如何防止新 STA 线程上的非模式窗口关闭

发布于 2024-11-11 16:23:34 字数 516 浏览 0 评论 0原文

我想打开一些非模型窗口 (WPF)，但此时我处于非 STA 线程上。所以我开始一个新线程并在那里打开它们。但一旦打开，它们就会再次关闭。（顺便说一句。这些窗口的行为应该独立于主窗口。因此没有设置所有者属性）

private void SomeMethod_OnA_NON_STA_Thread()
{
    // some other work here

    Thread ANewThread = new Thread(OpenSomeWindows);
    ANewThread.SetApartmentState(ApartmentState.STA);
    ANewThread.Start();
}


private void OpenSomeWindows()
{
    TestWindow T;

    for (int i = 0; i < 3; i++)
    {
        T = new TestWindow();
        T.Show();
    }
}

我在这里做错了什么？

原文

I want to open some non model windows (WPF) but at the point that this has to happen I am on a non STA thread. So I start a new thread and open them on there. But as soon as the are opened they close again. (By the way. the behaviour of these windows should be independent from the mainwindow. So no owner property is set)

private void SomeMethod_OnA_NON_STA_Thread()
{
    // some other work here

    Thread ANewThread = new Thread(OpenSomeWindows);
    ANewThread.SetApartmentState(ApartmentState.STA);
    ANewThread.Start();
}


private void OpenSomeWindows()
{
    TestWindow T;

    for (int i = 0; i < 3; i++)
    {
        T = new TestWindow();
        T.Show();
    }
}

What am I doing wrong here?

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开始看清了 2024-11-18 16:23:34

如果你希望你的窗口存活，你必须在创建它们后启动消息循环（否则你的线程就会退出，窗口没有机会渲染自己）：（

private void OpenSomeWindows()
{
    for (int i = 0; i < 3; i++)
    {
        TestWindow T = new TestWindow();
        T.Show();
    }
    Dispatcher.Run(); // <---------
}

在主线程中，消息循环是由框架创建的为你。）

PS：我不确定窗口是否可以被垃圾收集，所以我会在某个地方保留对它们的引用：

List<TestWindow> windows = new List<TestWindow>();
for (int i = 0; i < 3; i++)
{
    TestWindow t = new TestWindow();
    t.Show();
    windows.Add(t);
}
Dispatcher.Run();

PPS：也许你希望你的窗口在主线程中运行？其实你可以这样做。您只需要以下内容：

private void SomeMethod_OnA_NON_STA_Thread()
{
    // some other work here
    Application.Current.Dispatcher.Invoke(OpenSomeWindows);
}

private void OpenSomeWindows()
{
    for (int i = 0; i < 3; i++)
    {
        TestWindow T = new TestWindow();
        T.Show();
    }
    // this way, no Dispatcher.Run is needed
}

If you want your windows to live, you have to start the message loop after you created them (otherwise your thread just exits, and the windows have no chance to render themselves):

private void OpenSomeWindows()
{
    for (int i = 0; i < 3; i++)
    {
        TestWindow T = new TestWindow();
        T.Show();
    }
    Dispatcher.Run(); // <---------
}

(In main thread, the message loop is created by the framework for you.)

P.S.: I am not sure whether the windows can be garbage collected, so I would keep references to them somewhere:

List<TestWindow> windows = new List<TestWindow>();
for (int i = 0; i < 3; i++)
{
    TestWindow t = new TestWindow();
    t.Show();
    windows.Add(t);
}
Dispatcher.Run();

P.P.S.: Maybe you want your windows to run in the main thread? Actually you can do this. You need just the following:

private void SomeMethod_OnA_NON_STA_Thread()
{
    // some other work here
    Application.Current.Dispatcher.Invoke(OpenSomeWindows);
}

private void OpenSomeWindows()
{
    for (int i = 0; i < 3; i++)
    {
        TestWindow T = new TestWindow();
        T.Show();
    }
    // this way, no Dispatcher.Run is needed
}

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