3

我目前正在开发一个办公室插件,我需要显示一个显示进度的通知对话框,我正在使用Philipp Sumi 的 wpf-notifyicon

我需要从一个单独的线程显示通知图标,因为我有很多代码已经在主线程上执行,这会导致 wpf-notifyicon 阻塞并等待,因为 Windows 消息队列中的消息没有被处理。

我知道我宁愿在单独的线程上执行这个耗时的代码,并从主线程显示通知图标并相应地更新它,但不幸的是,这不是一个替代方案,因为整个解决方案是单线程的。

例子:

    private FancyPopup fancyPopup;

    private void button1_Click(object sender, EventArgs e)
    {
        notifyIcon = new TaskbarIcon();
        notifyIcon.Icon = Resources.Led;

        fancyPopup = new FancyPopup();

        Thread showThread = new Thread(delegate()
        {
            notifyIcon.ShowCustomBalloon(fancyPopup, System.Windows.Controls.Primitives.PopupAnimation.Fade, null);
        });

        showThread.Start();
    }

    private void button2_Click(object sender, EventArgs e)
    {
        fancyPopup.TextB.Text = "Doing something...";

        //Keep the main thread busy.
        Thread.Sleep(5000);

        fancyPopup.TextB.Text = "Done doing something...";
    }

更新 我已经能够使用这个更新的代码进一步取得进展:

我正在新线程上创建 TaskbarIcon 对象,并使用 Application.Run 处理该线程上的应用程序消息循环...

    private FancyPopup fancyPopup;

    private void button1_Click(object sender, EventArgs e)
    {
        Thread showThread = new Thread(delegate()
        {
            notifyIcon = new TaskbarIcon();
            notifyIcon.Icon = Resources.Led;

            fancyPopup = new FancyPopup();


            notifyIcon.ShowCustomBalloon(fancyPopup, System.Windows.Controls.Primitives.PopupAnimation.Fade, null);

            System.Windows.Forms.Application.Run();
        });

        showThread.SetApartmentState(ApartmentState.STA);
        showThread.Start();
    }

    private void button2_Click(object sender, EventArgs e)
    {
        fancyPopup.Dispatcher.Invoke(new Action(delegate
        {
            fancyPopup.TextB.Text = "Doing something...";
        }));

        //Keep the main thread busy.
        Thread.Sleep(5000);

        fancyPopup.Dispatcher.Invoke(new Action(delegate
        {
            fancyPopup.TextB.Text = "Done doing something..."; 
        }));
    }
4

1 回答 1

3

我已经解决了我的问题,我必须在单独的 STA 线程上初始化notifyIcon并使用Application.Run以便开始在该线程上发送 windows 消息。

        var myThread = new Thread(delegate()
        {
            notifyIcon = new NotifyIcon();

            Application.Run();
        });

        myThread.SetApartmentState(ApartmentState.STA);
        myThread.Start();

然后我只需要调用通知对话框的 UI。

于 2012-05-24T06:24:17.263 回答