Verify that user input is one of several allowed words

Question

So I am having trouble with this line of code, which is meant to check if user input matches the given options.

set /p "myVar=---> " 
echo %myVar%|findstr /ix "red:Red:blue:Blue">nul && ( 
 echo %myVar% matched 
 ) || ( 
 echo %myVar% not matched 
 ) 

Is there a way I can go it like the above or any other way?

Answer 1

You could make use of the fact that for /F returns an exit code of 1 in case of zero iterations:

set /P VAR="Enter something: " || ((echo Empty input!) & exit /B)
(for /F "delims=AaEeIiOoUuYy eol=y" %%K in ("%VAR%") do rem/) && (
    echo No match found!
) || (
    echo Match encountered.
)

If the input consists only of characters listed after delims= , the for /F loop does not iterate and returns an exit code of 1 ; the command behind && only executes in case the exit code is 0 , the command behind || only executes in case the exit code is not 0 .
If no input is provided, set /P sets the exit code to 1 .

Answer 2

Here is a basic example of what you want to do. I have not tested in DOS since I do not have any VM (or very old computer) right here, but IIRC this should work in plain DOS as well.

@echo off
if %var% == A GOTO :anything
if %var% == E GOTO :anything
if %var% == I GOTO :anything
if %var% == O GOTO :anything
if %var% == U GOTO :anything
if %var% == Y GOTO :anything
goto end
:anything
echo anything
:end

Answer 3

I think, choice is your best friend here. But if you want to do it without choice (it isn't available in all Windows versions):

set /p "myVar=---> "
echo %myVar%|findstr /ix "a e i o u y">nul && (
  echo %myVar% matched
) || (
  echo %myVar% not matched
)

It asks for input and checks if it is exactly one of the characters from the list, ignoring capitalization.