DBMS lab manual

LEIT116

Exp:1 Create and modifying the relations

Date: 30/7/2013

Aim:

To write a query on creation and modification of relations using library management system..

Queries:

SQL> create table library(book_id number(7),isbn_no number(7),subject char(25),language char(25),name char(25),copies number(7),available number(7),cost number(7),author char(25));

Table created.

SQL>desc library;

Name Null? Type

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

BOOK_ID NUMBER(7)

ISBN_NO NUMBER(7)

SUBJECT CHAR(25)

LANGUAGE CHAR(25)

NAME CHAR(25)

COPIES NUMBER(7)

AVAILABLE NUMBER(7)

COST NUMBER(7)

AUTHOR CHAR(25)

SQL> create table libissue as select book_id from library;

Table created.

SQL> ALTER table libissue add(student_id number(7),issue_datedate,due_datedate,return_datedate,fine number(7),issuers_id number(7));

Table altered

SQL>desclibissue

LEIT116

Name Null? Type

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

BOOK_ID NUMBER(7)

STUDENT_ID NUMBER(7)

ISSUE_DATE DATE

DUE_DATE DATE

RETURN_DATE DATE

FINE NUMBER(7)

ISSUERS_ID NUMBER(7)

SQL> create table libauthor as select author from library;

Table created.

SQL> alter table libauthor add(author_id number(7),edition number(7),publisher_name char(25),publications char(25),yr_of_pub number(7),address varchar(50));

Table altered.

SQL>desclibauthor;

Name Null? Type

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

AUTHOR CHAR(25)

AUTHOR_ID NUMBER(7)

EDITION NUMBER(7)

PUBLISHER_NAME CHAR(25)

PUBLICATIONS CHAR(25)

YR_OF_PUB NUMBER(7)

ADDRESS VARCHAR2(50)

SQL> create table student as select student_id from libissue;

Table created.

LEIT116

SQL> alter table student add(student_name char(25),acc_no number(7),reg_no number(7),department char(25),total_limit number(7),books_returned number(7),books_pending number(7));

Table altered.

SQL>desc student

Name Null? Type

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


STUDENT_NAME CHAR(25)

ACC_NO NUMBER(7)

REG_NO NUMBER(7)

DEPARTMENT CHAR(25)

TOTAL_LIMIT NUMBER(7)

BOOKS_RETURNED NUMBER(7)

BOOKS_PENDING NUMBER(7)

SQL> create table staffincharge as select issuers_id from libissue;

Table created.

SQL> alter table staffincharge add(staff_id number(7),name char(25),qualification char(25),designation char(25),date_of_joiningdate,contact_no number(7),address varchar(50));

Table altered.

SQL>descstaffincharge;

Name Null? Type

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


NAME CHAR(25)

QUALIFICATION CHAR(25)

DESIGNATION CHAR(25)

DATE_OF_JOINING DATE

LEIT116

CONTACT_NO NUMBER(7)

ADDRESS VARCHAR2(50)

SQL> insert into library values(1,100,'computerscience','english','java',200,50,500,'samba');

1 row created.

SQL> select * from library;

BOOK_ID ISBN_NO SUBJECT LANGUAGE NAME

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

1 100 computer science English java

COPIES AVAILABLE COST AUTHOR

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

200 50 500 samba

SQL> insert into student values(12,'ram',123,97,'it',100,5,15);

1 row created.

SQL> select * from student;

STUDENT_ID STUDENT_NAME ACC_NO REG_NO DEPARTMENT

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

12 ram 123 97 it

TOTAL_LIMIT BOOKS_RETURNED BOOKS_PENDING

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

100 5 15

SQL> insert into libauthorvalues('samba',1,10,'samba','samba publish',2013,'anna nagarchennai');

1 row created.

SQL> select * from libauthor;

AUTHOR AUTHOR_ID EDITION PUBLISHER_NAME

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

samba 1 10 samba

LEIT116

PUBLICATIONS YR_OF_PUB ADDRESS

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

samba publish 2013 annanagarchennai

SQL> insert into staffincharge values('2','sita','mba','librarian','222222','anna nagarmadurai');

1 row created.

SQL> select * from staffincharge;

ISSUERS_ID NAME QUALIFICATION DESIGNATION CONTACT_NO

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

2 sita mba librarian 222222

ADDRESS

-----------

annanagarmadurai

SQL> insert into libissue values('1','123','100','2','200713');

1 row created.

SQL> select * from libissue;

BOOK_ID STUDENT_ID FINE ISSUERS_ID ISSUE_DATE

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

1 123 100 2 200713

SQL> truncate table libissue;

Table truncated.

SQL>desclibissue;

Name Null? Type

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

BOOK_ID NUMBER(7)


FINE NUMBER(7)

LEIT116


ISSUE_DATE NUMBER(7)

SQL> select *from libissue;

no rows selected

SQL> delete from libissue where book_id=1;

0 rows deleted.

SQL> alter table libissue drop column issue_date;

Table altered.

SQL> alter table libissue drop column due_date;

Table altered.

SQL> alter table libissue drop column return_date;

Table altered.

SQL>desclibissue

Name Null? Type

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

BOOK_ID NUMBER(7)


FINE NUMBER


SQL> alter table libissue add issue_datenumber(7);

Table altered.

LEIT116

Result:

Thus the creation and modification of relation are done and successfully output was verified.

Exp:2 Integrity Constraint

LEIT116

Aim:To create and implement the usage of queries using constraint.

Queries:Primary key :

Column level:

SQL> create table book2(bid number(10),bnamevarchar(10),authnamevarchar(20),co

nstraint book2_bid_pk primary key(bid));

Table created.

SQL>desc book1;

Name Null? Type

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

BOOKNAME CHAR(20)

PRICE NOT NULL NUMBER(18)

Table level:

SQL> create table book2(bid number(10),bnamevarchar(10),authnamevarchar(20),co

nstraint book2_bid_pk primary key(bid));

Table created.

SQL>desc book2;

Name Null? Type

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

BID NOT NULL NUMBER(10)

BNAME VARCHAR2(10)

AUTHNAME VARCHAR2(20)

BID BNAME PID

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

435 maths 999

123 social 998

Unique:

LEIT116

Column level:SQL> create table bbbc(bid number(10) constraint bbbc_bid_ukunique,bnamevarchar(20),pid number(10));

Table created.

SQL> insert into bbbcvalues(435,'maths',999);

1 row created.

SQL> insert into bbbcvalues(123,'social',998);

1 row created.

SQL> insert into bbbcvalues(123,'science',997);

insert into bbbc values(123,'science',997)

*ERROR at line 1:

ORA-00001: unique constraint (SCOTT.BBBC_BID_UK) violated

SQL> select * from bbbc;

BID BNAME PID

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

435 maths 999

123 social 998

SQL>alter table student2 add rollnonumber(6) constraint student2_rollno_ck check(rollno>=55000);Table altered.SQL>insert into student2 values('12it70','Priya','Linux',52345);ORA-02290: check constraint (SYSTEM.STUDENT2_ROLLNO_CK) violated

Table level:SQL>create table book7(bid number(10),bnamevarchar(20),authnamevarchar(2),price number(3),constraint book7_price_cc check(price>0));Table created.SQL>insert into book7 values(192,'dsd','morrismano',0);ORA-02290: check constraint (SYSTEM.BOOK7_PRICE_CC) violated

Default:SQL>create table student3(rollno number(5),regnovarchar(7),department varchar(6) default 'IT');Table created.SQL>insert into student3(rollno,regno) values(456,'12it71');

LEIT116

1 row(s) inserted.SQL>select * from student3;

ROLLNO REGNO DEPARTMENT--------- --------- --------------------456 12it71 IT

Foreign keyTable level:SQL>create table publisher(pid number(10),pnamevarchar(20),phno number(10),constraint publisher_pid_pk primary key(pid));Table createdSQL>insert into book6 values(1232,'Linux',435,768,31245);ORA-02291: integrity constraint (SYSTEM.BOOK6_PID_FK) violated - parent key not foundSQL>insert into publisher values(111,'grewal',991123459)1 row(s) inserted.

SQL>select * from publisher;PID PNAME PHNO------ ----------- ----------111 Grewal 991123459

SQL>insert into book6 values(1241,'Linux',610,111,6122);1 row(s) inserted.

Column level:SQL>create table book01(bid number(10),bnamevarchar(20),authnamevarchar(20),pid number(10),constraint book02_pid_fk foreign key(pid) references publisher01(pid) );Table created.SQL>insert into book01 values(11,'Maths','Grewall',190);1 row(s) inserted.SQL>insert into book01 values(124,'CO','Zvonko',191);1 row(s) inserted.

Not null:SQL>alter table book6 add isbnnumber(10) constraint book6_isbn_nn NOT NULL;Table altered.SQL>insert into book6(bid,bname,price,pid) values(465,'Java2',99.50,65);ORA-01400: cannot insert NULL into ("SYSTEM"."BOOK6"."ISBN")

Cascade:

LEIT116

SQL>create table book02(bid number(10),bnamevarchar(20),authnamevarchar(20),pid number(10),constraint book02_pid_fk foreign key(pid) references publisher02(pid) on delete cascade);Table created.SQL>insert into book02 values(453,'Java','IraPohl',134);1 row(s) inserted.SQL>insert into book02 values(456,'dsd','Morrismano',135);1 row(s) inserted.

Set :SQL>create table book03(bid number(10),bnamevarchar(20),authnamevarchar(20),pid number(10),constraint book03_pid_fk foreign key(pid) references publisher03(pid) on delete set null);Table created.SQL>insert into book03 values(862,'Surveying','Punmia',637);1 row(s) inserted.SQL>insert into book03 values(852,'CO','Zvonko',737);1 row(s) inserted.

Result:

Thus the constraint was successfully completed and output was verified

Exp:3 Simple SQL Queries

LEIT116

Aim:

To execute the simple sql queries.

Queries:

Query 1

To display the DOB and address of the employee ‘John Smith’.

SQL> select bdate,address from employee where fname='John' and lname='Smith';

BDATE ADDRESS

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

18-AUG-90 Pudur, Madurai

Query 2

To select the name and address of the employee who works in ‘Research department’

SQL> select fname||' '||lname as Name,address from employee where ssn=(select mgrssn from department where dname='Research');

NAME ADDRESS

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

Raj Kumar Kodambakkam, Chennai

Query 3

To display the Project No, Department No. And Department manager’s last name,DOB and address for the projects located in Chennai.

SQL> select p.pnumber,p.dnumber,e.lname,e.bdate as DOB,e.address from employee e,projectp,department d where d.dnum=p.dnumber and p.plocation='Chennai' and d.mgrssn=e.ssn;

LEIT116

PNUMBER DNUMBER LNAME DOB ADDRESS

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

21 6 Smith 18-AUG-90 Pudur Madurai

Query 4

To retrieve the first and last name of all employees along with their Manager’s no.

SQL> select e.fname,e.lname,e.ssn,d.mgrssn from employee e,department d where e.ssn!=d.mgrssn and e.dno=d.dnum;

FNAME LNAME SSN MGRSSN

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

Raj Kumar 5 10

John Smith 2 4

Ram Kumar 1 9

Query 5

To select all employee’s name and SSN

SQL> select fname||’ ‘||lname as Name,ssn from employee;

NAME SSN

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

Ram Kumar 1

John Smith 2

Raj Kumar 5

Query 6:

To select all employee’s name, SSN and the department’s name.

LEIT116

SQL> select d.dname,d.dnum,e.fname||’’||e.lname as name,e.ssn from employee e,department d where d.mgrssn=e.ssn;

DNAME DNUM NAME SSN

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

Research 3 Raj Kumar 5

Technical 6 John Smith 2

Accounts 5 Ram Kumar 1

Query 7:

Retrieving the salary of all employees

SQL> select fname||' '||lname as Name_Of_Employee,salary from employee;

NAME_OF_EMPLOYEE SALARY

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

Ram Kumar 14000

John Smith 10500

Raj Kumar 10500

Query 8

To retrieve the distinct salary values.

SQL> select distinct salary from employee;

SALARY

----------

14000

10500

LEIT116

Query 9

To retrieve the name of the employee who is from Chennai.

SQL> select fname||' '||lname as name from employee where address like '%Theni';

NAME

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

Ram Kumar

Query 10

Employees who were born in 1990.

SQL> select fname||’ ‘||lname as name from employee where bdate like '%90';

NAME

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

John Smith

Query 11

To show the resulting salary and the name of the employee who is working on the project Web Design and is given 10% rise in salary.

SQL> select p.pname,e.fname,w.essn,e.salary+0.1*e.salary as salary from employee e,projectp,workson w where p.pname='Web Design' and w.pno=p.pnumber and w.essn=e.ssn;

PNAME FNAME ESSN SALARY

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

Web Design John 2 11550

Query 12

LEIT116

To retrieve all employees in the department 3 whose salary is in between 13K and 15K.

SQL> select fname,salary from employee where dno=3 and salary between 13000 and 15000;

FNAME SALARY

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

Ram 14500

Query 13

To retrieve the list of employees and the project they are working on ordered by the department number.

SQL> select distinct e.fname,p.pname,p.pnumber,e.dno from employee e,project p where e.dno=p.dnumber order by e.dno;

FNAME PNAME PNUMBER DNO

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

Raj Cryptography 42 3

Ram yyy 41 5

Ram zzz 35 5

John Web Design 21 6

Query 14

To retrieve the name of employee whose project is being controlled by department number 6

SQL> select e.fname||' '||e.lname as name,p.pnumber,e.dno from project p,employe e e,workson w where w.essn=e.ssn and w.pno=p.pnumber and p.dnumber=6;

NAME PNUMBER DNO

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

LEIT116

John Smith 21 6

Result:

Thus the simple sqlqueries was successfully executed and output was verified.

LEIT116

Exp:4 Basic simple SQL queries

20/8/2013

Aim:

To write a basic simple sql queries in library management system

Queries:

1. Find the id and title of all courses which do not require any prerequisities.

selectcourse_id,title from course where course.course_id not in (select course_id from prereq);

COURSE_ID TITLE

BIO-301 Genetics BIO-399 Computational BiologyCS-101 Intro.to Computer Science FIN-201 Investment Banking HIS-351 World History MU-199 Music Video Production PHY-101 Physical Principles

2. Write SQL update query to increase 10% salary to all instructors.SALARY6500090000400001950006000087000175000620008000072000

update instructor set salary=salary+(0.1*salary);

14 row(s) updated.

Select salary from instructor; SALARY

LEIT116

71500 99000 44000 104500 66000 95700 82500 68200 88000 79200

3.Write SQL update query to increase the total credits of all students who have taken the course title ‘Genetics’ by the no. Of credits associated with the course.

select * from student;

ID NAME DEPT_NAME TOT_CRED

00128 Zhang Comp. Sci. 102 12345 Shankar Comp. Sci. 32 19991 Brandt History 80 23121 Chavez Finance 110 44553 Peltier Physics 56 45678 Levy Physics 46 54321 Williams Comp. Sci. 54 55739 Sanchez Music 38 70557 Snow Physics 0 76543 Brown Comp. Sci. 5876653 Aoi Elec. Eng. 60 98765 Bourikas Elec. Eng. 98 98988 Tanaka Biology 120

update student120 set tot_cred=tot_cred+(select credits from course where course.title='Genetics') where id in(select student120.id from student120,course where student120.dept_name=course.dept_name and course.title='Genetics');

1 row(s) updated.

ID NAME DEPT_NAME TOT_CRED 00128 Zhang Comp. Sci. 102 12345 Shankar Comp. Sci. 32 19991 Brandt History 80

LEIT116

23121 Chavez Finance 110 44553 Peltier Physics 56 45678 Levy Physics 46 54321 Williams Comp. Sci. 54 55739 Sanchez Music 38 70557 Snow Physics 0 76543 Brown Comp. Sci. 58 76653 Aoi Elec. Eng. 60 98765 Bourikas Elec. Eng. 98 98988 Tanaka Biology 122

4.Find the names of students who have not taken any biology department courses.

select name from student120 where dept_name!='Biology';

NAMEZhangShankar Brandt Chavez PeltierLevy Williams Sanchez Snow Brown

5. Write SQL update query to list all the instructors who are advisor of atleast two students,increase the salary by 50000.

Srinivasan 71500

Wu 99000

Mozart 44000

Einstein 104500

Elsaid 66000

Gold 95700

Katz 82500

Califeri 68200

LEIT116

Singh 88000

Crick 79200

Brandt 101200

Sara 55000

Sandy 44000

Kim 193000

select count(advisor.s_id),instructor.name from advisor,instructor,student where student.id=advisor.s_id and instructor.id=advisor.i_id group by instructor.name;

COUNT(ADVISOR.S_ID) NAME

2 Einstein

1 Crick

1 Srinivasan

2 Kim

1 Singh

2 Katz

update instructor set salary=salary+50000 where name in (select name from instructor,advisor where instructor.id=advisor.i_id group by instructor.name having count(s_id) > 1);

3 row(s) updated.

NAME SALARY

Srinivasan 71500

Wu 99000

Mozart 44000

Einstein 214500

El Said 66000

Gold 95700

Katz 192500

Califieri 68200

Singh 88000

Crick 79200

Brandt 101200

Kim 198000

LEIT116

Sara 55000

Sandy 440006.Write SQL update query to set the credits to 2 for all courses which have less than 5 students taking them COURSE_ID TITLE DEPT_NAME

CREDITS

BIO-301 Genetics Biology 4

BIO-399 ComputationalBiology Biology 3

CS-101 Intro.to Computer Science Comp. Sci. 4

CS-190 Game Design Comp. Sci. 4

CS-315 Robotics Comp. Sci. 3

CS-319 Image Processing Comp. Sci. 3

CS-347 Database System Concepts Comp. Sci. 3

EE-181 Intro.to Digital Systems Elec. Eng. 3

FIN-201 Investment Banking Finance 3

HIS-351 World History History 3

MU-199 Music Video Production Music 3

PHY-101 Physical Principles Physics 4

update course set credits=2 where course_id in (select course.course_id from course,takes where course.course_id=takes.course_id group by course.course_id having count(takes.id)<5);

COURSE_ID TITLE DEPT_NAME CREDITS

BIO-301 Genetics Biology 2

BIO-399 Computational Biology Biology 3

CS-101 Intro.to Computer Science Comp. Sci. 4

CS-190 Game Design Comp. Sci. 2

CS-315 Robotics Comp. Sci. 2

LEIT116

CS-319 Image Processing Comp. Sci. 2

CS-347 Database System Concepts Comp. Sci. 2

EE-181 Intro.to Digital Systems Elec. Eng. 2

FIN-201 Investment Banking Finance 2

HIS-351 World History History 2

MU-199 Music Video Production Music 2

Result;

Thus the basic simple sql queries are successfully executed and the output was verified.

LEIT116

EXP:5 JOINS

27/8/13

Aim:

To write a simple sql queries on joins using our application.

Queries:

SQL> create table stu(id number(7),name varchar(20),dept_namevarchar(20),tot_credit number(3));

Table created.

SQL>descstu

Name Null? Type

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

ID NUMBER(7)

NAME VARCHAR2(20)

DEPT_NAME VARCHAR2(20)

TOT_CREDIT NUMBER(3)

SQL> insert into stuvalues(116,'maha','it',60);

1 row created.

SQL> insert into stuvalues(97,'subha','it',90);

1 row created.

SQL> insert into stuvalues(116,'sne','it',50);

1 row created.

SQL> create table tak(id number(7),cource_id number(7),sec_id number(7),semester number(3),year number(7),grade char(3));

Table created.

SQL>desctak

Name Null? Type

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

LEIT116

ID NUMBER(7)

COURCE_ID NUMBER(7)

SEC_ID NUMBER(7)

SEMESTER NUMBER(3)

YEAR NUMBER(7)

GRADE CHAR(3)

SQL> insert into takvalues(45,'32',5,3,2013,'A');

1 row created.

SQL> insert into takvalues(52,31,7,3,2012,'C');

1 row created.

SQL> insert into tak values(47,35,8,4,2010,'A');

1 row created.

SQL> insert into tak(id,sec_id,semester,year,grade) values(78,9,4,2009,'C');

1 row created.

SQL> insert into takvalues(116,45,6,3,2007,'B' );

1 row created.

SQL> create table ins(id number(7),name varchar(20),dept_namevarchar(20),salary number(10));

Table created.

SQL>descins;

Name Null? Type

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

ID NUMBER(7)

NAME VARCHAR2(20)

DEPT_NAME VARCHAR2(20)

SALARY NUMBER(10)

SQL> insert into ins values(78,'susila','cse',40000);

LEIT116

1 row created.

SQL> insert into ins values(95,'chells','maths',45000);

1 row created.

SQL> insert into ins values(64,'sankar','it',48000);

1 row created.

SQL> create table teach(id number(7),cource_id number(7),sec_id number(7),semester number(3),year number(7));

Table created.

SQL>desc teach

Name Null? Type

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

ID NUMBER(7)

COURCE_ID NUMBER(7)

SEC_ID NUMBER(7)

SEMESTER NUMBER(3)

YEAR NUMBER(7)

SQL> insert into teach values(118,33,7,3,2009);

1 row created.


1 row created.


1 row created.

Joins queries:

INNER JOINS:

SQL> select stu.name,tak.id from stu inner join tak on stu.id=tak.id;

NAME ID

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

LEIT116

maha 116

sne 116

OUTER JOINS:

Left outer join:

SQL> select * from stu left outer join tak on stu.id=tak.id;

ID NAME DEPT_NAME TOT_CREDIT ID

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

COURCE_ID SEC_ID SEMESTER YEAR GRA

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

97 subha it 90

116 sne it 50

Rigtht outer join:

SQL> select * from stu right outer join tak on stu.id=tak.id;


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


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

78

9 4 2009 C

52

31 7 3 2012 C

LEIT116

45

32 5 3 2013 A


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


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

47

35 8 4 2010 A

116 maha it 60

Full outer join:

SQL> select * from stu full outer join tak on stu.id=tak.id;


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


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

97 subha it 90

116 sne it 50

116 maha it 60

LEIT116


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


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

78

9 4 2009 C

52

31 7 3 2012 C

45

32 5 3 2013 A


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


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

47

35 8 4 2010 A

7 rows selected.

Interact:

LEIT116

SQL>select name from tak interact select name from stu;

NAME

-----------

Maha

Subha

Sne

Union:

SQL> select namedept_name from stu union select semester from tak where tot_cred=”90”;

DEPT_NAME SEMESTER

Subha 3

LEIT116

Result;

Thus the joins queries are successfully executed by using the library management application application.

LEIT116

EXP:6 creation and updation of views

10/9/13

Aim:

To write a simple queries by using views

Queries:

Create view:

SQL> create view lib_view as select subject,author from library;

View created.

Display:

SQL> select * from lib_view;

SUBJECT AUTHOR

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

computer science balagursamy

SQL>desclib_view;

Name Null? Type

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

SUBJECT CHAR(25)

AUTHOR CHAR(25)

SQL> select author from lib_view where subject='computer science';

AUTHOR

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

Balagursamy

SQL> create view lib_view1(name,dob,cellno) as select docname,docid,cellno from docdet where docid=12 ;

View created.

SQL>desc lib_view1;

Name Null? Type

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

LEIT116

NAME VARCHAR2(5)

DOB NUMBER(10)

CELLNO NUMBER(10)

Check option:

SQL> create view lib_view3(name,address,wardno) as select docname,docid,cellno from docdet where docid=112 with check option;

View created.

Update:

SQL>update lib_view3 libno=116 where name=’main’;

O Rows selected.

SQL> alter view lib_view3 compile;

View altered.

Read only option:

SQL> create view lib_view4(name,dob,cellno) as select docname,docid,cellno from docdet where docid=1WITH READ ONLY;

View created.

SQL>desc lib_view4;

Name Null? Type

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

NAME VARCHAR2(5)

DOB NUMBER(10)

CELLNO NUMBER(10)

Drop view:

SQL> drop view lib_view;

View dropped.

SQL> select * from lib_view;

no rows selected.

LEIT116

Result:

Thus the creation and updation of view was successfully executed and output was verified.

LEIT116

Expno:7 EXERCISES IN PL/SQL

17/09/2013

Aim:

To implement an query on pl/sql statement.

Pl/sql blocks:

Bind variable:

SQL> set serveroutput on

SQL> VARIABLE bindvar NUMBER

SQL> declare

2 v_num number(2);

3 BEGIN

4 v_num := 5;

5 :bindvar := v_num*2;

6 END;

7 /

PL/SQL procedure successfully completed.

SQL> PRINT bindvar;

G_DOUBLE

----------

10

SQL> VARIABLE g_double NUMBER

SQL> declare

2 v_num number(2);

3 begin

4 v_num := &p_num;

5 :g_double := v_num*2;

LEIT116

6 end;

7 /

Enter value for p_num: 20

old 4: v_num := &p_num;

new 4: v_num := 20;


SQL> print g_double;

G_DOUBLE

----------

40

SQL> variable num number


SQL> declare

2 double number;

3 begin

4 :num := 5;

5 double := :num*2;

6 dbms_output.put_line(('double of'||to_char(:num) || 'is' || to_char(double)

));

7 end;

8 /

double of5is10


Anchor declaration

SQL> variable b number

LEIT116

SQL> declare

2 v_num number(2);

3 v_num2 v_num%type;

4 begin

5 v_num:=4;

6 v_num2:=8;

7 :b:=v_num*v_num2;

8 end;

9 /


SQL> print b

B

----------

32

Control statements

Decode function

SQL> select fname,lname,

2 decode(101,1000,

3 102,2000,

4 103,3000)from emp;

FNAME LNAME DECODE(101,1000,102,2000,103,3000)

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

thrigaya 3000

nathpriy 3000

kumararun 3000

LEIT116

if..then..elseif..endif

SQL> declare

2 m number:=&mark;

3 grade char;

4 begin

5 if m>=190 and m<=200then

6 grade:='a';

7 elseif m>=160 and m<=190 then

8 grade:='b';

9 elseif m>=120 and n<=160 then

10 grade:='u';

11 end if;

12 dbms_output.put_line(grade);

13 end;

14 /

Enter value for mark: 199

old 2: m number:=&mark;

new 2: m number:=199;

a


Searched case:

SQL> declare

2 v_num number:=&any_num;

3 v_res number;

4 begin

5 v_res:=v_num

LEIT116

6 casev_res

7 when 0 then dbms_output.put_line(v_num||'is even');

8 elsedbms_output.put_line(v_num||'is odd');

9 end case;

10 end;

11./

Enter value for any_num: 3

old 2: v_num number:=&any_num;

new 2: v_num number:=3;

3is odd


Case structure

SQL> declare

2 v_num number:=&any_num;

3 v_res number;

4 begin

5 v_res:=v_num

6 casev_res

7 when 0 then dbms_output.put_line(v_num||'is even');

8 elsedbms_output.put_line(v_num||'is odd');

9 end case;

10 end;

11 /

Enter value for any_num: 4

old 2: v_num number:=&any_num;

new 2: v_num number:=4;

LEIT116

4is even


Nested if


SQL> declare

2 v_name char:='&name';

3 v_id number(2):='&id';

4 v_age number(2):='&age';

5 begin

6 if(v_name='gayu' and v_id=11)then

7 v_age:=23;

8 end if;

9 if(v_name='anu' and v_id=12)then

10 v_age:=24;

11 end if;

12 if(v_name='priya' and v_id=13)then

13 v_age:=25;

14 end if;

15 if(v_name='kris' and v_id=14)then

16 v_age:=26;

17 end if;

18 dbms_output.put_line(v_name);

19 dbms_output.put_line(to_char(v_id));

20 dbms_output.put_line(to_char(v_age));

21 en

2 /

LEIT116

Enter value for name: anu

old 2: v_name char:='&name';

new 2: v_name char:='anu';

Enter value for id: 12

old 3: v_id number(2):='&id';

new 3: v_id number(2):='12';

Enter value for age: 24

old 4: v_age number(2):='&age';

new 4: v_age number(2):='24';

anu

12

24


If..then.end..if

SQL> declare

2 v_num number(5):=112;

3 v_id number(5):=115;

4 employee number(5);

5 begin

6 ifv_num>113 then

7. employee:=114;

8. employee:=112;

9. end if;

10. end;

11./


LEIT116

If..then..else..endif

SQL> declare

2 v_num number(5):=112;

3 v_id number(5):=115;

4 employee number(5);

5 begin

6 ifv_num>113 then

7 employee:=114;

8 else

9 employee:=112;

10 end if;

11 end;

12 /


If..then..elseif..endif

SQL> declare

2 m number(3):=&mark;

3 grade char;

4 begin

5 if m>=190 and m<=200then

6 grade:='good';

7 elseif m>=170 and m<=190;

8 grade:='excellent';

9 elseif m>=140 and m<=170;

10 grade:='fair';

LEIT116

11 endif;

12 dbms_output.put_line(grade);

13 end;

14 /

Enter value for mark: 170

old 2: m number(3):=&mark;

new 2: m number(3):=170;

Looping statements

for loop

SQL> declare

2 v_countnumber(2);

3 v_sumnumber(2):=0;

4 v_avgnumber(3,1);

5 begin

6 for v_count in 1..10 loop

7 v_sum:=v_sum+v_count;

8 end loop;

9 v_avg:=v_sum/10;

10 dbms_output.put_line(to_char(v_avg));

11 end;

12 /

5.5


While loop:


LEIT116

SQL> declare

2 v_count number(2);

3 v_sum number(2):=0;

4 v_avg number(3,1);

5 begin

6 v_count:=1;

7 whilev_count<=10 loop


9 v_count:=v_count+1;

10 end loop;


12 v_avg:=v_sum/(v_count-1);

13 dbms_output.put_line(to_char(v_avg));

14 end;

15 /

6.6


Basic loop:

SQL> setserveroutput on

SQL> declare

2 numnumber(5);

3 id number(5);

4 begin

5 loop

6 num:=num+1;

7 id:=id+1;

LEIT116

8 exit when num=2;

9 end loop;

10 end;

11 /

10

2


Update:


SQL> declare

2 cnumber number:=&value;

3 begin

4 update employee

5 set salary=salary*(1+value)

6 whereemployeeif=&emp_id;

7 commit;

8 end;

9 /

Enter value for value: 3

old 2: cnumber number:=&value;

new 2: cnumber number:=3;

Enter value for emp_id: 3

old 6: where employeeif=&emp_id;

new 6: where employeeif=3;


Insert:

LEIT116

SQL> declare

2 v_nameemployee.name%type;

3 begin

4 insert into employee(name,id,salary) values('gayu',112,1000);

5 commit;

6 end;

7 /


Delete:

SQL> declare

2 v_nameemployee.name%type;

3 begin

4 select id from employee where name='gayu';

5 delete from employee

6 where name=v_name;

7 commit;

8 end;

9 /


Comment line:

Singleline:


SQL> variable a number

SQL> declare

2 v_num number(10);


LEIT116

4 begin

5 v_num:=5;

6 v_num2:=10;

7 :a:=v_num*v_num2;

8 end;

9 --PROGRAM ENDED

10 /


Multiline:

SQL> variable b number

SQL> declare

2 v_num number(5);


4 begin

5 v_num:=5;

6 v_num2:=10;

7 :a:=v_num*v_num2;

8 end;

9 /*

10 */

11 /


LEIT116

Result:

Thus the pl/sql program was executed successfully.

LEIT116

EXP:08 CURSOR MANAGEMENT

01/10/13

Aim:

To implement the cursor management on library management system.

Queries:

Explicit cursor attributes:

SQL> set serveroutput on;

SQL> declare

2 -- declare the variables

3 c_namestudent.name%type;

4 c_tot_credstudent.tot_cred%type;

5 --declare the cursor

6 cursor student_cur is

7 select name,tot_cred

8 from student

9 where id=76543;

10 begin

11 if not student_cur%isopen then

12 --open the cursor

13 open student_cur;

14 end if;

15 loop

16 --fetch the rows from the cursor

17 fetch student_cur

18 into c_name,c_tot_cred;

19 exit when not student_cur%found;

LEIT116

20 dbms_output.put_line(c_name||' '||c_tot_cred);

21 end loop;

22 dbms_output.put_line(student_cur%rowcount||'student(s) found');

23 --close the cursor

24 closestudent_cur;

25 end;

26 /

Brown 58

1student(s) found


Cursor for loop:

SQL> declare

2 cursorstudent_cur is

3 selectname,tot_cred

4 from

5 student;

6 begin

7 forstu_rec in student_cur loop

8 ifstu_rec.tot_cred>90 then

9 dbms_output.put_line(stu_rec.name||' ');

10 dbms_output.put_line(stu_rec.tot_cred||' ');

11 end if;

12 end loop;

13 end;

LEIT116

14 /

Zhang

102

Chavez

110

Bourikas

98

Tanaka

120


Create a cursor for update:

SQL> declare

2 cursor student_cur is

3 select * from student

4 for update of tot_cred;

5 begin

6 for stu_rec in student_cur

7 loop

8 update student

9 set tot_cred=(stu_rec.tot_cred/10)

10 --Where current of clause

11 where current of student_cur;

12 end loop;

13 end;

LEIT116

14 /


Cursor for loop using a subquery:

begin

forlib_rec in

(selectname,reg_no,dept,no_of_book,fine

from library

wheredept='it')loop

dbms_output.put_line

(lib_rec.name||''||lib_rec.reg_no||'$'||to_char(lib_rec.no_of

_book+NVL(lib_rec.fine,0)));

end loop;

end;

/


Cursor with parameter

SQL> declare

2 cursorc_lib is select * from library;

3 begin

4 forlib_no in c_lib loop

5 dbms_output.put_line('lib_no.id:'||lib_no.id);

6 end loop;

7 commit;

8 end;

9 /

LEIT116

lib_no.id:112

lib_no.id:114

lib_no.id:115

Cursor with REF:

SQL> declare

2 typebook_type is ref cursor return book%rowtype;

3 v_bookbook_type;

4 n_bookbook%rowtype;

5 begin

6 openv_book for select * from library where id=112;

7 fetchv_book into n_book;

8 dbms_output.put_line(v_book.number||' is '||n_book.name);

9 closev_book;

10 end;

11 /

java

112

Exception:

TOO_MANY_ROWS:


SQL> declare

2 v_namelibrary.id%type;

3 begin

4 dbms_output.put_line('first'||v_name);

5 select id into v_name from library;

LEIT116

6 dbms_output.put_line('second'||v_name);

7 exception

8 whentoo_many_rows then

9 dbms_output.put_line('third'||v_name);

10 dbms_output.put_line('exception'||v_name);

11 end;

12 /

first

third112

exception112


User defined exception:

SQL> declare

2 v_name exception;

3 v_no exception;

4 v_namelibrary.name%type;

5 begin

6 select name into v_name from libaray

7 wherelibraryid=&v_id;

8 ifv_name<0 then

9 raisev_name;

10 elseifv_name is null then

11 raisev_id;

12 else

13 dbms_output.put_line(to_char(v_name);

14 endif;

LEIT116

15 exception

16 whenno_data_found

17 dbms_output.put_line('no_data_found');

18 end;

19 /

Enter value for v_id: 2

old 7: where libraryeid=&v_id;

new 7: where libraryid=2

Result:

Thus the cursor was successfully executed and output was verified.

LEIT116

EXP:9 PROCEDURE,FUNCTION AND PACKAGE

DATE:15/10/13

PL/SQL BLOCKS:

Procedures:

SQL> create or replace procedure search_lib

2 (i_libid in number,

3 o_last out varchar,

4 o_first out varchar)

5 is

6 begin

7 selectid,name

8 intoo_last,o_first

9 from wow

10 where wow.id=i_empid;

11 exception

12 when others then

13 dbms_output.put_line('name is'||o_last);

14 endsearch_emp;

15 /

Procedure created.


SQL> declare

2 v_lastwow.name%type;

3 v_firstwow.category%type;

4 v_idwow.id%type:=&mp_id;

5 begin

6 search_emp(v_id,v_last,v_first);

LEIT116

7 ifv_last is not null then

8 dbms_output.put_line('library'||v_id);

9 dbms_output.put_line('name'||v_last||','||v_first);

10 end if;

11 end;

12 /

Enter value for mp_id: 12

old 4: v_idwow.id%type:=&mp_id;

new 4: v_idwow.id%type:=12;

library 12

name12,maha


Function:

SQL> create or replace function val_get

2 (i_id in number)

3 returnvarchar

4 is

5 v_namevarchar(10);

6 begin

7 select name into v_name

8 from wow

9 where id=i_id;

10 returnv_name;

11 endval_get;

12 /

Function created.

SQL> declare

LEIT116

2 v_idwow.id%type:=&id;

3 v_namewow.name%type;

4 begin

5 select id

6 intov_id from wow

7 where wow.id=v_id;

8 v_name:=val_get(v_id);

9 dbms_output.put_line('the name is'||v_name);

10 exception

11 when others then

12 dbms_output.put_line(v_id||'not found');

13 end;

14 /

Enter value for id: 12

old 2: v_idwow.id%type:=&id;

new 2: v_idwow.id%type:=12;

the name is maha


SQL> create or replace procedure displayquantity is

2 temp_quantitynumber(10);

3 begin

4 select quantity into temp_quantity from book

5 where bookno=12;

6 if temp_quantity>100 then

7 dbms_output.put_line('success');

8 else

9 dbms_output.put_line('not');

LEIT116

10 end if;

11 exception

12 when NO_DATA_FOUND then

13 dbms_output.put_line('medicene not found');

14 end displayquantity;

15 /

Procedure created.

Execute:

SQL> execute displayquantity



SQL> execute displayquantity

success


SQL>drop procedure displayquantity;

Procedure dropped.

SQL> create or replace function retrievequantity

2 return number

3 is

4 v_quantity number(10);

5 begin

6 select quantity into v_quantity

7 from book

8 wherebookno=12;

9 returnv_quantity;

10 endretrievequantity;

11 /

LEIT116

Function created.

SQL>varv_qty number;

SQL>EXEC :v_qty := retrievequantity;


SQL> print v_qty;

V_QTY

----------

150

Drop a function:

SQL>drop function retrievequantity;

Function dropped.

Package:

SQL> create or replace PACKAGE book AS

2 PROCEDUREfindbook(

3 medno IN book.bkno%type,

4 description IN book.description%type,

5 cost IN book.cost%type);

6 v_bknoNOTFOUND EXCEPTION;

7 FUNCTIONgoodidentifier(

8 bkno IN book.bkno%type)

9 return BOOLEAN;

10 end book;

11 /

Package created.

SQL> create or replace PACKAGE BODY book AS

2 PROCEDURE findbook(

3 v_medno IN book.bkno%type,

LEIT116

4 v_description OUT book.description%type,v_cost OUT book.cost%type) IS

5 begin

6 select description,cost

7 into v_description,v_cost

8 from

9 book where medno=v_medno;

10 if SQL%ROWCOUNT = 0 then

11 RAISE v_bkno NOTFOUND;

12 end if;

13 end findbook;

14 FUNCTION goodidentifier(

15 v_medno IN book.bkno%type)

16 return BOOLEAN

17 IS

18 v_id_count number;

19 begin

20 select count(*) into v_id_count

21 from book

22 where bkno=v_bkno;

23 return (1=v_id_count);

24 EXCEPTION

25 WHEN OTHERS THEN

26 RETURN FALSE;

27 END goodidentifier;

28 end med;

29 /

LEIT116

Package Body created.

SQL> declare

2 v_descriptionbook.description%type;

3 v_costbook.cost%type;

4 v_mednobook.bkno%type;

5 begin

6 book.findbook(v_bkno,v_description,v_cost);

7 dbms_output.put_line('the book is found');

8 EXCEPTION

9 WHEN OTHERS THEN

10 DBMS_OUTPUT.PUT_LINE('cannot find book');

11 end;

12 /

the book is found


Result:

Thus the pl/sql in procedure,function and packages was successfully executed and the output was verified.

LEIT116

EXP:10 TRIGGERS

22.10.2013

BEFORE TRIGGERS:

SQL> SET SERVEROUTPUT ON

SQL> CREATE OR REPLACE TRIGGER LIB_BI_TRIGGER

2 BEFORE INSERT ON LIB

3 FOR EACH ROW

4 DECLARE

5 V_BKID NUMBER(9);

6 V_BKNAME VARCHAR(12);

7 BEGIN

8 SELECT BKID,BKNAME INTO V_BKID,V_BKNAME FROM LIB;

9 :NEW.BKID:=V_BKID;

10 :NEW.BKNAME:=V_BKNAME;

11 END;

12 /

Trigger created.

AFTER TRIGGERS:

SQL> CREATE OR REPLACE TRIGGER LIB_ADU_TRIGGER

2 AFTER DELETE OR UPDATE ON LIB

3 DECLARE

4 V_LIBNAME VARCHAR(9);

5 BEGIN

6 IF DELETING THEN

7 V_LIBNAME:='DELETE';

8 ELSIF UPDATING THEN

LEIT116

9 V_LIBNAME:='UPDATE';

10 END IF;

11 INSERT INTO LIB VALUES('AVIL',78,'RASES');

12 END;

13 /

Trigger created.

DELETE:

SQL> DELETE FROM LIB1 WHERE MEDNAME='PARACIT';

1 row deleted.

SQL> SELECT * FROM LIB1;

BKNAME BKTYPE BKID

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

JAVA ENGLISH 80

UPDATE:

SQL> UPDATE LIB1 SET BKID=BKID+10 WHERE BKNAME='JAVA';

1 row updated.

SQL> SELECT * FROM LIB1;

BKNAME BKTYPE LIBID

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

JAVA ENGLISH 90

INSTEAD OF TRIGGERS:

NO DATA MANIPULATION THROUGH COMPLEX VIEW:

SQL> CREATE TABLE DOC1(DOCNAME VARCHAR(10),DOCID NUMBER(9),CELLNO NUMBER(10));

Table created.

SQL> INSERT INTO DOC1 VALUES('SIVA',789,9876543456);

1 row created.

LEIT116

SQL> INSERT INTO DOC1 VALUES('GUNA',779,9878967251);

1 row created.

SQL> SELECT * FROM DOC1;

DOCNAME DOCID CELLNO

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

SIVA 789 9876543456

GUNA 779 9878967251

SQL> CREATE OR REPLACE VIEW HOST ASSELECT DOCNAME,CELLNO FROM DOC1 WHERE DOCID=779;

View created.

DELETE:

SQL> DELETE FROM HOST WHERE DOCNAME='GUNA';

1 row deleted.

SQL> SELECT * FROM HOST;

no rows selected

SQL> SELECT * FROM DOc1;

DOCNAME DOCID CELLNO

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

SIVA 789 9876543456

DATA MANIPULATION AND THE INSTEAD OF TRIGGER:

SQL> CREATE TABLE HOS4(DOCNAME VARCHAR(9),DOCID NUMBER(9),CELLNO NUMBER(10));

Table created.

SQL> INSERT INTO HOS4 VALUES('RAVI',897,9876543212);

1 row created.

SQL> INSERT INTO HOS4 VALUES('KAVI',899,9876543218);

1 row created.

LEIT116

SQL> CREATE OR REPLACE VIEW DOCHOS AS

2 SELECT DOCNAME,CELLNO FROM HOS4

3 WHERE DOCID=899;

View created.

SQL> CREATE OR REPLACE TRIGGER GOVTHOS_DELETE_IOD

2 INSTEAD OF DELETE ON DOCHOS

3 FOR EACH ROW

4 BEGIN

5 DELETE FROM HOS4

6 WHERE DOCID=899;

7 END;

8 /

Trigger created.

DELETE:

SQL> DELETE FROM DOCHOS WHERE DOCNAME='RAVI';

0 rows deleted.

QUERIES:

WRITE A PL/SQL TRIGGER TO ENFORCE THE FOLLOWING RULES :

A.) AN EMPLOYEE CAN’T HAVE A SALARY HIGHER THAN HIS/HER DEPT MANAGER:

B.) AN EMPLOYEE CAN’T WORK ON MORE THAN TWO PROJECTS CONTOLLED BY A SINGLE DEPT:

SQL> create or replace trigger tri1 before insert on employee

2 for each row

3 declare

4 p number(20);

LEIT116

5 s number(10);

6 begin

7 select salary into s from employee e,department d

8 wheree.dno=d.dno;

9 selectmgrssn into p from employee e,department d

10 wheree.dno=d.dno;

11 if p>s then

12 raise_application_error(-20078,'salary of the employee shld not be greater

than the manager');

13 end if;

14 end;

15 /

Trigger created.

SQL> create or replace trigger pro1 before insert or update on project

2 for each row

3 declare

4 no number(10);

5 begin

6 select count(pname) into no from project where dnum=:new.dnum;

7 if no>2 then

8 raise_application_error(-20011,'an employee cant work on more than two projects');

9 end if;

10 end;

11 /

Trigger created.

LEIT116

SQL> insert into project values('p1',15,'ngl',13);

1 row created.

SQL> insert into project values('p1',16,'ngl',13);

1 row created.

RESULT:

Thus the queries are implemented by using triggers.

LEIT116

EXP:11 CLASSIFICATION AND CLUSTERING USING WEKA

29.10.13

Aim :

To implement clustering and classification by using weka.

CLASSIFICATION:

=== Run information ===

Scheme: weka.classifiers.rules.ZeroR

Relation: weather

Instances: 14

Attributes: 5

outlook

temperature

humidity

windy

play

Test mode: 10-fold cross-validation

=== Classifier model (full training set) ===

ZeroR predicts class value: yes

Time taken to build model: 0 seconds

=== Stratified cross-validation ===

=== Summary ===

Correctly Classified Instances 9 64.2857 %

LEIT116

Incorrectly Classified Instances 5 35.7143 %

Kappa statistic 0

Mean absolute error 0.4762

Root mean squared error 0.4934

Relative absolute error 100 %

Root relative squared error 100 %

Total Number of Instances 14

=== Detailed Accuracy By Class ===

TP Rate FP Rate Precision Recall F-Measure ROC Area Class

1 1 0.643 1 0.783 0.178 yes

0 0 0 0 0 0.178 no

Weighted Avg. 0.643 0.643 0.413 0.643 0.503 0.178

=== Confusion Matrix ===

a b <-- classified as

9 0 | a = yes

5 0 | b = no

LEIT116

@relation weather

@attribute outlook {sunny, overcast, rainy}

@attribute temperature real

@attribute humidity real

@attribute windy {TRUE, FALSE}

@attribute play {yes, no}

@data

sunny,85,85,FALSE,no

sunny,80,90,TRUE,no

overcast,83,86,FALSE,yes

rainy,70,96,FALSE,yes


rainy,65,70,TRUE,no

overcast,64,65,TRUE,yes

sunny,72,95,FALSE,no

LEIT116

sunny,69,70,FALSE,yes


sunny,75,70,TRUE,yes

overcast,72,90,TRUE,yes

overcast,81,75,FALSE,yes

rainy,71,91,TRUE,no

CLUSTERING:

=== Run information ===

Scheme: weka.clusterers.EM -I 100 -N -1 -M 1.0E-6 -S 100

Relation: weather

Instances: 14

Attributes: 5

outlook

temperature

humidity

windy

play

Test mode: split 50% train, remainder test

=== Clustering model (full training set) ===

EM

LEIT116

==

Number of clusters selected by cross validation: 1

Cluster

Attribute 0

(1)

======================

outlook

sunny 6

overcast 5

rainy 6

[total] 17

temperature

mean 73.5714

std. dev. 6.3326

humidity

mean 81.6429

std. dev. 9.9111

windy

TRUE 7

FALSE 9

[total] 16

play

yes 10

LEIT116

no 6

[total] 16

=== Model and evaluation on test split ===

EM

==

Number of clusters selected by cross validation: 1

Cluster

Attribute 0

(1)

======================

outlook

sunny 3

overcast 3

rainy 4

[total] 10

temperature

mean 72.2857

std. dev. 6.227

humidity

mean 79.5714

std. dev. 11.4998

LEIT116

windy

TRUE 6

FALSE 3

[total] 9

play

yes 5

no 4

[total] 9

Clustered Instances

0 7 (100%)

Log likelihood: -9.73408

LEIT116

LEIT116

Result:

Thus the implementation of classification and clustering was successfully executed.

Education

DBMS lab manual