Powershell - String Manipulation - getting dir from full file-path - string

Question

Keep in mind I'm new to this and be gentle.
I have a full file path for a document "C:\folder1\folder2\01.03.2017 - FileName.csv" and I want to manipulate it to return the dir that the file is stored in (C:\folder1\folder2), minus the filename (01.03.2017 - FileName.csv).
I'm trying to make this modular so that it will work regardless of the amount of sub-folders a file sits in; we also won't know the FileName in advance, so again this needs to be modular and remove up to and including the last "\"
For background info on how this is currently built, I nicked a bit of code from a previous question I saw on StackOverflow:
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 = “All files (*.*)| *.*”
$OpenFileDialog.ShowDialog() | Out-Null
$OpenFileDialog.filename
} #end function Get-FileName
# *** Entry Point to Script ***
$originalData = Get-FileName -initialDirectory “c:\” | Out-String
Write-Host $originalData
$originalDir = $originalData.Split('\')
$originalDir
Running this currently prompts for an "open dialog box" you would see in Windows. You select a folder and the output is currently:
C:\folder1\folder2\01.03.2017 - FileName.csv
C:
folder1
folder2
01.03.2017 - FileName.csv
I've tried a few different -join attempts but none successful.
We will have an input of C:\folder1\folder2\01.03.2017 - FileName.csv as a variable $originalData.
We want the output to be C:\folder1\folder2 as a variable $originalDir.

Answer 1

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 = “All files (*.*)| *.*”
$OpenFileDialog.ShowDialog() | Out-Null
$OpenFileDialog.filename
} #end function Get-FileName
$originalData = Get-FileName -initialDirectory “c:\”
Write-Host $originalData
$originalDir = (Get-ChildItem $originalData).DirectoryName
you can use it like so, so take result or your function and use it with get-childitem.
edit: notice there's no | Out-String on the third to last line