How can I get the current PowerShell executing file?

Question

Note: PowerShell 1.0
I'd like to get the current executing PowerShell file name. That is, if I start my session like this:

powershell.exe .\myfile.ps1

I'd like to get the string ".\myfile.ps1" (or something like that). EDIT: "myfile.ps1" is preferable.
Any ideas?

Answer 1

I've tried to summarize the various answers here, updated for PowerShell 5:

• If you're only using PowerShell 3 or higher, use $PSCommandPath

• If want compatibility with older versions, insert the shim:

  if ($PSCommandPath -eq $null) { function GetPSCommandPath() { return $MyInvocation.PSCommandPath; } $PSCommandPath = GetPSCommandPath; }

  This adds $PSCommandPath if it doesn't already exist.

  The shim code can be executed anywhere (top-level or inside a function), though $PSCommandPath variable is subject to normal scoping rules (eg, if you put the shim in a function, the variable is scoped to that function only).

Details

There's 4 different methods used in various answers, so I wrote this script to demonstrate each (plus $PSCommandPath):

function PSCommandPath() { return $PSCommandPath; }
function ScriptName() { return $MyInvocation.ScriptName; }
function MyCommandName() { return $MyInvocation.MyCommand.Name; }
function MyCommandDefinition() {
    # Begin of MyCommandDefinition()
    # Note: ouput of this script shows the contents of this function, not the execution result
    return $MyInvocation.MyCommand.Definition;
    # End of MyCommandDefinition()
}
function MyInvocationPSCommandPath() { return $MyInvocation.PSCommandPath; }

Write-Host "";
Write-Host "PSVersion: $($PSVersionTable.PSVersion)";
Write-Host "";
Write-Host "`$PSCommandPath:";
Write-Host " *   Direct: $PSCommandPath";
Write-Host " * Function: $(PSCommandPath)";
Write-Host "";
Write-Host "`$MyInvocation.ScriptName:";
Write-Host " *   Direct: $($MyInvocation.ScriptName)";
Write-Host " * Function: $(ScriptName)";
Write-Host "";
Write-Host "`$MyInvocation.MyCommand.Name:";
Write-Host " *   Direct: $($MyInvocation.MyCommand.Name)";
Write-Host " * Function: $(MyCommandName)";
Write-Host "";
Write-Host "`$MyInvocation.MyCommand.Definition:";
Write-Host " *   Direct: $($MyInvocation.MyCommand.Definition)";
Write-Host " * Function: $(MyCommandDefinition)";
Write-Host "";
Write-Host "`$MyInvocation.PSCommandPath:";
Write-Host " *   Direct: $($MyInvocation.PSCommandPath)";
Write-Host " * Function: $(MyInvocationPSCommandPath)";
Write-Host "";

Output:

PS C:\> .\Test\test.ps1

PSVersion: 5.1.19035.1

$PSCommandPath:
 *   Direct: C:\Test\test.ps1
 * Function: C:\Test\test.ps1

$MyInvocation.ScriptName:
 *   Direct:
 * Function: C:\Test\test.ps1

$MyInvocation.MyCommand.Name:
 *   Direct: test.ps1
 * Function: MyCommandName

$MyInvocation.MyCommand.Definition:
 *   Direct: C:\Test\test.ps1
 * Function:
    # Begin of MyCommandDefinition()
    # Note this is the contents of the MyCommandDefinition() function, not the execution results
    return $MyInvocation.MyCommand.Definition;
    # End of MyCommandDefinition()


$MyInvocation.PSCommandPath:
 *   Direct:
 * Function: C:\Test\test.ps1

Notes:

• Executed from C:\, but actual script is C:\Test\test.ps1.
• No method tells you the passed invocation path (.\Test\test.ps1)
• $PSCommandPath is the only reliable way, but was introduced in PowerShell 3
• For versions prior to 3, no single method works both inside and outside of a function

Answer 2

While the current Answer is right in most cases, there are certain situations that it will not give you the correct answer. If you use inside your script functions then:

$MyInvocation.MyCommand.Name 

Returns the name of the function instead name of the name of the script.

function test {
    $MyInvocation.MyCommand.Name
}

Will give you "test" no matter how your script is named. The right command for getting the script name is always

$MyInvocation.ScriptName

this returns the full path of the script you are executing. If you need just the script filename than this code should help you:

split-path $MyInvocation.PSCommandPath -Leaf

Answer 3

If you only want the filename (not the full path) use this:

$ScriptName = $MyInvocation.MyCommand.Name