Batch Extract path and filename from a variable

Question

How can I extract path and filename from a variable?

Setlocal EnableDelayedExpansion
set file=C:\\Users\\l72rugschiri\\Desktop\\fs.cfg

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Answer 1

Late answer, I know, but for me the following script is quite useful - and it answers the question too, hitting two flys with one flag ;-)

The following script expands SendTo in the file explorer's context menu:

@echo off
cls
if "%~dp1"=="" goto Install

REM change drive, then cd to path given and run shell there
%~d1
cd "%~dp1"
cmd /k
goto End

:Install
rem No arguments: Copies itself into SendTo folder
copy "%0" "%appdata%\Microsoft\Windows\SendTo\A - Open in CMD shell.cmd"

:End

If you run this script without any parameters by double-clicking on it, it will copy itself to the SendTo folder and renaming it to "A - Open in CMD shell.cmd". Afterwards it is available in the "SentTo" context menu.

Then, right-click on any file or folder in Windows explorer and select "SendTo > A - Open in CMD shell.cmd"

The script will change drive and path to the path containing the file or folder you have selected and open a command shell with that path - useful for Visual Studio Code, because then you can just type "code ." to run it in the context of your project.

How does it work?

%0 - full path of the batch script
%~d1 - the drive contained in the first argument (e.g. "C:")
%~dp1 - the path contained in the first argument
cmd /k - opens a command shell which stays open

Not used here, but %~n1 is the file name of the first argument.

I hope this is helpful for someone.