog c

where a.aid=c.aid and

c.bid=(select bid

from odetails

group by bid

having sum(qty) = (select max(sum(qty))

from odetails

group by bid));

AID NAME

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

1 mik

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

update catalog

set price=1.1*price

where pid in (select pid

from publisher

where name='abc');

select * from catalog;

BID TITLE AID PID CID YEAR PRICE

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

111 lib1 1 101 1001 2002 500

112 lib2 2 102 1002 2000 800

113 lib3 3 103 1003 2003 200

114 lib4 4 104 1001 2006 385

115 lib5 5 105 1004 2007 100

116 lib6 2 103 1005 2007 600

117 lib7 2 105 1002 2007 450

118 lib8 1 101 1001 2002 500



CHAPTER 7

BANK DATABASE

Consider the following database for a banking enterprise

BRANCH (bname: string, bcity: string, assets: real)

ACCOUNT (accno: int, bname: string, balance: real)

DEPOSITOR (cname: string, accno: int)

BANK_CUST (cname: string, cstreet: string, ccity: string)

LOAN (loan_no: int, bname: string, amount: real)

BORROWER (cname: string, loan_no: int)

Schema-Diagram:

BRANCH

bname bcity assets

ACCOUNT

accno bname balance

DEPOSITOR

cname accno

BANK_CUST

cname cstreet ccity

LOAN

loan_no bname amount

BORROWER

cname loan_no



Creation of Tables:

create table branch

(

bname varchar(10) primary key,

city varchar(10),

assets real

);

create table account

(

accno int primary key,

bname varchar(10),

balance real,

foreign key(bname) references branch(bname)

);

create table cust

(

cname varchar(10) primary key,

cstreet varchar(10) not null,

city varchar(10) not null

);

create table depositor

(

cname varchar(10),

accno int,

primary key(cname,accno),

foreign key(accno) references account(accno) on delete cascade,

foreign key(cname) references cust(cname) on delete cascade

);

create table loan

(

lno int primary key,

bname varchar(10) not null,

amt real,

foreign key(bname) references branch(bname) on delete cascade

);



create table borrower

(

cname varchar(10),

lno int,

primary key(cname,lno),

foreign key(cname) references cust(cname) on delete cascade,

foreign key(lno) references loan(lno) on delete cascade

);


insert into branch values('abc','bang',1200000);

insert into branch values('def','che',2000000);

insert into branch values('abn','mum',330000);

insert into branch values('xyz','hyd',555555);

insert into branch values('mno','bang',9999999);

insert into account values(1,'abc',25000);

insert into account values(2,'def',12000);

insert into account values(3,'def',1000);

insert into account values(4,'abn',10000);

insert into account values(5,'mno',600000);

insert into account values(6,'xyz',50000);

insert into cust values('mik','ab','bang');

insert into cust values('muj','cd','bang');

insert into cust values('maj','ef','che');

insert into cust values('waj','xy','del');

insert into cust values('prad','lm','mum');

insert into cust values('now','op','hyd');

insert into depositor values('mik',2);

insert into depositor values('muj',1);

insert into depositor values('muj',5);

insert into depositor values('prad',4);

insert into depositor values('maj',3);

insert into depositor values('waj',6);

insert into depositor values('mik',3);

insert into loan values(1,'abc',5000);

insert into loan values(2,'def',1500);

insert into loan values(3,'abn',10000);

insert into loan values(4,'xyz',3500);

insert into loan values(5,'mno',20000);



insert into borrower values('mik',2);

insert into borrower values('muj',1);

insert into borrower values('prad',3);

insert into borrower values('maj',4);

insert into borrower values('waj',5);

select *from branch;

BNAME CITY ASSETS

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

abc bang 1200000

def che 2000000

abn mum 330000

xyz hyd 555555

mno bang 9999999

select *from account;

ACCNO BNAME BALANCE

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

1 abc 25000

2 def 12000

3 def 1000

4 abn 10000

5 mno 600000

6 xyz 50000

select *from cust;

CNAME CSTREET CITY

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

mik ab bang

muj cd bang

maj ef che

waj xy del

prad lm mum

now op hyd

select *from depositor;

CNAME ACCNO

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

mik 2

muj 1

muj 5

prad 4

maj 3

waj 6

mik 3



select *from loan;

LNO BNAME AMT

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

1 abc 5000

2 def 1500

3 abn 10000

4 xyz 3500

5 mno 20000

select *from borrower;

CNAME LNO

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

mik 2

muj 1

prad 3

maj 4

waj 5

Queries:

iii. Find all the customers who atleast two accounts at the MAIN branch.

select cname

from account a, depositor d

where a.accno=d.accno and

bname='def'

group by cname

having count(*)>1;

CNAME

----------

mik

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

select cname

from cust c

where not exists (select bname

from branch

where city='bang')

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='bang');

CNAME

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

muj



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

delete from account

where bname in (select bname

from branch

where city='che');

2 rows deleted.

select *from account;

ACCNO BNAME BALANCE

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

1 abc 25000

4 abn 10000

5 mno 600000

6 xyz 50000

chapter 1 · a mini project should be implemented by the students in teams. the maximum size of a...

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