Invoke-Command with dynamic function name

Question

I found this awesome post: Using Invoke-Command -ScriptBlock on a function with arguments

I'm trying to make the function call (${function:Foo}) dynamic, as in I want to pass the function name.

I tried this:

$name = "Foo"
Invoke-Command -ScriptBlock ${function:$name}

but that fails. I also tried various escape sequences, but just can't get the function name to be dynamic.

EDIT: For clarity I am adding a small test script. Of course the desired result is to call the ExternalFunction.

Function ExternalFunction()
{
  write-host "I was called externally"
}

Function InternalFunction()
{
    Param ([parameter(Mandatory=$true)][string]$FunctionName)
    #working: Invoke-Command -ScriptBlock ${function:ExternalFunction}
    #not working: Invoke-Command -ScriptBlock ${invoke-expression $FunctionName}
    if (Test-Path Function:\$FunctionName) {
    #working,but how to use it in ScriptBlock?
    }
}

InternalFunction -FunctionName "ExternalFunction"

mjolinor · Accepted Answer · 2013-02-09T19:56:37.350

6

Alternate solution:

function foo {'I am foo!'}

$name = 'foo'

$sb = (get-command $name -CommandType Function).ScriptBlock
invoke-command -scriptblock $sb

I am foo!

edited Feb 09 '13 at 19:56

answered Feb 09 '13 at 19:42

mjolinor

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I didn't mean that you should add a working example. I was happy that the example you provided worked! :-) Thanks – Dennis G Feb 09 '13 at 20:00
2

Just for future reference, this will actually fail if you use it with an advanced function that has a Begin, Process or End block. You'll get an exception about containing more than one clause. You can get around it by using the call operator instead of Invoke-Command though. So something like this: & $sb "Some Function Argument" – user2233949 Jun 28 '17 at 15:25

score 2 · Answer 2 · answered Feb 09 '13 at 15:52

2

as simple as :

invoke-expression  $name

or if you want to keep invoke-commande for remoting for example

Invoke-Command -ScriptBlock { invoke-expression  $name}

answered Feb 09 '13 at 15:52

Loïc MICHEL

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`invoke-expression` works outside a function. But when using it together with invoke command within a function I get a `"Cannot validate argument on parameter 'ScriptBlock'. The argument is null. Supply a non-null argument and try the"` – Dennis G Feb 09 '13 at 19:37
you have an unwanted $ before your scriptblock – Loïc MICHEL Feb 09 '13 at 19:54

Frode F. · Answer 3 · 2013-02-09T19:40:31.957

2

You could try the following. It tests if the name specified is a valid function before it attempts to run it:

$myfuncnamevar = "Foo"
Invoke-Command -ScriptBlock {
    param($name)
    if (Test-Path Function:\$name) { 
        #Function exists = run it
        & $name
    }
} -ArgumentList $myfuncnamevar

edited Feb 09 '13 at 19:40

answered Feb 09 '13 at 15:54

Frode F.

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Test-Path does work, but how would I call it within the Invoke-Command? – Dennis G Feb 09 '13 at 19:36
1

I put a line at the bottom to explain it. Updated with solution now. Remember that the function needs to exist on the remote computer. If this is a local funciton, you need to pass it in the scriptblock. – Frode F. Feb 09 '13 at 19:38

Invoke-Command with dynamic function name

3 Answers3