I have found the need to report an error level to a program that calls batch scripts.
The script will create a CSV log file that I would like to check in order to display whether the script was run successfully.
What I would like to say is something like
IF NOT FIND "[ERR" "test.csv"
or to say, that if the string "[ERR" is not in the output ERRORLEVEL = 0 ELSE ERRORLEVEL = 1
Hope this makes sense. I'm sure it is simple but setting error levels seems to be a royal pain in the a$se!
FIND "[ERR" "text.csv" sets ERRORLEVEL to 0 if at least one [ERR is found, and 1 if no [ERR is found. You need a way to invert the logic.
If all you want to do is immediately return 0 if no [ERR found, or 1 if at least one [ERR found, then I would use the following:
find "[ERR" "test.csv" >nul && exit /b 1 || exit /b 0
If you want to capture the result in a variable to be returned later, then:
find "[ERR" "test.csv" >nul && set "err=1" || set "err=0"
or
find "[ERR" "test.csv" >nul
set /a "err=!%errorlevel%"
When you are ready to return the result
exit /b %err%
Below is my original answer that was accepted, yet was embarrassingly wrong :-(
Thanks Corey for pointing out the logic error. As he says in his comment, the faulty code below will only report an error if all of the lines have the [ERR text.
FIND will set the ERRORLEVEL to 0 if found, 1 if not found. To reverse the logic, simply use the /V option.
find /v "[ERR" "test.csv" >nulIf you have a variable set that indicates the errorlevel, then you can use EXIT /B to set the ERRORLEVEL
set err=1 exit /b %err%