Question

I wrote a simple batch file to run Frequently Used websites based on a number selection. Here’s the code I have. I am trying to set it so if someone inputs a number 6 or greater it will go to :N but whenever I type 6 the batch file exits. I have tried if %input% > 6 goto :N but it just tells me I am going to Google.

@echo off
:Start2 
cls
goto Start
:Start
title Frequently Used Websites
echo Please select a website from the list
echo with the corresponding key
echo --------------------------------------
echo [1] Google
echo [2] Wikipedia
echo [3] Facebook
echo [4] Youtube
echo [5] Yahoo
set input=
set /p input= Choice:
if %input%==1 goto Z if NOT goto Start2
if %input%==2 goto X if NOT goto Start2
if %input%==3 goto C if NOT goto Start2
if %input%==4 goto V if NOT goto Start2
if %input%==5 goto B if NOT goto Start2
if %input%>=6 goto N

:Z
cls
echo You have selected Google
pause
start www.google.com
exit
:X
cls
echo You have selected Wikipedia
pause
start www.wikipedia.com
exit
:C
cls
echo You have selected Facebook
pause
start www.facebook.com
exit
:V
cls
echo You have selected Youtube
pause
start www.youtube.com
exit
:B
cls
echo You have selected Yahoo
pause
start www.Yahoo.com
exit
:N
cls
echo Invalid Selection! Try again
pause
goto :start2

Answer 1

try this:

if 3 gtr 2 @echo "biggger"

This outputs:

"biggger"

The other operators are:

EQU – equal
NEQ – not equal
LSS – less than
LEQ – less than or equal
GTR – greater than
GEQ – greater than or equal

Reference

• Microsoft TechNet | Windows Server 2012 | Command-line Reference: If (Archived here)
• SS64: If (Archived here)