Extract and replace string from filename in Batch Script

Question

I try to extract the filename from a list of files, replace specific string and feed that into a process. However I got stuck at replace. Any idea?

@echo on
setlocal ENABLEDELAYEDEXPANSION

for %%f in (*.txt) do (

            REM echo %%~nf
            set filename=%%~nf
            echo %@filename%
            set Replaced=replaced
            set @ver=!%filename:ToBeReplace=%Replaced%!
            echo %@ver%    
            )

The echo from ver is not what I expected. Thanks in advance!

when you are using delyaed expansion you need to use ! instead of % you can combine it with % when you are replaceing a string - set @ver=!filename:ToBeReplace=%Replaced%! if %replaced% is defined before the brackets block — npocmaka
– npocmaka, Commented Jan 11, 2018 at 13:04
It still echoed empty, I still can't figure out what's wrong :/ — Gregor Isack
– Gregor Isack, Commented Jan 11, 2018 at 13:16

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

I have a file called something.txt in the same folder as the following script:

@echo off
setlocal ENABLEDELAYEDEXPANSION

set "old=some"
set "new=else"
for %%f in (*.txt) do (

            REM echo %%~nf
            set "filename=%%~nf"
            echo !filename!
            rem set Replaced=replaced
            set "@ver=!filename:%old%=%new%!"
            echo !@ver!    
)

the optuput after execution is:

something

elsething

edited Jun 20, 2020 at 9:12

CommunityBot

11 silver badge

answered Jan 11, 2018 at 13:58

npocmaka

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1 Comment

Gregor Isack Over a year ago

Thank you. I realized I should use !.

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