How to get the current directory of the cmdlet being executed
-
27-10-2019 - |
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, when I execute C:\temp\myscripts\mycmdlet.ps1
which has a settings file at C:\temp\myscripts\settings.xml
I would like to be able to store C:\temp\myscripts
in a variable within mycmdlet.ps1
.
This is one solution which works (although a bit cumbersome):
$invocation = (Get-Variable MyInvocation).Value
$directorypath = Split-Path $invocation.MyCommand.Path
$settingspath = $directorypath + '\settings.xml'
Another one suggested this solution which only works on our test environment:
$settingspath = '.\settings.xml'
I like the latter approach a lot and prefer it to having to parse the filepath as a parameter each time, but I can't get it to work on my development environment. What should I to do? Does it have something to do with how PowerShell is configured?
Solution
The reliable way to do this is just like you showed $MyInvocation.MyCommand.Path
.
Using relative paths will be based on $pwd, in PowerShell, the current directory for an application, or the current working directory for a .NET API.
PowerShell v3+:
Use the automatic variable $PSScriptRoot
.
OTHER TIPS
Yes, that should work. But if you need to see the absolute path, this is all you need:
(Get-Item -Path ".\").FullName
The easiest method seems to be to use the following predefined variable:
$PSScriptRoot
about_Automatic_Variables
and about_Scripts
both state:
In PowerShell 2.0, this variable is valid only in script modules (.psm1). Beginning in PowerShell 3.0, it is valid in all scripts.
I use it like this:
$MyFileName = "data.txt"
$filebase = Join-Path $PSScriptRoot $MyFileName
You can also use:
(Resolve-Path .\).Path
The part in brackets returns a PathInfo
object.
(Available since PowerShell 2.0.)
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("\"));
}
}
Try :
(Get-Location).path
or:
($pwd).path
Get-Location
will return the current location:
$Currentlocation = Get-Location
I like the one-line solution :)
$scriptDir = Split-Path -Path $MyInvocation.MyCommand.Definition -Parent
Try this:
$WorkingDir = Convert-Path .
In Powershell 3 and above you can simply use
$PSScriptRoot
You would think that using '.\' as the path means that it's the invocation path. But not all the time. Example, if you use it inside a job ScriptBlock. In which case, it might point to %profile%\Documents.
For what it's worth, to be a single-line solution, the below is a working solution for me.
$currFolderName = (Get-Location).Path.Substring((Get-Location).Path.LastIndexOf("\")+1)
The 1 at the end is to ignore the /
.
Thanks to the posts above using the Get-Location cmdlet.
To expand on @Cradle 's answer: you could also write a multi-purpose function that will get you the same result per the OP's question:
Function Get-AbsolutePath {
[CmdletBinding()]
Param(
[parameter(
Mandatory=$false,
ValueFromPipeline=$true
)]
[String]$relativePath=".\"
)
if (Test-Path -Path $relativePath) {
return (Get-Item -Path $relativePath).FullName -replace "\\$", ""
} else {
Write-Error -Message "'$relativePath' is not a valid path" -ErrorId 1 -ErrorAction Stop
}
}
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