Extract the filename from a path

问题

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)

回答1:

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

回答2:

Use .net:

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

回答3:

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

回答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.

回答5:

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

回答6:

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

回答7:

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

来源：https://stackoverflow.com/questions/35813186/extract-the-filename-from-a-path