Replace string % with %% using batch file

Question

I have a text like below in input.txt file.

sdf%5Ddfssdsd%2Ddfdf

I would like to replace "%" with "%%" in the output file. so the text should look like

sdf%%5Ddfssdsd%%2Ddfdf

Appreciate your help!!

Potential duplicate: http://stackoverflow.com/questions/14718591/escape-percent-signs-in-given-variables — David Starkey, Apr 15 '14 at 17:56

score 2 · Accepted Answer · answered Apr 15 '14 at 18:00

2

@echo off

setlocal ENABLEDELAYEDEXPANSION

for /f "tokens=1* delims=" %%i in (input.txt) do (
  set _line=%%i
  set _line=!_line:%%=%%%%!
  echo !_line! >> output.txt
)

endlocal

answered Apr 15 '14 at 18:00

James L.

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Replace string % with %% using batch file

1 Answers1