powershell - extract file name and extension

Question

I need to extract file name and extension from e.g. my.file.xlsx. I don't know the name of file or extension and there may be more dots in the name, so I need to search the string from the right and when I find first dot (or last from the left), extract the part on the right side and the part on the left side from that dot.

Maybe there is better solution, but I did'n find anything here or anywhere else. Thank you

Answer 1

If the file is coming off the disk and as others have stated, use the BaseName and Extension properties:

PS C:\> dir *.xlsx | select BaseName,Extension

BaseName                                Extension
--------                                ---------
StackOverflow.com Test Config           .xlsx  

If you are given the file name as part of string (say coming from a text file), I would use the GetFileNameWithoutExtension and GetExtension static methods from the System.IO.Path class:

PS C:\> [System.IO.Path]::GetFileNameWithoutExtension("Test Config.xlsx")
Test Config
PS H:\> [System.IO.Path]::GetExtension("Test Config.xlsx")
.xlsx

Answer 2

PS C:\Windows\System32\WindowsPowerShell\v1.0>split-path "H:\Documents\devops\tp-mkt-SPD-38.4.10.msi" -leaf
tp-mkt-SPD-38.4.10.msi

PS C:\Windows\System32\WindowsPowerShell\v1.0> $psversiontable

Name                           Value
----                           -----
CLRVersion                     2.0.50727.5477
BuildVersion                   6.1.7601.17514
PSVersion                      2.0
WSManStackVersion              2.0
PSCompatibleVersions           {1.0, 2.0}
SerializationVersion           1.1.0.1
PSRemotingProtocolVersion      2.1

Answer 3

If is from a text file and and presuming name file are surrounded by white spaces this is a way:

$a = get-content c:\myfile.txt

$b = $a | select-string -pattern "\s.+\..{3,4}\s" | select -ExpandProperty matches | select -ExpandProperty value

$b | % {"File name:{0} - Extension:{1}" -f $_.substring(0, $_.lastindexof('.')) , $_.substring($_.lastindexof('.'), ($_.length - $_.lastindexof('.'))) }

If is a file you can use something like this based on your needs:

$a = dir .\my.file.xlsx # or $a = get-item c:\my.file.xlsx 

$a
    Directory: Microsoft.PowerShell.Core\FileSystem::C:\ps


Mode           LastWriteTime       Length Name
----           -------------       ------ ----
-a---      25/01/10    11.51          624 my.file.xlsx


$a.BaseName
my.file
$a.Extension
.xlsx

Answer 4

Check the BaseName and Extension properties of the FileInfo object.

Answer 5

PS C:\Users\joshua> $file = New-Object System.IO.FileInfo('file.type')
PS C:\Users\joshua> $file.BaseName, $file.Extension
file
.type

Answer 6

Use Split-Path

$filePath = "C:\PS\Test.Documents\myTestFile.txt";
$fileName = (Split-Path -Path $filePath -Leaf).Split(".")[0];
$extension = (Split-Path -Path $filePath -Leaf).Split(".")[1];

Answer 7

just do it:

$file=Get-Item "C:\temp\file.htm"
$file.Name
$file.Extension

Answer 8

This is an adaptation, if anyone is curious. I needed to test whether RoboCopy successfully copied one file to multiple servers for its integrity:

   $Comp = get-content c:\myfile.txt

ForEach ($PC in $Comp) {
    dir "\\$PC\Folder\Share\*.*" | Select-Object $_.BaseName
}

Nice and simple, and it shows the directory and the file inside it. If you want to specify one file name or extension, just replace the *'s with whatever you want.

    Directory: \\SERVER\Folder\Share

Mode                LastWriteTime     Length Name                                                                                                                                             
----                -------------     ------ ----                                                                                                                                             
-a---         2/27/2015   5:33 PM    1458935 Test.pptx