WPF应用程序如何自定义启动？

Question

When a new WPF Application project is created, MainWindow.xaml , App.xaml and their corresponding code behind classes are automatically generated. In the App.xaml there is an attribute that defines which window is going to be run initially and by the default it's StartupUri="MainWindow.xaml"

I have created a new Dispatcher class in the same project. At startup, I want the instance of that class Dispatcher to be constructed and then one of its method to run. That method would actually create and show the MainWindow window. So how do I modify the App.xaml or App.xaml.cs in order to make it happen? Or, if it cannot be done by App , how should I implement it? Thanks.

Answer 1

You can remove the StartupUri attribute from the App.xaml.

Then, by creating an override for OnStartup() in the App.xaml.cs, you can create your new instance of your Dispatcher class.

Here's what my quick app.xaml.cs implementation looks like:

public partial class App : Application
{
    protected override void OnStartup(StartupEventArgs e)
    {
      base.OnStartup(e);

      new MyClassIWantToInstantiate();
    }
  }
}

Update

I recently discovered this workaround for a bug if you use this method to customize app startup and suddenly none of the Application-level resources can be found.

Answer 2

Try to use the Startup event (class Application) - MSDN .

You can show MainWindow in this event handler - after you create a Dispatcher instance.

Answer 3

1.In App.xaml, To replace the StartupUri with a subscription to the Startup event.

2. Use the event in App.xaml.cs.

For instance,

Startup="Application_Startup" in.xaml.

public partial class App : Application
{
    private void Application_Startup(object sender, StartupEventArgs e)
    {
        // Create the startup window
        MainWindow wnd = new MainWindow();
        // Do stuff here, e.g. to the window
        wnd.Title = "Something else";
        // Show the window
        wnd.Show();
    }
}