ddl & dml command

7/27/2019 Ddl & Dml Command

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SQL


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Data definition language

SQL includes commands to create

Database objects such as tables, indexes, and views

Commands to define access rights to those database

objects

Data manipulation language

Includes commands to insert, update, delete, andretrieve data within the database tables


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SQL Data Types


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SQL> create table product(p_price number(7,2));

Table created.

SQL> insert into values('&p_price);

Enter value for p_price: 1234567

old 1: insert into values('&p_price)

new 1: insert into values('1234567)ERROR:

ORA-01756: quoted string not properly terminated


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SQL> insert into product values('&p_price');


old 1: insert into product values('&p_price')

new 1: insert into product values('1234567')

insert into product values('1234567')

*

ERROR at line 1:

ORA-01438: value larger than specified precision allows

for this column


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SQL> /



new 1: insert into product values('12345')

1 row created.

SQL> /

Enter value for p_price: 123.1234567


new 1: insert into productvalues('123.1234567')

1 row created.

SQL> /

Enter value for p_price: .123456

old 1: insert into product values('&p_price')new 1: insert into product values('.123456')

1 row created.

SQL> /

Enter value for p_price: +1234.12345


new 1: insert into product

values('+1234.12345')1 row created.

SQL> /

Enter value for p_price: -123.123456


new 1: insert into product values('-

123.123456')

1 row created.

//////////////output/////////////////////

SQL> select * from product;

P_PRICE

----------12345

123.12

.12

1234.12

-123.12


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Number data typeSQL> create table product1(p_price number,p_cost

number(3));

Table created.

SQL> insert into product1 values(123456789,12);

SQL> /

1 row created.

/*********output**********/

SQL> select * from product1;

P_PRICE P_COST

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

123456789 12

SQL> insert into product1 values(1234,123);

1 row created.

SQL> insert into product1 values(1234,1234)

2 ;

insert into product1 values(1234,1234)

*

ERROR at line 1:

ORA-01438: value larger than specified

precision allows for this column


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DATE TIMEData type Description

DATE() A date. Format: DD-MM-YYYY

Note: The supported range is from '1000-01-01' to '9999-12-31'

DATETIME() *A date and time combination. Format: DD-MM-YYYY HH:MM:SS

Note: The supported range is from '1000-01-01 00:00:00' to '9999-12-

31 23:59:59'

TIMESTAMP() *A timestamp. TIMESTAMP values are stored as the number of

seconds since the Unix epoch ('1970-01-01 00:00:00' UTC). Format:

DD-MM-YYYY HH:MM:SSNote: The supported range is from '1970-01-01 00:00:01' UTC to

'2038-01-09 03:14:07' UTC

TIME() A time. Format: HH:MM:SS

Note: The supported range is from '-838:59:59' to '838:59:59'

*Even if DATETIME and TIMESTAMP return the same format, they work very differently. In an INSERT

or UPDATE query, the TIMESTAMP automatically set itself to the current date and time. TIMESTAMP

also accepts various formats, like YYYYMMDDHHMMSS, YYMMDDHHMMSS, YYYYMMDD, or

YYMMDD.


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DATESQL> create table employee(d_O_b date);

Table created.

SQL> insert into employee values('01-jan-99');1 row created.

SQL> insert into employee values('01-feb-1999');

SQL> /

1 row created.SQL> select * from employee;

D_O_B

-------------01-JAN-99

01-FEB-99


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timestamp

SQL> create table employee4(e_name

varchar(20),s_date TIMESTAMP(7))SQL> insert into employee4 values('deep','17-jun-

87 02.22.22')

1 row created.E_NAME S_DATE

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

Deep 17-JUN-87 02.22.22.0000000 AM


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SQL> create table employee5(e_name varchar(20),s_dateTIMESTAMP(4))

SQL> /

Table created.

1* insert into employee5 values('deep','17-jun-87 12.22.22')

SQL> /1 row created.

E_NAME S_DATE

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

deep 17-JUN-87 12.22.22.0000 PM


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MySQL Date FunctionsFunction Description

NOW() Returns the current date and time

CURDATE() Returns the current dateCURTIME() Returns the current time

DATE() Extracts the date part of a date or date/timeexpression

EXTRACT() Returns a single part of a date/time

DATE_ADD() Adds a specified time interval to a date

DATE_SUB() Subtracts a specified time interval from a date

DATEDIFF() Returns the number of days between two dates

DATE_FORMAT() Displays date/time data in different formats


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Different use of Alter tableCREATE TABLE EMP1( ecode number(3)ename varchar2(50) );

/*********************Add column ************

Alter table emp1 add ( age number(3));

/********drop column**********

Alter table emp1 drop column age;

/*********add constraint *********

Alter table emp1 add primary key (ename);

Alter table emp1 add primary key (ecode,ename);

//sim..

Alter table emp1 add UNIQUE(ename);

//change column name (rename)

Alter table emp1 rename column ecode to e_id ;

/******************** ******************


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/********************Modify column ******************

SQL> desc emp1;

Name Null? Type

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

ECODE NUMBER(3)

ENAME VARCHAR2(50)AGE NUMBER(3)

SQL> Alter table emp1 modify ( ename varchar2(100));

Table altered.

SQL> desc emp1;

Name Null? Type----------------------------------------- -------- ---------------

ECODE NUMBER(3)

ENAME VARCHAR2(100)

AGE NUMBER(3)

SQL> Alter table emp1 modify ( ename int);

Table altered.

SQL> desc emp1;

Name Null? Type

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

ECODE NUMBER(3)

ENAME NUMBER(38)AGE NUMBER(3)


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Drop column

emp_id emp_name e_post e_salary e_age----------- ---------- ---------- ----------- --------- ---------

111 shraddha manager 150000 33

222 sagar clerk 5000 24

333 sarvesh security 5000 33

444 raj cashier 50000 32

/* drop cloumn */

alter table employee drop column e_age


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/* output of above query

emp_id emp_name e_post e_salary

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

111 shraddha manager 150000222 sagar clerk 5000

333 sarvesh security 5000

444 raj cashier 50000

delete from employee where emp_name='sagar;select * from employee



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

111 shraddha manager 150000




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Diff. clause (Logical,relational.)create table employee(emp_id int primary key,emp_name varchar(10),e_post

varchar(10),e_salary int);


----------- ---------- ---------- -----------111 shraddha manager 150000

222 sagar clerk 5000



/* and clause */

select emp_name,e_salary from employee where e_salary>500 ande_salary


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/* as clause */

select e_salary as e_s1 from employee


e_s1-----------

150000

5000

5000

50000


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/* distinct (eliminate duplicates) */

select distinct e_salary from employee


e_salary

-----------

5000

50000

150000

/* St i ti */


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/* String operations */create table employee(emp_id int primary key,emp_name

varchar(10),e_post varchar(10),e_salary int);

EMP_ID EMP_NAME E_POST E_SALARY---------- ---------- ---------- ----------





SQL> select * from employee where emp_name like 'sa%;

EMP_ID EMP_NAME E_POST E_SALARY---------- ---------- ---------- ----------




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select * from employee where emp_name like '%sa%

EMP_ID EMP_NAME E_POST E_SALARY

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



select * from employee where emp_name like _ _ _


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


select * from employee where emp_name like '_ _ _ _'

no rows selected

//

select * from employee where emp_name like '_ _ _%'


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





select * from employee where emp name like ' a%'


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select * from employee where emp_name like _a%


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




1* select * from employee where emp_name like '_ _ _ _ _a%'

SQL> /

no rows selected

1* select * from employee where emp_name like '_ _ _a_%'

SQL> /


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



1* select * from employee where emp_name like '_ _ _ _a_%'SQL> /

no rows selected


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Not like

select * from employee where emp_name not

like '%s% ;/* output of above query

emp_id emp_name e_post e_salary----------- ---------- ---------- -----------



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/* aggregation function */emp_id emp_name e_post e_salary e_age

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

111 shraddha manager 150000 33

222 sagar clerk 5000 24

333 sarvesh security 5000 33

444 raj cashier 50000 32

/* aggregation function sum */

select sum(e_salary) from employe

/*output of above query-----------

210000

/* aggregation function avg */

select avg(e_Salary) from employe

/*output of above query

-----------52500

/* aggregation function max,min */

select max(e_Salary),min(e_salary) from employe

/*output of above query

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

8640 40

/* f */


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/* aggregation function count */

//it count all rowin table

select count(*)from employe-----------

4

//it count all rows in table that have non nullvalue for column name

select count(e_salary)from employe /* rows

table have not null value */-----------

4


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Employee table

EMPNO ENAME JOB MGR SAL COMM DEPTNO

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

7369 SMITH CLERK 7902 800 20

7499 ALLEN SALESMAN 7698 1600 300 30

7521 WARD SALESMAN 7698 1250 500 307566 JONES MANAGER 7839 297 1 20

14 rows selected.


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Write down the SQL (SELECT..from..where) queries based on emp & dept

table

a) Write a query to display ename and salary of employees whose salaryis

greater than or equal to 2200.

b) Write a query to display employee name and salary of those employees

who do not have their salary in range of 2500 to 4000.

c) Write a query to display details of employee who are not getting

commission from table emp.

d) Write a query to display name, job, tittle and salary of employee who donot have manager.

e) Write a query to display name of employee whose name contain A as third

alphabet.

f) Write a query to display name of employee whose name contain T as last

alphabet.g) Write a query to display name of employee whose name contain M as

first alphabet and L as third alphabet.


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h) List employee name and their department number for employee number

between 4000 and 5000.

i) List the minimum salary and maximum salary of each job type.j) Show the average salary for all departments with more then three people

for a job.

k) Display only the jobs with minimum salary greater than or equal to 3000.

l) Find out the number of employees having manager as job.

m) Find the average salary for each job type, remember that salesman earnscommission.

n) Show the average salary for all departments for more then three people for

a job.

o) List the count of employee grouped by department number.

p) List the sum of employees, salaries grouped by department.

q) List the maximum salary of employee grouped by their department

number.


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a) Write a query to display ename and

salary of employees whose salary is greater

than or equal to 2200.


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Query :

SQL> select ename,sal from emp where sal>=2200;

ENAME SAL

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

JONES 2975

BLAKE 2850

CLARK 2450

SCOTT 3000

KING 5000FORD 3000


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b) Write a query to display employee name

and salary of those employees who do not

have their salary in range of 2500 to 4000.


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Query:

SQL> select ename,sal from emp where sal not between 2500 and 4000;

ENAME SAL

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

SMITH 800

ALLEN 1600

WARD 1250

MARTIN 1250

CLARK 2450

KING 5000

TURNER 1500

ADAMS 1100

JAMES 950

MILLER 1300


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c) Write a query to display details of

employee who are not getting commission

from table emp.


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Query:

SQL> select * from emp where comm is NULL;

EMPNO ENAME JOB MGR HIREDATE SAL COMM

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

DEPTNO

----------

7369 SMITH CLERK 7902 17-DEC-80 800

20

7566 JONES MANAGER 7839 02-APR-81 2975

20

7698 BLAKE MANAGER 7839 01-MAY-81 2850

30


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d) Write a query to display name, job, tittle

and salary of employee who do not have

manager.


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Query:

SQL> select empno,ename,job,sal from emp where mgr is null;

EMPNO ENAME JOB SAL

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

7839 KING PRESIDENT 5000


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e) Write a query to display name of employee whose name contain A as third

alphabet.

Query:

SQL> select ename from emp where ename like '__A%';

ENAME

----------

BLAKE

CLARK

ADAMS


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f) Write a query to display name of employee whose name contain T as last alphabet.

Query:

SQL> select ename from emp where ename like '%T';

ENAME

----------

SCOTT

g) Write a query to display name of employee whose name contain M as first

alphabet and L as third alphabet.

Query:

SQL> select ename from emp where ename like 'M_L%';

ENAME

----------

MILLER

ddl & dml command

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