.bat current folder name

Question

In my bat script, I\'m calling another script and passing it a string parameter

cscript log.vbs \"triggered from folder  by Eric\"
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Answer 1

If you want the directory where you're currently at, you can get that from %cd%. That's your current working directory.

If you're going to be changing your current working directory during the script execution, just save it at the start:

set startdir=%cd%

then you can use %startdir% in your code regardless of any changes later on (which affect %cd%).

---

If you just want to get the last component of that path (as per your comment), you can use the following as a baseline:

    @setlocal enableextensions enabledelayedexpansion
    @echo off
    set startdir=%cd%
    set temp=%startdir%
    set folder=
:loop
    if not "x%temp:~-1%"=="x\" (
        set folder=!temp:~-1!!folder!
        set temp=!temp:~0,-1!
        goto :loop
    )
    echo.startdir = %startdir%
    echo.folder   = %folder%
    endlocal && set folder=%folder%

This outputs:

    C:\Documents and Settings\Pax> testprog.cmd
    startdir = C:\Documents and Settings\Pax
    folder   = Pax

It works by copying the characters from the end of the full path, one at a time, until it finds the \ separator. It's neither pretty nor efficient, but Windows batch programming rarely is :-)

EDIT

Actually, there is a simple and very efficient method to get the last component name.

for %%F in ("%cd%") do set "folder=%~nxF"

Not an issue for this situation, but if you are dealing with a variable containing a path that may or may not end with \, then you can guarantee the correct result by appending \.

for %%F in ("%pathVar%\.") do set "folder=%~nxF"