Extract the filename from a path

99
votes

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)
10
powershellpowershell-2.0

10 Answers

187
votes

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
73
votes

Use :

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

27
votes

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

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

$filepath.BaseName

Basic-English-Grammar-1

$filepath.Name

Basic-English-Grammar-1.pdf
4
votes

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.

4
votes
$(Split-Path "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv" -leaf)
4
votes
Get-ChildItem "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv"
|Select-Object -ExpandProperty Name
3
votes

Just to complete the answer above that use .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 Extension:
$fileNameOnly = [System.IO.Path]::GetFileNameWithoutExtension("$FilePath")
Write-Host "fileNameOnly :" $fileNameOnly

# Get the Extension:
$fileExtensionOnly = [System.IO.Path]::GetExtension("$FilePath")
Write-Host "fileExtensionOnly :" $fileExtensionOnly
3
votes

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
2
votes

Find a file using wildcard and getting filename:

Resolve-Path "Package.1.0.191.*.zip" | Split-Path -leaf
-1
votes
$file = Get-Item -Path "c:/foo/foobar.txt"
$file.Name

Works with both relative an absolute paths