Getting just the lowest-level directory name for a file from split-path using PowerShell

Question

I need to get just the last part of the path name for a file.

Example:

c:\dir1\dir2\dir3\file.txt

I need to get dir3 into a variable.

I have been trying with Split-Path, but it gives me the whole path.

Answer 1

This takes two invocations of Split-Path AFAICT:

PS> Split-Path (Split-Path c:\dir1\dir2\dir3\file.txt -Parent) -Leaf
dir3

Answer 2

This question is specifically asking for split-path it seems, but some other ways are:

If the file exists, I find it is much nicer to do:

(Get-Item c:\dir1\dir2\dir3\file.txt).Directory.Name

If the file does not exist, this won't work. Another way in that case is to use the .NET API, for example:

$path = [System.IO.Path];
$path::GetFileName($path::GetDirectoryName("c:\dir1\dir2\dir3\file.txt"))

Answer 3

If you want to keep it simple and the path is going to be in normal form, you can use String.Split():

"c:\dir1\dir2\dir3\file.txt".split("\")[-2]

Answer 4

Another option using System.Uri:

PS> ([uri]"c:\dir1\dir2\dir3\file.txt").segments[-2].trim('/')
dir3

And if the file exists on disk:

PS> (dir c:\dir1\dir2\dir3\file.txt).directory.name

Answer 5

In general, if you want the name of the directory you are in I used this (thanks to Shay Levi for the original idea):

PS> (dir).directory.name[0]