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 Answers

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

1

Use .NET:

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

3

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

0

Find a file using a wildcard and get the filename:

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

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

0

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

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

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

1

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.

2

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

Works with both relative and absolute paths

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