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"?