'Using Invoke-Command -ScriptBlock on a function with arguments

I'm writing a PowerShell script that will execute commands on a remote host using Invoke-Command and its -ScriptBlock parameter. For example,

function Foo {
    ...
    return "foo"
}
$rv = Invoke-Command --Credential $c --ComputerName $fqdn -ScriptBlock ${function:Foo}

This works fine. What I'd like to do now is the same thing, but call a function with local arguments. For example,

function Bar {
    param( [String] $a, [Int] $b )
    ...
    return "foo"
}
[String] $x = "abc"
[Int] $y = 123
$rv = Invoke-Command --Credential $c --ComputerName $fqdn -ScriptBlock ${function:Foo($x,$y)}

But this does not work:

Invoke-Command : Cannot validate argument on parameter 'ScriptBlock'. The argument is null. Supply a non-null argument and try the command again.

How can I use Invoke-Command with a -ScriptBlock that is a local function with arguments?

I realize that I can wrap the entire function and the parameters in a big code block, but that is not a clean way of doing it, in my opinion.



Solution 1:[1]

I think you want:

function Foo ( $a,$b) {
    $a
    $b
    return "foo"
}

$x = "abc"
$y= 123

Invoke-Command -Credential $c -ComputerName $fqdn -ScriptBlock ${function:Foo} -ArgumentList $x,$y

Solution 2:[2]

You can wrap the functions in a block and pass the block;

$a = {
  function foo{}
  foo($args)
}

$a.invoke() // Locally

$rv = Invoke-Command --Credential $c --ComputerName $fqdn -ScriptBlock $a //remotely

It's hardly elegant though.

Solution 3:[3]

This also works:

function foo
{
    param([string]$hosts, [string]$commands)
    $scriptblock = $executioncontext.invokecommand.NewScriptBlock($commands)
    $hosts.split(",") |% { Invoke-Command -Credential $cred -ComputerName $_.trim() -Scriptblock $scriptblock }
}

Solution 4:[4]

$Log = "PowerShellCore/Operational"

Invoke-Command -ComputerName Server01 -ScriptBlock {Get-WinEvent -LogName $Using:Log -MaxEvents 10}\

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 manojlds
Solution 2 reconbot
Solution 3 8DH
Solution 4 levymtmr