Extract the Filename from a Path

Use .NET:

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

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

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)

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

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.

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

Works with both relative and absolute paths