I need to replace all occurrences of several characters in an argument to batch file:
helper.bat
@echo off echo %1 call:replace_newline_characters_and_additional_quotes arg %%1 echo %arg% goto:eof :replace_newline_characters_and_additional_quotes set in=%2 set tmp=%in:\n=|% set %1=%tmp:""=\"% goto:eof Run it as
helper "1""2\n3"
Output
"1""2\n3"
"3"" is not recognized as an internal or external command, operable program or batch file
What am i doing wrong?
set str=%1 set str=%str:""=\"% set str=%str:\n=^|% echo %str:|=^|% pause Just worked for me with your test case.
You get the "3"" is not recognized error because of the pipe | character, as it means that CMD is trying to pass the text before the | to the 'program' after it, which is why you'll notice I've used ^| to 'escape' the pipe and stop it from having it's special meaning
You should switch to delayed expansion, as it's safe to use special characters.
setlocal EnableDelayedExpansion set "param1=%1" echo !param1! call :replace_newline_characters_and_additional_quotes arg param1 echo !arg! exit /b :replace_newline_characters_and_additional_quotes set "in=!%2!" set "tmp=!in:\n=|!" set "%1=!tmp:""=\"!" exit /b The only problematic line is set "param1=%1", as it would fail in the case of calling
helper "&"^& And it's not easy to avoid this, but there exists a solution at
SO: Get list of passed arguments in Windows batch script
|?"1\"2|3"?