CODEWITHSUNDEEP
powershell
RipRapRob
RipRapRob

Reputation: 65

PowerShell: "The given path's format is not supported" when path does not contain folders. Using Split-Path

I have a function that returns a folder name with full path, like C:\Folder\Subfolder.

I then extract the name of the subfolder using Split-Path.

It works like expected when its folders or subfolders like C:\Folder or C:\Folder\Subfolder, but not when a drive is selected

Here's my code. The "Get-folder" function is just a file dialogue where the user can select a folder:

$FolderNavn = Get-Folder

$FolderNavnKort = Split-Path -Path $Foldernavn -Leaf

Here's what I'm expecting:

if the folder is 'C:\Folder\Subfolder' it should return "Subfolder" (works)

If the folder is just 'C:\Folder' it should return "Folder" (works)

If C:\ is selected (and thus not a folder) is should return "", but instead I get:

Out-File : The given path's format is not supported.

When I'm testing, $FolderNavn returns "C:\".

Any suggestions?

Edit, content of Get-Folder:

function Get-Folder {
[CmdletBinding()]
param (
    [Parameter(Mandatory=$false, ValueFromPipeline=$true, ValueFromPipelineByPropertyName=$true, Position=0)]
    [string]$Message = "Please select a directory.",

    [System.Environment+SpecialFolder]$InitialDirectory = [System.Environment+SpecialFolder]::MyComputer,

    [switch]$ShowNewFolderButton
)
Add-Type -AssemblyName System.Windows.Forms
$dialog = New-Object System.Windows.Forms.FolderBrowserDialog
$dialog.Description = 'Select a directory.'
$dialog.RootFolder  = $InitialDirectory
$dialog.ShowNewFolderButton = if ($ShowNewFolderButton) { $true } else { $false }
$selected = $null

# force the dialog TopMost
# Since the owning window will not be used after the dialog has been 
# closed we can just create a new form on the fly within the method call
$result = $dialog.ShowDialog((New-Object System.Windows.Forms.Form -Property @{TopMost = $true }))
if ($result -eq [Windows.Forms.DialogResult]::OK){
    $selected = $dialog.SelectedPath
}
# clear the FolderBrowserDialog from memory
$dialog.Dispose()
# return the selected folder
$selected

}

Upvotes: 0

Views: 1784

Answers (1)

Guy S
Guy S

Reputation: 472

Based on some research, it seems this is likely due to an extra colon in the file path. Without the code relevant to building the path / file ($KanalerStiNavn), my guess is that it's attempting to create a folder with the leaf subfolder you're retrieving earlier in this line.

$FolderNavnKort = Split-Path -Path $Foldernavn -Leaf

When you have a subfolder (such as "Subfolder", "Folder" as in your example) the returned value can be created as a folder in your desired directory in the Out-File command. Unfortunately, when (in the above line) your $foldernavn is the root, it returns "C:\" as the folder, and you try to create a folder in another directory including that

# Assuming the root export folder is D:\newfolder
# When $foldernavn is C:\folder (and therefore $FolderNavnKort is "folder")
$KanalerStiNavn could be "D:\newfolder\folder"

# When $foldernavn is C:\ (and therefore $FolderNavnKort is "C:\")
$KanalerStiNavn could be "D:\newfolder\C:\"

The second example is an invalid filename, hence the error.

To correct this you need to handle the case that the returned value of the Split-Path function contains a colon. Could be as simple as:

if ($FolderNavnKort -like "*:*")
{
    $FolderNavnKort = ""
}

Not overly beautiful, but should do the job.

Upvotes: 1

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