Create table Project and Employee relationship between many to many with descriptive attribute start_date.





Consider the following database 

Project (pno int, pname char (30), ptype char (20), duration int) 

Employee (empno int, ename char (20), joining_date date) 

The relationship between Project and Employee is many to many with descriptive attribute start_date. 




Create Table Project 


Project (pno int, pname char (30), ptype char (20), duration int) 


create table project1 (pno int primary key , pname varchar(50),ptype varchar (20), duration integer  );


Insert values in Project table


insert  into project1 values(1,'aa','game    ' , 25   );

insert  into project1 values(2,'bb','appliaction ' , 30    );

insert  into project1 values(3,'cc','website     ' , 45   );

insert  into project1 values(4,'dd','robtotic    ' , 35    );


Output of Table


Acode3=# select * from project;


o\p


Acode3=# select * from project;

 pno | pname |    ptype     | duration 

-----+-------+--------------+----------

   1 | aa    | game         |       25

   2 | bb    | appliaction  |       30

   3 | cc    | website      |       45

   4 | dd    | robtotic     |       35

(4 rows)


Create tabel Employee


Employee (empno int, ename char (20), joining_date date) 


create table employee3 (eno int primary key, ename varchar(50), salary float,dateJ date );


Insert values in Employee

insert into employee3 values (1,'Acode',15000.00 ,'2020-04-28');

insert into employee3 values (2,'mujju',20000.00,'2020-05-21' );

insert into employee3 values (3,'adharsh', 25000,'2020-04-21');

insert into employee3 values (4,'rohan',26000,'2019-04-12');


Output of table

Acode3=# select * from employee3;


o\p

Acode3=# select * from employee3;

 eno |  ename  | salary |   datej    

-----+---------+--------+------------

   1 | Acode   |  15000 | 2020-04-28

   2 | mujju   |  20000 | 2020-05-21

   3 | adharsh |  25000 | 2020-04-21

   4 | rohan   |  26000 | 2019-04-12

(4 rows)



Create table(Pro) M-M relationship


create table pro ( eno int references employee3 (eno),pno int references project1 (pno));



Insert values in Pro table


insert into pro values ( 1,1 );

insert into pro values ( 1,2 );

insert into pro values ( 1,3 );

insert into pro values ( 1,4 );

insert into pro values ( 2,1 );

insert into pro values ( 2,3 );

insert into pro values ( 2,2 );

insert into pro values ( 2,4 );

insert into pro values ( 3,1 );

insert into pro values ( 3,3 );

insert into pro values ( 3,4 );

insert into pro values ( 4,4 );

insert into pro values ( 4,3 );

insert into pro values ( 4,2 );

insert into pro values ( 4,1 );


Output of tabel

Acode3=# select * from pro;


o\p

Acode3=# select * from pro;

 eno | pno 

-----+-----

   1 |   1

   1 |   1

   1 |   2

   1 |   3

   1 |   4

   2 |   1

   2 |   3

   2 |   2

   2 |   4

   3 |   1

   3 |   3

   3 |   4

   4 |   4

   4 |   3

   4 |   2

   4 |   1

(16 rows)



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