Using variables outside a function

Question

I am currently writing my first script in Powershell and I am already facing the first problem. I would like to read the value from a variable in a function so that I can use this variable in another cmd-let later. The problem now is that the variable is only recognized inside the function block and not outside. How do I get this to work?

Thanks for the help :-)

function Write-Log([string]$logtext, [int]$level=0)
{
  if($level -eq 0)
{
    $logtext = "[INFO] " + $logtext
    $text = "["+$logdate+"] - " + $logtext
    Write-Host $text
}
}

Send-MailMessage -To "<[email protected]>" -Subject "$text" -Body "The GPO backup creation was completed with the following status: `n $text" -SmtpServer "[email protected]" -From "[email protected]"

I would like to submit $text

Can you show us the code you currently have? Variables outside function scope are visible inside the function by default — Mathias R. Jessen
– Mathias R. Jessen, Commented Apr 4, 2020 at 15:17
Does this answer your question? Why won't Variable update? or Powershell "private" scope seems not useful at all — iRon
– iRon, Commented Apr 4, 2020 at 15:47
If I create the variable at the top with $script: or $global:, it still won't work :( — Julian05
– Julian05, Commented Apr 4, 2020 at 16:06

Mathias R. Jessen · Accepted Answer · 2020-04-04 16:11:41Z

This has to do with variable scoping behavior in PowerShell.

By default, all variables in the caller's scope is visible inside the function. So we can do:

function Print-X
{
  Write-Host $X
}

$X = 123
Print-X # prints 123
$X = 456 
Print-X # prints 456

So far, so good. But when we start writing to variables outside the function itself, PowerShell transparently creates a new variable inside the function's own scope:

function Print-X2
{
  Write-Host $X   # will resolve the value of `$X` from outside the function
  $X = 999        # This creates a new `$X`, different from the one outside
  Write-Host $X   # will resolve the value of the new `$X` that new exists inside the function
}

$X = 123
Print-X2       # Prints 123, and 999
Write-Host $X  # But the value of `$X` outside is still 123, unchanged

So, what to do? You could use a scope modifier to write to the variable outside the function, but the real solution here is to return the value from the function instead:

function Write-Log([string]$logtext, [int]$level=0, [switch]$PassThru = $true)
{
    if($level -eq 0)
    {
        $logtext = "[INFO] " + $logtext
        $text = "["+$logdate+"] - " + $logtext
        Write-Host $text
        if($PassThru){
            return $text
        }
    }
}

$logLine = Write-Log "Some log message" -PassThru

Send-MailMessage -Subject $logLine ...

Mohamed Shabeer kp · Accepted Answer · 2020-04-04 15:11:33Z

2

if you need to access a variable outside a function in Powershell you might need to use the global variable.

$global:myglobalvariable="This is a PowerShell global variable"

or if its a null

$global:myglobalvariable2 = $null

answered Apr 4, 2020 at 15:11

Mohamed Shabeer kp

8928 silver badges15 bronze badges

1 Comment

postanote Over a year ago

Using global is not prudent for this use case. Not saying it should not be used, just know when to use it. - sapien.com/blog/2013/03/06/… - Ideally, Items defined when PowerShell opens are set at the global scope. These items include system-created objects like PowerShell drives and also anything you have defined in a PowerShell profile since your profile runs at startup. The most common use of PowerShell global variables is to use PowerShell global variables between scripts, otherwise, the practice leans toward using a script. scope.

Peter L · Accepted Answer · 2024-03-08 21:07:03Z

0

The old array trick
I ended up using an array to get around this problem:

$lastRun = @($null) # array so function can edit
function Delay() {
    if ($lastRun[0] -ne $null) {
        ...
    }

    $lastRun[0] = Get-Date
}

answered Mar 8, 2024 at 21:07

Peter L

3,4691 gold badge35 silver badges34 bronze badges

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