What does% ~ 1 do in this batch file?

Question

I found this code, but there are parts that I do not understand.

This is my code:

Main.bat:

@echo off set "CallCount=0" set "Mood=" set /P "Mood=Your mood is: " call Receive.bat "%Mood%" rem *Random stuff* set "Food=" set /P "Food=The food you want is: " call Receive.bat "%Food%" set "CallCount=" 

receive.bat:

 @echo off set /A CallCount+=1 if "CallCount"=="2" goto Call2 if not "%~1"=="" echo %1 <---- *Random Stuff* | goto :EOF |---What is %~1 doing in this area? :Call2 | if not "%~1"=="" echo %1 <---- rem Commands for second call. 

Edit: This is a file that uses the call command twice.

Answer 1

%1 is the first argument to the form of the calling command line. If the argument passed has quotes around it, %1 contains quotation marks. Where as, %~1 provides the argument value with quotes removed.

Useful link here .