Find and replace in a file

[sri2002]

I have script which generates SQL's with variables.

@echo off

set "YYYY=2013"
set "YY=13"
set "MM=03"
set "LOCATION=D://DCOUMENTS/SQLSCRIPT/"

select count(*) from table1 where filedirectory='XXXXX' and file_name like 'ABC_percentage_%YYYY%%MM%__.txt'; >> %LOCATION%Generated_Script.sql

select count(*) from table2 where filedirectory='XXXXX' and file_name like 'ABC_percentage_%YYYY%%MM%__.txt'; >> %LOCATION%Generated_Script.sql

select count(*) from table3 where filedirectory='XXXXX' and file_name like 'ABC_percentage_%YYYY%%MM%__.txt'; >> %LOCATION%Generated_Script.sql

--After loading all sql's INTO %LOCATION%Generated_Script.sql. I want to find "percentage" and replace with "%" character.

Can someone help me to create a script finding "percentage" and replace with "%". I tried to create but its not working may be because of % character.

Appreciate your help in this. thanks in advance

[foxidrive]

echo %% will create a single % character

Show us the code that you have tried.

[sri2002]

echo %% will create a single % character

Show us the code that you have tried.

Thanks foxidrive.

Your tip help me to resolve the issue. Now I dont need to find and replace it.

@echo off

set "YYYY=2013"
set "YY=13"
set "MM=03"
set "LOCATION=D://DCOUMENTS/SQLSCRIPT/"

select count(*) from table1 where filedirectory='XXXXX' and file_name like 'ABC_%%_%YYYY%%MM%__.txt'; >> %LOCATION%Generated_Script.sql

select count(*) from table2 where filedirectory='XXXXX' and file_name like 'ABC_%%_%YYYY%%MM%__.txt'; >> %LOCATION%Generated_Script.sql

select count(*) from table3 where filedirectory='XXXXX' and file_name like 'ABC_%%_%YYYY%%MM%__.txt'; >> %LOCATION%Generated_Script.sql

[foxidrive]

Thanks foxidrive.

Your tip help me to resolve the issue. Now I dont need to find and replace it.

That's great. It's better to solve the issue in the first instance.