Extract the filename from a path

Question

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

User · Answer

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

User · Answer

Use  net    System IO Path   GetFileName  c  foo txt   returns foo txt   System IO Path   GetFileNameWithoutExtension  c  foo txt   returns foo

User · Answer

Split-Path  D  Server User CUST MEA Data In Files CORRECTED CUST MEAFile csv  -leaf

User · Answer

file   Get-Item -Path  quot c  foo foobar txt quot   file Name  Works with both relative an absolute paths

User · Answer

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

User · Answer

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

User · Answer

Get-ChildItem  quot D  Server User CUST MEA Data In Files CORRECTED CUST MEAFile csv quot   Select-Object -ExpandProperty Name

User · Answer

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

User · Answer

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

User · Answer

Find a file using wildcard and getting filename   Resolve-Path  Package 1 0 191   zip    Split-Path -leaf

[powershell] Extract the filename from a path

Examples related to powershell

Examples related to powershell-2.0