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

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

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

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

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

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

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

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