Question

i’m relative new to MVVM. My current problem is a modular dialog which should “autostart” at the beginning.

I’ve followed the example of WAFs Email Client for modular dialogs. Is it right that the only important thing is to set the Owner Property of the dialog to the instance of the main window of the application (and of course show the window with ShowDialog() instead of Show()?

If you close this dialog without configuration the application will shutdown. But now, if I open the main window in visual studios designer mode the configuration dialog comes up and if I close it visual studio crashes.
This is because I call the ShowDialog() of the configuration dialog in the constructor of my main windows view model.

To avoid this i can check for DesignerProperties.IsInDesignTool Property, but this is more a workaround as good code style, right?

Do you have any suggestions? Thanks.

Answer 1

The problem here is that you are showing a dialog in the constructor of a class. That’s something you don’t want to do.
I would solve it like this:
Don’t specify a StartupUri in your app.xaml but override OnStartup. There you check whether the configuration dialog should be shown or not. If it should be shown, show it and after it has closed with OK, show you main window.

Something like this:

override void OnStartup(...)
{
    if(configurationNotComplete)
    {
        ConfigDialog cfg = new ConfigDialog();
        if(!(cfg.ShowDialog() ?? false))
        {
            Shutdown();
            return;
        }
    }
    MainWindow window = new MainWindow();
    window.Show();
}

You have another problem with your current approach: Your ViewModel shows a modal dialog. This means it knows at least about one View: That of the modal dialog. MVVM is one way: The View knows about the ViewModel, the ViewModel knows about the Model. There should be no connection in the other direction.