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Asked: 2026-05-11T02:41:18+00:00

I have tried with the following, but it just says & was unexpected at

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I have tried with the following, but it just says ‘& was unexpected at this time.’

@echo off :enter-input echo Please enter a number between 1 and 15: echo 1 = Selection one echo 2 = Selection two echo 4 = Selection three echo 8 = Selection four echo x = Quit  set INPUT= set /P INPUT=Type number: %=%  if '%INPUT%' == '' goto enter-input if '%INPUT%' == 'x' goto end if '%INPUT%' == 'X' goto end  set /A %INPUT% if %INPUT% & 1 == 1 echo Selection one if %INPUT% & 2 == 2 echo Selection two if %INPUT% & 4 == 4 echo Selection three if %INPUT% & 8 == 8 echo Selection four  echo Done :end
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1 Answer

  1. I found a way of doing this.

    @echo off :enter-input echo Please enter a number between 1 and 15: echo 1 = Selection one echo 2 = Selection two echo 4 = Selection three echo 8 = Selection four echo x = Quit  set /P INPUT=Type number:  if '%INPUT%' == '' goto enter-input if '%INPUT%' == 'x' goto end if '%INPUT%' == 'X' goto end  set /A isOne = '(%INPUT% & 1) / 1' set /A isTwo = '(%INPUT% & 2) / 2' set /A isThree = '(%INPUT% & 4) / 4' set /A isFour = '(%INPUT% & 8) / 8'  if %isOne% == 1 echo Selection one if %isTwo% == 1 echo Selection two if %isThree% == 1 echo Selection three if %isFour% == 1 echo Selection four  echo Done :end
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