SQL Most Important Query

Ramani Hitesh (iOS Software Engineer) — Mon, 26 Apr 2021 06:56:05 +0000

MySql

To see the databases

1)show databases;

to create database

2)create database cognizantstud;

to goto the database

3)use cognizantstud;

to drop database

4)drop database cognizantstud;

to create a table

5)create table student

( studid int not null auto_increment,
name varchar(30) not null,
address varchar(30) not null,
primary key(studid));

6)to insert record

insert into student values(101,'tarun','chennei');

7)to view the records

select * from student;

8)update the record

update student set address='Bangalore' where name='tarun';

9)to create a view (view will not contain any data.it will contain only the query)
create view stud as select studid,name from student;

we can do insert,update and delete of the view it will replect to the base table.

10)to see all the tables and views

show tables;

11)where condition :-

select * from student where name='tarun';

12)distinct :-(unique data will be displayed)

select distinct address from student;

13)order by:-
select * from student order by name;
select * from student order by name asc;

select * from student order by name desc;

14)group by

select address, count(*) from student group by address;

15)aggregate function

count(),sum(),avg(),min(),max(),first(),last()

i)select count(*) from student;
first record :- select name from student limit 1;

last record :-select name from student order by name desc limit 1;

16)alter table:-
alter table student add salary int not null;
to add data to salary column :-
update student set salary=45000 where studid=101;

update student set salary=35000 where studid in (102,103,104);

17)select sum(salary) from student;
18)select avg(salary) from student;
19)select min(salary) from student;

20)select max(salary) from student;

between and example:-

21)select * from student where salary between 30000 and 46000;

like % (wildcard)

22)select * from student where name like 'p%';(word starting with p will display)

select * from student where name like '%r';(word ending with r will display)

23)truncate table:-
it will delete all the data from the table.

truncate table student;

24)drop table;
it will delete the data as well as the table also.

drop table student;

25)join

create table marks
(studid int(4),
phy int(4),
chem int(4),
maths int(4),
tot int(4),avg int(4));

i)insert into marks(studid,phy,chem,maths) values(101,56,67,78);
ii)update marks set tot=(phy+chem+maths) where studid in (101,102,103);

iii)update marks set avg=tot/3 where studid in (101,102,103);

joining 2 tables:-

select a.studid,a.name,a.address,a.salary,b.phy,b.chem,b.maths,b.tot,b.avg from student a,marks b where a.studid=b.studid;

DEV Community: Ramani Hitesh (iOS Software Engineer)