How do I get the directory of a file path in PowerShell?

Run Split-Path -Path $path -Parent. -Parent is the default, so Split-Path $path works too. The Split-Path documentation lists every switch.

How do I get just the folder name from a path?

Use Split-Path (Split-Path $path) -Leaf, or (Get-Item $path).Directory.Name when the file exists. Both return only the last folder, such as Q4.

What is the difference between Split-Path and GetDirectoryName?

Both return the folder as a string. Split-Path ignores a trailing backslash and returns an empty string at the root; GetDirectoryName keeps the last folder and returns $null at the root.

Does Split-Path work if the file doesn't exist?

Yes. Split-Path works on the text only. Add -Resolve, or use Get-Item, when you want PowerShell to confirm the item exists; both fail on a missing path.

How do I get the folder of the current script?

Use $PSScriptRoot inside the script. It holds the script's parent folder, as described in about_Automatic_Variables . It's empty when you type it at the prompt.

Extract Directory from File Path in PowerShell (7 and 5.1)

To extract the directory from a file path in PowerShell, pass the path to Split-Path. Its default -Parent switch drops the file name and returns the folder that holds it:

Split-Path -Path 'C:\Shared\Finance\Invoice-1042.pdf' -Parent    # C:\Shared\Finance

Below I’ll show the folder name on its own, parent folders, Get-Item, the .NET method, relative and network paths, the running script’s folder, and where Windows PowerShell 5.1 answers differently.

This Tutorial Covers:

Set up a sample folder to test file paths

Most examples run against real files, so let’s create a few. Open PowerShell 7 (pwsh) or Windows PowerShell, paste this script and press Enter:

$root = 'C:\psfaqs\t\extractdir'
$null = New-Item "$root\Finance\2026\Q4", "$root\Scripts" -ItemType Directory -Force
'Invoice 1042'      | Set-Content "$root\Finance\2026\Q4\Invoice-1042.pdf"
'Vendor,City,State' | Set-Content "$root\Finance\2026\Q4\Vendor-List.csv"
'Item,Amount'       | Set-Content "$root\Finance\2026\Budget-2026.xlsx"

Get-ChildItem $root -Recurse -File | Sort-Object FullName |
    ForEach-Object { $_.FullName.Substring($root.Length + 1) }

Output:

Finance\2026\Budget-2026.xlsx
Finance\2026\Q4\Invoice-1042.pdf
Finance\2026\Q4\Vendor-List.csv

You now have a Finance folder with a 2026 subfolder and a Q4 folder inside it. Change $root to any folder you like; the examples below use the same layout.

This post works backward from a full path you already have. If you still need that path, start with how to get the full path of a file first.

Get the directory of a file with Split-Path -Parent

Split-Path cuts a path string into parts. With -Parent, or with no switch at all, it returns everything before the last backslash:

$file   = 'C:\psfaqs\t\extractdir\Finance\2026\Q4\Invoice-1042.pdf'
$folder = Split-Path -Path $file -Parent

"Folder:  $folder"
"Default: $(Split-Path $file)"
"Type:    $($folder.GetType().Name)"

Output:

Folder:  C:\psfaqs\t\extractdir\Finance\2026\Q4
Default: C:\psfaqs\t\extractdir\Finance\2026\Q4
Type:    String

PowerShell Split-Path -Parent returning the folder of a file path — Split-Path returns the folder of Invoice-1042.pdf, with or without -Parent (PowerShell 7)

Both lines print the same folder because -Parent is the default. The result is a plain String, not a folder object, so you can join it, compare it or pass it to any cmdlet.

Split-Path also takes paths from the pipeline, and it doesn’t care whether they exist. Here are three at once, including a bare file name:

$paths = 'C:\Shared\Finance\Invoice-1042.pdf',
         'D:\Data\Denver\Sales-2026.csv',
         'Report.docx'
$paths | Split-Path | ForEach-Object { "[$_]" }

Output:

[C:\Shared\Finance]
[D:\Data\Denver]
[]

The bare file name returned an empty string, because it has no folder part. Need the other half of the path? Here’s how to get the file name from a path with -Leaf.