Issue with script

[message1]

I am trying to replace '12345' with '56789' in a text file.
For this I am calling the following command in run_rename.bat file:

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for /f "usebackq delims==" %%a IN (`dir /b *4869*.txt`) do call renameFile.bat %%a

And the renamefile.bat has the following code:

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'net use P:\

CD C:\temp

REM
REM

setLocal DisableDelayedExpansion

REM Rename File to X.xml to enable change of text 
if exist X.xml del X.xml
REM ren email_4869_12345.txt X.xml
ren %1 X.xml

REM @echo off > XXX 
setLocal DisableDelayedExpansion

for /f "tokens=* delims= " %%G in (X.xml) do ( set str=%%G


setLocal EnableDelayedExpansion

REM set the string "12345 " to be "56789 "
set str=!str:12345 =56789 !

REM generate a new file with the rename in place
REM >> email_4869_12345.txt echo(!str!) 
>> %1 echo(!str!) 

del X.xml

This code is chopping off '!' from the text file and replacing nothing and taking the character off.
For example:
if the file has 'when!', the script is deleting '!' and taking the position off in the generated file.
Can anybody suggest where is it going or suggest me how to suppress this?


Thanks.

[abc0502]

so all what you are trying to do is to change a part of a file name with another part, is that right?

[message1]

Yes.
12345 with 56789.


Thanks.

[abc0502]

I guess this code will do that for the renaming part,

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For /F "tokens=1,2,3* delims=_" %%a in ("email_4869_12345.txt") Do Ren "email_4869_12345.txt" "%%a_56789_%%c"

[message1]

Sorry, I made you understand the issue wrongly.


I am trying to process the data in the .txt file and i am processing it and not trying to rename the file name just.


Thanks

[abc0502]

Then what you are trying to do is "replace" text in a file with another text

[message1]

Yes

[abc0502]

IF so, have alook at this
http://www.dostips.com/?t=Batch.FindAndReplace
and also this is somthing simmelar to what you was trying to do
http://www.computing.net/answers/programming/batch-file-to-replace-a-word/15770.html

and this
http://www.computing.net/answers/programming/batch-file-to-replace-a-word/15770.html

[foxidrive]

Here's a WSH method.

Write your batch file to call this one.

@echo off
call SAR.BAT "input.txt" "output.txt" "12345" "56789"

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::Search and replace
@echo off
if "%~3"=="" (
echo.Search and replace
echo Syntax: 
echo %0 "filein.txt" "fileout.ext" "regex" "replace_text" [first]
echo.
echo. if [first] is present only the first occurrence is changed
goto :EOF
)
if "%~5"=="" (set global=true) else (set global=false)
set s=regex.replace(wscript.stdin.readall,"%~4")
 >_.vbs echo set regex=new regexp
>>_.vbs echo regex.global=%global%
>>_.vbs echo regEx.IgnoreCase=True            
>>_.vbs echo regex.pattern="%~3"
>>_.vbs echo wscript.stdOut.write %s%
cscript /nologo _.vbs <"%~1" >"%~2"
del _.vbs