How to remove trailing and leading whitespace for user-provided input in a batch file?

Question

I know how to do this when the variable is pre-defined. However, when asking for the user to enter in some kind of input, how do I trim leading and trailing whitespace? This is what I have so far:

@echo off  set /p input=: echo. The input is %input% before  ::trim left whitespace for /f "tokens=* delims= " %%a in ("%input%") do set input=%%a ::trim right whitespace (up to 100 spaces at the end) for /l %%a in (1,1,100) do if "!input:~-1!"==" " set input=!input:~0,-1!   echo. The input is %input% after  pause 

Answer 1

The solution below works very well for me.

Only 4 lines and works with most (all?) characters.

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Solution:

:Trim SetLocal EnableDelayedExpansion set Params=%* for /f "tokens=1*" %%a in ("!Params!") do EndLocal & set %1=%%b exit /b 

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Test:

@echo off  call :Test1 & call :Test2 & call :Test3 & exit /b  :Trim SetLocal EnableDelayedExpansion set Params=%* for /f "tokens=1*" %%a in ("!Params!") do EndLocal & set %1=%%b exit /b  :Test1 set Value=   a b c    set Expected=a b c call :Trim Actual %Value% if "%Expected%" == "%Actual%" (echo Test1 passed) else (echo Test1 failed) exit /b  :Test2 SetLocal EnableDelayedExpansion set Value=   a \ / : * ? " ' < > | ` ~ @ # $ [ ] & ( ) + - _ = z     set Expected=a \ / : * ? " ' < > | ` ~ @ # $ [ ] & ( ) + - _ = z call :Trim Actual !Value! if !Expected! == !Actual! (echo Test2 passed) else (echo Test2 failed) exit /b  :Test3 set /p Value="Enter string to trim: " %=% echo Before: [%Value%] call :Trim Value %Value% echo After : [%Value%] exit /b