Get-Content error: "Cannot bind argument to parameter 'Path' because it is null."

Question

I've found this function that I use to get the full path to a file:

Function Get-Filename($initialDirectory="")

{

    [System.Reflection.Assembly]::LoadWithPartialName(“System.windows.forms”) | Out-Null
 
    $OpenFileDialog = New-Object System.Windows.Forms.OpenFileDialog
    $OpenFileDialog.initialDirectory = $initialdirectory
    # $OpenFileDialog.filter = "TXT (*.txt)| *.txt"
    $OpenFileDialog.ShowHelp = $true
    $OpenFileDialog.ShowDialog() | out-null
    $OpenFileDialog.filename
  
    return $filnavn 
}

$InputFil = Get-FileName

$InputFil

$OurFilesData = Get-Content $InputFil -Encoding UTF8

I get this error message:

Get-Content : Cannot bind argument to parameter 'Path' because it is null.
At line:27 char:29
+ $OurFilesData = Get-Content $InputFil -Encoding UTF8
+                             ~~~~~~~~~
    + CategoryInfo          : InvalidData: (:) [Get-Content], ParameterBindingValidationException
    + FullyQualifiedErrorId : ParameterArgumentValidationErrorNullNotAllowed,Microsoft.PowerShell.Commands.GetContentCommand

What I know / has tried:

I'm running the code locally on my own machine
$InputFil returns the correct full path and filename (example: C:\Temp\E-Drevet_KANALNAVNE.CSV)
If it set the $InputFil variable manually ($InputFil = "C:\Temp\E-Drevet_KANALNAVNE.CSV") I don't get the same error
Power Shell Version 5.1.19041.546

Why do I get this error when the input seems to be the same as when I manually set the $InputFil variable?

The error in your post doesn't seem to match the title, can you double check and clarify? — jonrsharpe
– jonrsharpe, Commented Nov 10, 2020 at 14:33

Mathias R. Jessen · Accepted Answer · 2020-11-10 14:39:17Z

4

The error is thrown because one of the elements in $InputFil is $null.

Let's take a look at the last two statements in the function:

    $OpenFileDialog.filename
  
    return $filnavn

PowerShell outputs everything, not just whatever expression follows the return statement - so here, PowerShell first outputs the string value of $OpenFileDialog.filename, and then it subsequently outputs $filnavn - but $filnavn it never assigned to, so it resolve to $null.

The result is that $InputFil now holds an array, basically $InputFil = @("C:\actual\file\path.ext",$null)

Change your function definition to:

function Get-Filename($initialDirectory="")
{

    [System.Reflection.Assembly]::LoadWithPartialName(“System.windows.forms”) | Out-Null
 
    $OpenFileDialog = New-Object System.Windows.Forms.OpenFileDialog
    $OpenFileDialog.initialDirectory = $initialdirectory
    # $OpenFileDialog.filter = "TXT (*.txt)| *.txt"
    $OpenFileDialog.ShowHelp = $true

    if($OpenFileDialog.ShowDialog() -eq 'OK'){
      return $OpenFileDialog.filename
    }
}

answered Nov 10, 2020 at 14:39

Mathias R. Jessen

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3 Comments

RipRapRob Over a year ago

Worked like a champ. Thank you so much. I've upvoted you, but since I'm new it's not visible (?!?)

Mathias R. Jessen Over a year ago

@RobertGrøndahlWinther That's great, you're welcome! FWIW you need at least 15 rep to upvote, don't sweat it :)

notjustme Over a year ago

Upvote by proxy at your service. Have a great day. :)

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