How to ensure Forms are the top window - forms

Question

I have a PS script that implements System.Windows.Forms in order to query technicians for some data.
I create the forms and set both .Topmost and .TopLevel to true in an attempt to have them show up over the Powershell window, but they continue to (for some reason inconsistently) appear behind the Powershell window. This slows down the process and is confusing in its inconsistency.
If anyone knows how to ensure these windows stay top without a mountain of code larger than the script itself that would be incredibly useful. I'll include the code I use to build one of the basic forms below.
Any simple solution that will allow these Forms to appear over the Powershell window is appreciated. It could even just minimize the PS window, but I don't want to launch without the window as we need it open. Thanks.
$form.Text = 'Computer Name Entry'
$form.Size = New-Object System.Drawing.Size(550,400)
$form.StartPosition = 'CenterScreen'
$okButton = New-Object System.Windows.Forms.Button
$okButton.Location = New-Object System.Drawing.Point(75,300)
$okButton.Size = New-Object System.Drawing.Size(75,23)
$okButton.Text = 'OK'
$okButton.DialogResult = [System.Windows.Forms.DialogResult]::OK
$form.AcceptButton = $okButton
$form.Controls.Add($okButton)
$label = New-Object System.Windows.Forms.Label
$label.Location = New-Object System.Drawing.Point(10,20)
$label.Size = New-Object System.Drawing.Size(400,40)
$label.Text = 'Text is here:'
$form.Controls.Add($label)
$textBox = New-Object System.Windows.Forms.TextBox
$textBox.Location = New-Object System.Drawing.Point(10,70)
$textBox.Size = New-Object System.Drawing.Size(400,20)
$form.Controls.Add($textBox)
$form.Topmost = $true
$form.TopLevel = $true
$form.Add_Shown({$textBox.Select()})
$result = $form.ShowDialog()
if ($result -eq [System.Windows.Forms.DialogResult]::OK)
{
Do-Stuff
}

Answer 1

The easiest way I know of forcing the form to be topmost is to open it with a new temporary form that is TopMost as parameter for ShowDialog().
First, from your code remove the lines $form.Topmost = $true and $form.TopLevel = $true
Next, show your form like this:
# force the dialog TopMost by creating a temporary parent window for this form
$result = $form.ShowDialog((New-Object System.Windows.Forms.Form -Property #{TopMost = $true }))
Another way of doing this is to use a piece of C# to return a windowhandle which implements the IWin32Window interface.
Then use this handle as the owner window for this form in the .ShowDialog() method of the form.
For this method, also remove the lines $form.Topmost = $true and $form.TopLevel = $true from your original code.
$iWin32Code = #"
using System;
using System.Windows.Forms;
public class Win32Window : IWin32Window {
public Win32Window(IntPtr handle) {
Handle = handle;
}
public IntPtr Handle { get; private set; }
}
"#
if (-not ([System.Management.Automation.PSTypeName]'Win32Window').Type) {
Add-Type -TypeDefinition $iWin32Code -ReferencedAssemblies System.Windows.Forms.dll
}
Now, using that code, create a handle for the currently running PowerShell process
# get the owner handle from this PowerShell process
$ownerHandle = New-Object Win32Window -ArgumentList ([System.Diagnostics.Process]::GetCurrentProcess().MainWindowHandle)
# and use that in the ShowDialog method as argument
$result = $form.ShowDialog($ownerHandle)
P.S. do not forget to clear your form from memory after you are done with it by calling
$form.Dispose()