Opening a Window Using Its Name

RogerSchlueter-7899 1,196

From my MainWindow I am trying to open another window knowing only its name. In my MainWindow ViewModel, I have:

Private Declare Function FindWindow Lib "user32" Alias "FindWindowA" (lpClassName As String, lpWindowName As String) As Long
Private Declare Function ShowWindow Lib "user32" (HWND As Long, ncmdShow As Long) As Boolean

Private ReadOnly Property _OpenWindow As New RelayCommand(Of Object)(AddressOf DoOpenWindow)
Public ReadOnly Property OpenWindow As RelayCommand(Of Object)
	Get
		Return _OpenWindow
	End Get
End Property
Private Sub DoOpenWindow(obj As Object)
	Dim WinName As String = CStr(obj)
	Dim XhWnd As Long = FindWindow(WinName, WinName)
	Debug.WriteLine(XhWnd)
	Dim x As Boolean = ShowWindow(XhWnd, 5)
	Debug.WriteLine(x)
End Sub

For testing purposes the Class, Name and Title of the window to be opened are all set to "NewWindow". The first Debug statement returns 0 so the function FindWindow is not working. Where is the error?

Additional questions:

Since this is a 64 bit application, is the use of 32 bit functions an issue?
More broadly, is there a better way to do this? I have been unable to find one.

Viorel 112.1K Reputation points

2024-03-26T10:10:12.4+00:00

If the caption is “NewWindow”, then try FindWindow(Nothing, "NewWindow"). Check if XhWnd is not zero.

Try IntPtr instead of Long. If ShowWindow does not work, then use 9 instead of 5 and call the SetForegroundWindow function.

However, there are other methods if the windows belong to your application.
RLWA32 40,286 Reputation points

2024-03-26T10:42:22.4966667+00:00

If the window you are searching for belongs to your application it is unlikely that its window class name is "NewWindow". WPF generates its own names for window classes. Following image from Spy++ shows the windows of a minimal WPF application. Note the window class names that begin with HwndWrappter -

WPF app details from Spy++ -

WPF app main window -
RogerSchlueter-7899 1,196 Reputation points

2024-03-26T15:10:21.77+00:00

@Viorel Using nothing still gives XhWnd = 0. Yes, NewWindow is part of my application so I am open (anxious?) for any other approach.
Viorel 112.1K Reputation points

2024-03-26T15:19:52.12+00:00
If it is part of your application and its class is called NewWindow, maybe you can do this:

Dim w = New NewWindow w.Show()
RogerSchlueter-7899 1,196 Reputation points

2024-03-26T15:20:02.52+00:00
@RLWA32 Some of the XAML for NewWindow:

<Window x:Class="NewWindow" x:Name="NewWindow" Title="NewWindow" ....> </Window>

Of course, I don't have access to any internal names that the compiler might use. The functions defined in the "user32" library use user-defined names, not internal ones.
RLWA32 40,286 Reputation points

2024-03-26T15:49:59.2066667+00:00

What you think of as class is not the same as the Windows API window class name. Look again at the output I posted from Spy++. Passing "NewWindow" as the class name to the Windows API function FindWindow makes it a certainty that the function will fail to find the target window.
RogerSchlueter-7899 1,196 Reputation points

2024-03-27T15:43:50.3233333+00:00

@Viorel While I am not likely to be inducted into the William Henry Gates Windows Programmers Hall of Fame, I am not that dense. The class is not called NewWindow, the Name (a string) of the Window is called NewWindow. Please review my OP to see why.
RogerSchlueter-7899 1,196 Reputation points

2024-03-27T15:59:21.6033333+00:00
@RLWA32 That blank Window that you showed: Add "x:Name="ABC123XYZ" to the XAML. Now, assuming that that is Window1, add the following lines to the MainWindow code-behind:

Private Sub MainWindow_Loaded(sender As Object, e As RoutedEventArgs) Handles Me.Loaded Dim win As New Window1 Debug.WriteLine(win.Name) End Sub

Tell me what you get.

That is the name I am talking about. Your introduction of names that the compiler uses are totally irrelevant to this discussion and a distraction to this thread. In fact, ALL names that a compiler generates are never of interest to the programmer because you do not know what they might be until after a compile takes place.
RLWA32 40,286 Reputation points

2024-03-27T17:08:01.53+00:00

You have totally missed the point that I'm trying to make. If you want to use the windows API FindWindowA function then the class name that you pass to the function must be the window class name that is recognized by the windows API. That class name parameter is documented as "The class name or a class atom created by a previous call to the RegisterClass or RegisterClassEx function." That is why I pointed out that what you view as the class name will never succeed if it is passed to that Windows API function. It's that simple.
RogerSchlueter-7899 1,196 Reputation points

2024-03-27T18:46:24.0733333+00:00

@RLWA32 I most certainly DO NOT want to call the windows API functions. There has to be a better way. But using them has just been one of the avenues I have tried.

KOZ6.0 4,810

Do you want to do something like this?

Private Sub DoOpenWindow(title As String)
    For Each window As Window In Application.Current.Windows
        If window.Title = title Then
            window.Activate()
            Exit For
        End If
    Next
End Sub

RogerSchlueter-7899 1,196 Reputation points

2024-03-28T01:17:58.98+00:00

@KOZ6.0 That doesn't work because only already instantiated windows are enumerated by that code.
RogerSchlueter-7899 1,196 Reputation points

2024-03-28T16:54:28.84+00:00

Because I did not know how to do what you show in your code snippet.

Thank you very much for that answer. That is precisely what I was looking for.
KOZ6.0 4,810 Reputation points

2024-03-28T20:11:10.8633333+00:00

I'm glad the problem was resolved. I have changed the comment to an answer, so please accept it.

1 answer

KOZ6.0 4,810

This is a sample of opening a window by specifying its class name.

Imports System.Reflection

Private Sub DoOpenWindow(className As String)
    For Each type In Assembly.GetExecutingAssembly().GetTypes()
        If type.IsSubclassOf(GetType(Window)) AndAlso type.Name = className Then
            Dim window = CType(Activator.CreateInstance(type), Window)
            window.Show()
            Exit For
        End If
    Next
End Sub