How to get the current directory of the cmdlet being executed

Question

This should be a simple task, but I have seen several attempts on how to get the path to the directory where the executed cmdlet is located with mixed success. For instance,

Answer 1

Path is often null. This function is safer.

function Get-ScriptDirectory
{
    $Invocation = (Get-Variable MyInvocation -Scope 1).Value;
    if($Invocation.PSScriptRoot)
    {
        $Invocation.PSScriptRoot;
    }
    Elseif($Invocation.MyCommand.Path)
    {
        Split-Path $Invocation.MyCommand.Path
    }
    else
    {
        $Invocation.InvocationName.Substring(0,$Invocation.InvocationName.LastIndexOf("\"));
    }
}

Answer 2

You can also use:

(Resolve-Path .\).Path

The part in brackets returns a PathInfo object.

(Available since PowerShell 2.0.)

Answer 3

Get-Location will return the current location:

$Currentlocation = Get-Location

Answer 4

Most answers don't work when debugging in the following IDEs:

• PS-ISE (PowerShell ISE)
• VS Code (Visual Studio Code)

Because in those the $PSScriptRoot is empty and Resolve-Path .\ (and similars) will result in incorrect paths.

Freakydinde's answer is the only one that resolves those situations, so I up-voted that, but I don't think the Set-Location in that answer is really what is desired. So I fixed that and made the code a little clearer:

$directorypath = if ($PSScriptRoot) { $PSScriptRoot } `
    elseif ($psise) { split-path $psise.CurrentFile.FullPath } `
    elseif ($psEditor) { split-path $psEditor.GetEditorContext().CurrentFile.Path }

Answer 5

I like the one-line solution :)

$scriptDir = Split-Path -Path $MyInvocation.MyCommand.Definition -Parent

Answer 6

If you just need the name of the current directory, you could do something like this:

((Get-Location) | Get-Item).Name

Assuming you are working from C:\Temp\Location\MyWorkingDirectory>

Output

MyWorkingDirectory