staline

• Jun 28th, 2005

 

SELECT * FROM TABLE1 WHERE COL1 LIKE ''%' ESCAPE ''

siri

• Jul 7th, 2005

 

correct syntax is 
SELECT * FROM TABLE WHERE COL LIKE '%9!%%' ESCAPE '!' 
THIS RETURNS THE VALUES THAT CONTAIN 9% 
THE FIRST AND THE LAST PERCENT SIGNS ARE WILDCARD CHARACTERS AND " ! " SIGN IS THE ECAPE CHARACTER.

carry

• Aug 31st, 2005

 

select * from table where col1 like '%%%' ESCAPE ''

rajeshkoripalli

• Sep 14th, 2005

 

suppose consider a table named as symbol containing

SQL> select * from symbol;

CHARACTER  NAME
---------- ----------
%          modfun
#          hash

SQL> select * from symbol where character='%';

CHARACTER  NAME
---------- ----------
%          modfun

virgilkumar

• Oct 11th, 2005

 

use escape key word.Example a string is like this aru%n.In this u want to find the %.Then use like 'aru#%#%'escape#.Try this

Rajesh

• Nov 8th, 2005

 

select * from dept where 1=decode(instr(loc,'%'),0,0,1)

Jai

• Nov 8th, 2005

 

select * from tab_name where col_name like '%\%%' escape '\';

Atul Takiar

• Nov 9th, 2005

 

select * from table_name where col_name like '% \% %' escape '\';

Indrajit Adhya

• Nov 9th, 2005

 

We have to use ESCAPE character.

If job column if emp table has '%' ex.('vb%dev') then

SELECT * FROM EMP where job like '%/%%' escape '/'

Here '/' ESCAPE character

pradeep

• Feb 9th, 2006

 

select * from emp where instr('ename','%')!=0

ankitagujrati28

 Profile Answers by ankitagujrati28 

• May 3rd, 2010

 

SELECT *
FROM table_nm
WHERE col_nm LIKE '%/%' escape'/';

dj_dj_dj

 Profile Answers by dj_dj_dj 

• May 6th, 2010

 

select *
from table_name
where code in
                      (select code
                        from table_name
                        where instr(column_name,'%') 
                        and instr(column_name,'%') > 0);



Regards
Dharmendra Jaiswal.

yagneswarasankar

 Profile Answers by yagneswarasankar 

• Nov 22nd, 2010

 

SQL> select * from name;

NAME
--------------------
Girija%Sankar
Girija,Sankar

SQL> select * from name where name like '%%%' escape '';

NAME
--------------------
Girija%Sankar

yasodha

• Aug 29th, 2011

 

consider that a person have a name like this 'k%ng'
He can be found by


last_name like'k@%ng'

leelakrishna302

 Profile Answers by leelakrishna302 

• Jun 15th, 2012

 

the query is to find the name of the person starting with a
here is an example
select ename from emp where ename like a%;

abhishek

• Jul 17th, 2012

 

select * from emp where ename like ki\%ng escape ;

Nazeera Jaffar

• Sep 30th, 2012

 

The below code will select the row having column as %

Code
 

Nazeera Jaffar

• Oct 6th, 2012

 

The below code selects the column with a substring %
(e.g.) reg%istration

Code