How to set the EQU operator to work properly? - batch-file

Question

Everything works fine in this program except when user enters the correct number generated by the program. For example, it replies "too low" if you enter 13, and "too high" if you enter 14. How do you fix this so the 3rd if statement functions properly?
#echo off
title Game
set num1=%random%%101
:start
set /p num2=Enter a number:
if %num2% gtr %num1% (echo too high
pause
cls
goto :start
)
if %num2% lss %num1% (echo too low
pause
cls
goto :start
)
if %num2% equ %num1% (echo Bingo!)
pause
cls
exit

Answer 1

This batch file works fine:
#echo off
title Game
setlocal EnableDelayedExpansion
set /A num1=%random% %% 101
:Retry
cls
set "num2="
set /P "num2=Enter a number (%num1%): "
if !num2! GTR %num1% (
echo too high
pause
goto Retry
)
if !num2! LSS %num1% (
echo too low
pause
goto Retry
)
if !num2! EQU %num1% (
endlocal
echo Bingo!
pause
exit /B
)
goto Retry
The line
set num1=%random%%101
could be also used in the batch file above, but creates
a number with a random part and
appended the first argument passed to the batch file if called with a parameter at all and
appended the digits 0 and 1.
I suppose like SomethingDark that a random number between 0 and 100 is wanted here.
start is a standard Windows command and therefore it is advisable not to use start as label although possible.
Delayed expansion is enabled and used for num2 to avoid an exit of the batch file caused by a syntax error when the batch user enters nothing or by mistake not a number.
Also the environment variable num2 is always unset before prompting the user as otherwise the batch user could just hit RETURN or ENTER to continue with whatever variable num2 currently has as value.
Note: The random number is output in prompt text to see what to enter for testing.