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up vote -4 down vote favorite

I have list of files like first.js#19 edit, traval.jsp#233 edit, run.css#25 edit. I want to remove part from #, I just want file name ie first.js, traval.jsp, run.css How can do it using batch programming. Please share your code. Thanks for help

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up vote 1 down vote

try:

for /f "tokens=1*delims=#" %%a in ('dir /b /a-d') do ren "%%~a#%%~b" "%%~a"

this doesn't work, if the first # is followed by another #.

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up vote 0 down vote

Try this:

for /r %%a in (*) do (for /f "delims=#" %%z in ("%%~a") do (ren %%a %%z)

And that should work. Not sure about spaces, but your example didn't include any.

Mona

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up vote 0 down vote

Try this code (assuming that you the batch script is located in the same directory as the file):

@echo off
SetLocal EnableDelayedExpansion

for /f "delims=# tokens=1*" %%a in ('dir /b') do (
    set "oldName=%%a#%%b"
    set "newName=%%a"

    ren "!oldName!" "!newName!"
)

Hope this helps.

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Thanks for your answers. But I am new to batch programming. I have a text file which consist of all these file names. I want to travels that text file and show file name one by one – user2918831 Oct 25 '13 at 8:42
    
can you show me the text file that you are working on so that I will be able to help you better. – Dale Oct 25 '13 at 8:51
up vote 0 down vote

You can do that directly from the command line:

for /f "delims=#" %a in (theTextFile.txt) do echo %a

This works as long as the file names are placed in individual lines of the text file. If you want to make a Batch file with this line, double the percents.

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