myWindow.Activate();

试图将窗口带到前台并激活它。

这应该做的把戏，除非我误解了，你想要永远在顶端的行为。在这种情况下，你需要:

myWindow.TopMost = true;

我知道这是一个很晚的答案，也许对研究人员有帮助

 if (!WindowName.IsVisible)
 {
     WindowName.Show();
     WindowName.Activate();
 }

我已经找到了一个解决方案，将窗口带到顶部，但它表现为一个正常的窗口:

if (!Window.IsVisible)
{
    Window.Show();
}

if (Window.WindowState == WindowState.Minimized)
{
    Window.WindowState = WindowState.Normal;
}

Window.Activate();
Window.Topmost = true;  // important
Window.Topmost = false; // important
Window.Focus();         // important

如果用户正在与另一个应用程序交互，则可能无法将您的应用程序放在前面。作为一般规则，只有当进程已经是前台进程时，该进程才能期望设置前台窗口。(微软在SetForegroundWindow() MSDN条目中记录了限制。)这是因为:

The user "owns" the foreground. For example, it would be extremely annoying if another program stole the foreground while the user is typing, at the very least interrupting her workflow, and possibly causing unintended consequences as her keystrokes meant for one application are misinterpreted by the offender until she notices the change. Imagine that each of two programs checks to see if its window is the foreground and attempts to set it to the foreground if it is not. As soon as the second program is running, the computer is rendered useless as the foreground bounces between the two at every task switch.

我想到了一个变通办法。我从一个用于实现热键的键盘钩子进行调用。如果我把它放入一个暂停的BackgroundWorker中，调用就会像预期的那样工作。这是一个拼凑，但我不知道为什么它不工作最初。

void hotkey_execute()
{
    IntPtr handle = new WindowInteropHelper(Application.Current.MainWindow).Handle;
    BackgroundWorker bg = new BackgroundWorker();
    bg.DoWork += new DoWorkEventHandler(delegate
        {
            Thread.Sleep(10);
            SwitchToThisWindow(handle, true);
        });
    bg.RunWorkerAsync();
}

问题可能是线程从钩子调用你的代码还没有被运行时初始化，所以调用运行时方法不起作用。

也许您可以尝试执行Invoke将代码编组到UI线程，以调用将窗口带到前台的代码。