Using FolderBrowserDialog in Powershell - returns wrong data

Question

I have seen the post "How to properly use the FolderBrowserDialog in Powershell"

I am having an issue getting just the path selected to return from the function.

At the end of the script I "write-host $a" but instead of getting just the directory I selected (C:\Temp) I get System.Windows.Forms, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089 C:\Temp

Function Get-Folder($initialDirectory)

{ [System.Reflection.Assembly]::LoadWithPartialName("System.windows.forms")

$foldername = New-Object System.Windows.Forms.FolderBrowserDialog
$foldername.rootfolder = "MyComputer"

if($foldername.ShowDialog() -eq "OK")
{
    $folder += $foldername.SelectedPath
}
return $folder

}

$a = Get-Folder Write-Host $a

I was told this is obsolete and to use Add-Type. Not getting just the path with the following script.

   Add-Type -AssemblyName System.Windows.Forms
   $FolderBrowser = New-Object System.Windows.Forms.FolderBrowserDialog
   [void]$FolderBrowser.ShowDialog()
   $FolderBrowser.SelectedPath
   Write-Host "FolderBrowser= "$FolderBrowser

You probably have `$folder` defined outside the scope of the function, and the function is modifying that. Close PowerShell and re-launch it. Also, change your `$folder +=` to just `$folder =`. — TheMadTechnician, Aug 05 '16 at 18:11

score 2 · Answer 1 · edited May 23 '17 at 11:51

2

You are getting that result because this line also produces output:

[System.Reflection.Assembly]::LoadWithPartialName("System.windows.forms")

PowerShell returns all output from a function so your results are actually an array containing the output from the loading of the assembly and the folder name.

Adding [void] in front of the assembly loading operation like this will omit that extra output and give you the results you are expecting:

[void][System.Reflection.Assembly]::LoadWithPartialName("System.windows.forms")

Or as Anthony Stringer mentions in the comments, you can use Add-Type instead which does not produce any output and would probably be the preferred method:

Add-Type -AssemblyName System.Windows.Forms

Also, TheMadTechnician is correct that you don't need the +=, just the = for the $folder variable.

This answer explains the behavior of returning output from a PowerShell function in more detail.

edited May 23 '17 at 11:51

Community

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1

answered Aug 05 '16 at 18:15

Jon Dechiro

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because of this (obsolete): https://msdn.microsoft.com/en-us/library/12xc5368%28v=vs.110%29.aspx?f=255&MSPPError=-2147217396 i recommend this: `Add-Type -AssemblyName System.Windows.Forms` – Anthony Stringer Aug 05 '16 at 18:20
OK, obsolete is not what i want. tried Add-Type and not getting the result i need. Add-Type -AssemblyName System.Windows.Forms $FolderBrowser = New-Object System.Windows.Forms.FolderBrowserDialog [void]$FolderBrowser.ShowDialog() $FolderBrowser.SelectedPath Write-Host "FolderBrowser= "$FolderBrowser – Shannon Aug 05 '16 at 18:33
Looks like you added [void] in front of $FolderBrowser.ShowDialog(). You don't want to do that because you just basically sent the results to null. You only need to use [void] or pipe something to Out-Null when you want to discard its output. – Jon Dechiro Aug 05 '16 at 18:46
Jon removing the void does not help. in the original script is did replace [void][System.Reflection.Assembly]::LoadWithPartialName("System.windows.forms") with Add-Type -AssemblyName System.Windows.Forms and that worked. Thanks rock star! – Shannon Aug 05 '16 at 18:49

Using FolderBrowserDialog in Powershell - returns wrong data

1 Answers1