is it possible to use a variable to reference a .NET static class in powershell?

Question

I'm doing a lot of System.IO.Path operations and I was curious if it was possible to store reference to that static class in a variable so it is shorter?

Instead of writing these long winded namespace.class paths:

[System.IO.Path]::Combine([System.IO.Path]::GetDirectoryName($targetFile), [System.IO.Path]::GetFileName("newfile_$targetFile"))

It would be great to write this:

$path = [System.IO.Path]
$path.Combine($path.GetDirectoryName($targetFile), $path.GetFileName("newfile_$targetFile"))

Is there a way to do this in powershell?

latkin · Accepted Answer · 2015-07-31 21:20:02Z

5

Yes. Your suggested code is close, you just need use the :: static invocation syntax:

$path = [System.IO.Path]
$path::Combine( ... )

answered Jul 31, 2015 at 21:20

latkin

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2 Comments

Martin Brandl Over a year ago

nice, didnt know that!

Thraka Over a year ago

Ahh awesome. I tried this but the ISE didn't give me any intellesense so I just assumed it was invalid. Thanks!

Martin Brandl · Accepted Answer · 2015-07-31 21:32:34Z

1

PowerShell has built-in cmdlets for some path operations:

PS C:\Windows\system32> get-command -Noun Path
CommandType Name         ModuleName                     
----------- ----         ----------                     
Cmdlet      Convert-Path Microsoft.Powershell.Management
Cmdlet      Join-Path    Microsoft.Powershell.Management
Cmdlet      Resolve-Path Microsoft.Powershell.Management
Cmdlet      Split-Path   Microsoft.Powershell.Management
Cmdlet      Test-Path    Microsoft.Powershell.Management

Your example implemented with native PowerShell cmdlets:

Join-Path (Split-Path $targetFile) (Split-Path $targetFile -Leaf)

edited Jul 31, 2015 at 21:32

answered Jul 31, 2015 at 21:13

Martin Brandl

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2 Comments

Thraka Over a year ago

I'm still interested in if you can or can't do this, but I adjusted my example to pull filename also. Perhaps you adjust your example?

Martin Brandl Over a year ago

I edited my answer. Split-Path with -Leaf parameter gives you the file name

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