c# - 我如何在同一个程序/应用程序中重新打开 WPF 窗口？

Question

各位

我想要一个可以多次控制窗口打开和关闭的应用程序。比如按1，打开一个窗口；按 2，关闭窗口。

我找到了我的代码，它只能执行一次。

有人可以帮我改正吗？非常感谢。

public static void Main(string[] args)
    {
        var appthread = new Thread(new ThreadStart(() =>
        {
            app = new testWpf.App();
            app.ShutdownMode = ShutdownMode.OnExplicitShutdown;
            app.Run();
        }));
        appthread.SetApartmentState(ApartmentState.STA);
        appthread.Start();

        while (true)
        {

            var key = Console.ReadKey().Key;
            // press 1 to open
            if (key == ConsoleKey.D1)
            {
                DispatchToApp(() => new Window().Show());
            }
            // Press 2 to exit
            if (key == ConsoleKey.D2)
            {
                DispatchToApp(() => app.Shutdown());
            }
        }
    }

    static void DispatchToApp(Action action)
    {
        app.Dispatcher.Invoke(action);
    }

Answer 1

当您通过调用 app.Shutdown() 关闭您的应用程序时，您需要先再次运行它以打开它的窗口。

您应该保存对窗口的引用并关闭它，而不是关闭整个应用程序。为此，您需要一个能够返回窗口引用的调度方法:

static void Main(string[] args)
{
    var appthread = new Thread(new ThreadStart(() =>
    {
        app = new testWpf.App();
        app.ShutdownMode = ShutdownMode.OnExplicitShutdown;
        app.Run();
    }));
    appthread.SetApartmentState(ApartmentState.STA);
    appthread.Start();

    Window win = new Window();
    while (true)
    {
        var key = Console.ReadKey().Key;
        // press 1 to open
        if (key == ConsoleKey.D1)
        {
            win = DispatchToApp<Window>(() =>
            {
                var myWindow = new Window();
                myWindow.Show();
                return myWindow;
            });
        }
        // Press 2 to exit
        if (key == ConsoleKey.D2)
        {
            DispatchToApp(() => win.Close());
        }
    }
}

static TReturn DispatchToApp<TReturn>(Func<TReturn> action)
{
    return app.Dispatcher.Invoke<TReturn>(action);
}

static void DispatchToApp(Action action)
{
    app.Dispatcher.Invoke(action);        
}