Extract the filename from a path
PowershellPowershell 2.0Powershell Problem Overview
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)
Powershell Solutions
Solution 1 - Powershell
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
Solution 2 - Powershell
Use [tag:.NET]:
[System.IO.Path]::GetFileName("c:\foo.txt") returns foo.txt.
[System.IO.Path]::GetFileNameWithoutExtension("c:\foo.txt") returns foo
Solution 3 - Powershell
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
Solution 4 - Powershell
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.
Solution 5 - Powershell
$(Split-Path "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv" -leaf)
Solution 6 - Powershell
Get-ChildItem "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv"
|Select-Object -ExpandProperty Name
Solution 7 - Powershell
You can try this:
[System.IO.FileInfo]$path = "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv"
# Returns name and extension
$path.Name
# Returns just name
$path.BaseName
Solution 8 - Powershell
Find a file using a wildcard and getting the filename:
Resolve-Path "Package.1.0.191.*.zip" | Split-Path -leaf
Solution 9 - Powershell
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
Solution 10 - Powershell
$file = Get-Item -Path "c:/foo/foobar.txt"
$file.Name
Works with both relative and absolute paths