Batch - Compare variable with regular expression
batch-file, comparison, if-statement, regex
Solution
Batch/cmd does not support regexes directly. You have to use `findstr`for that, for example:
`echo %node% | findstr /r "[vV][0-9.]*" >nul 2>&1 && (echo contains) || (echo does not contain)` or
`echo %node% | findstr /r "[vV][0-9.]*" >nul 2>&1 if errorlevel 1 (echo does not contain) else (echo contains)`
This trick delegates comparison to `findstr` and than uses return code (errorlevel) from it. (please note that regexes `findstr` supports are also quite limited and has some quirks, more info http://ss64.com/nt/findstr.html)
Problem
I'm doing a batch script that has to check if there are some programs installed on the computer. For that, I execute `programName --version` and I store the output in a variable. The problem is when I try to compare with a regular expression (only to know if this program exists in the machine). I'm trying this code, but does't work ``` >output.tmp node --version <output.tmp (set /p hasNode= ) if "%hasNode%" == "[vV][0-9.]*" (echo Has node) else (echo You have to install node) ``` If I change the regular expression for the output of this command works properly, so I suppose that I'm doing a bad use of the regular expression (I've checked it and it's fine for the command's output) Thanks four your help guys