Get only filename from full path of a file

Question

I want to split the path and just save the file name test.xls in a new variable

$namearray = \"C:\\Users\\z003m\\Desktop\\Service_Tickets\\automatio         


        

Answer 1

You can also use the .Net implementation

[System.IO.Path] is 10 times faster than the split-path cmdlet

[System.IO.Path]::GetFileName('c:\myFile.txt')
# result myFile.txt

[System.IO.Path]::GetFileNameWithoutExtension('c:\myFile.txt') 
# result myFile

Performance comparision: 50.000 items

[System.IO.Path]::GetFileName(...)    Average: 12,84143 

Split-Path                            Average: 113,537884