Check directory in Windows batch

dir.cmd

@echo off

echo %~1

::::::::::::::::::: has argument ? :::::::::::::::::::
IF ["%~1"]==[""] (
	echo Missing argument!!!
	echo Syntax:
	echo 	setup ^<main_repo^>
	exit /B 1;
)

::::::::::::::::::: argument exist ? :::::::::::::::::::
IF NOT EXIST %~s1 (
	echo "%~1" not exist
	exit /B 1
)

::::::::::::::::::: is directory ? :::::::::::::::::::
IF NOT EXIST %~s1\NUL (
	echo "%~1" is a file. Require directory!
	exit /B 1
)

• in %~1 - the ~ removes any wrapping " or '.
• in %~s1 - the s makes the path be DOS 8.3 naming, which is a nice trick to avoid spaces in file-name while checking stuff (and this way no need to wrap the resource with more "s.
• the ["%~1"]==[""] "can not be sure" if the argument is a file/directory or just a generic string yet, so instead the expression uses brackets and the original unmodified %1 (just without the " wrapping, if any..)
  if there were no arguments of if we've used shift and the arg-list pointer has passed the last one, the expression will be evaluated to [""]==[""].
• this is as much specific you can be without using more tricks (it would work even in windows-95's batch-scripts...)

Source