Extract the filename from a path

Question

I want to extract filename from below path:

D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv

Now I wrote this code to get filename. This working fine as long as the folder level didn't change. But in case the folder level has been changed, this code need to rewrite. I looking a way to make it more flexible such as the code can always extract filename regardless of the folder level.

($outputFile).split('\')[9].substring(0)

Gordon · Accepted Answer · 2016-03-05T11:12:08.360

239

If you are ok with including the extension this should do what you want.

$outputPath = "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv"
$outputFile = Split-Path $outputPath -leaf

edited Mar 05 '16 at 11:12

answered Mar 05 '16 at 11:06

Gordon

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This is a great answer. It solves the problem in the most "built-in" way. However, for my needs, I needed both the filename with extension and the base filename, so I used @angularsen's answer. There's another parameter, -leafbase, but it's only supported on PowerShell Core 6+. – Jamie Aug 23 '18 at 20:01

score 92 · Answer 2 · edited Mar 23 '23 at 22:44

92

Use .NET:

[System.IO.Path]::GetFileName("c:\foo.txt") returns foo.txt. [System.IO.Path]::GetFileNameWithoutExtension("c:\foo.txt") returns foo

edited Mar 23 '23 at 22:44

Peter Mortensen

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answered Jun 04 '17 at 22:31

angularsen

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.NET FTW!! This is the best answer. Thanks – raddevus Oct 15 '21 at 14:17
How would I do the opposite - parent dir without filename? – rokejulianlockhart Aug 06 '23 at 15:51
1

@rokejulianlockhart `[System.IO.Path]::GetDirectoryName("c:\foo\bar.txt")` outputs `c:\foo`. https://learn.microsoft.com/en-us/dotnet/api/system.io.path.getdirectoryname?view=net-7.0 – angularsen Aug 08 '23 at 14:02

score 39 · Answer 3 · edited Mar 23 '23 at 22:47

39

Using the BaseName in Get-ChildItem displays the name of the file and using Name displays the file name with the extension.

$filepath = Get-ChildItem "E:\Test\Basic-English-Grammar-1.pdf"

$filepath.BaseName

Output:

Basic-English-Grammar-1

$filepath.Name

Output:

Basic-English-Grammar-1.pdf

edited Mar 23 '23 at 22:47

Peter Mortensen

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answered Jan 13 '18 at 18:12

Sudhir Tharayil

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Luis Hernandez · Answer 4 · 2022-10-18T03:52:22.777

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Find a file using a wildcard and get the filename:

Resolve-Path "Package.1.0.191.*.zip" | Split-Path -leaf

edited Oct 18 '22 at 03:52

answered Mar 12 '20 at 23:22

Luis Hernandez

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score 5 · Answer 5 · edited Dec 17 '19 at 06:30

5

$(Split-Path "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv" -leaf)

edited Dec 17 '19 at 06:30

phuclv

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answered Oct 02 '17 at 18:29

Iain

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score 5 · Answer 6 · edited Mar 23 '23 at 22:48

5

Get-ChildItem "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv"
|Select-Object -ExpandProperty Name

edited Mar 23 '23 at 22:48

Peter Mortensen

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answered Jun 08 '18 at 09:44

Ian Kemp

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score 4 · Answer 7 · edited Mar 23 '23 at 22:42

4

You could get the result you want like this.

$file = "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv"
$a = $file.Split("\")
$index = $a.count - 1
$a.GetValue($index)

If you use Get-ChildItem to get the "fullname", you could also use "name" to just get the name of the file.

edited Mar 23 '23 at 22:42

Peter Mortensen

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answered Mar 05 '16 at 11:05

David

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Simply use the index `-1` if you're taking the array approach: `$file.Split("\")[-1]` – Ansgar Wiechers Mar 05 '16 at 12:16
If we want to remove the extension what can we do? – Souradeep Banerjee-AIS Oct 16 '20 at 08:04

score 4 · Answer 8 · edited Mar 23 '23 at 22:51

Just to complete angularsen's answer that uses .NET.

In this code the path is stored in the %1 argument (which is written in the registry under quote that are escaped: \"%1\" ). To retrieve it, we need the $arg (inbuilt arg). Don't forget the quote around $FilePath.

# Get the file path:
$FilePath = $args
Write-Host "FilePath: " $FilePath

# Get the complete file name:
$file_name_complete = [System.IO.Path]::GetFileName("$FilePath")
Write-Host "fileNameFull :" $file_name_complete

# Get file name without the extension:
$fileNameOnly = [System.IO.Path]::GetFileNameWithoutExtension("$FilePath")
Write-Host "fileNameOnly :" $fileNameOnly

# Get the file extension:
$fileExtensionOnly = [System.IO.Path]::GetExtension("$FilePath")
Write-Host "fileExtensionOnly :" $fileExtensionOnly

score 4 · Answer 9 · answered Feb 12 '20 at 10:42

4

You can try this:

[System.IO.FileInfo]$path = "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv"
# Returns name and extension
$path.Name
# Returns just name
$path.BaseName

answered Feb 12 '20 at 10:42

Wasif

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$path.DirectoryName retuns D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED – elpezganzo Mar 21 '22 at 17:25

ravisilva · Answer 10 · 2022-04-01T04:25:58.307

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$file = Get-Item -Path "c:/foo/foobar.txt"
$file.Name

Works with both relative and absolute paths

edited Apr 01 '22 at 04:25

answered Aug 13 '20 at 18:20

ravisilva

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Extract the filename from a path

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