如果你需要窗口在第一次加载时在前面，那么你应该使用以下方法:

private void Window_ContentRendered(object sender, EventArgs e)
{
    this.Topmost = false;
}

private void Window_Initialized(object sender, EventArgs e)
{
    this.Topmost = true;
}

或者重写方法:

protected override void OnContentRendered(EventArgs e)
{
    base.OnContentRendered(e);
    Topmost = false;
}

protected override void OnInitialized(EventArgs e)
{
    base.OnInitialized(e);
    Topmost = true;
}

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.

这是上面几个简单有效的建议的结合。它只在那些事件触发时才出现在前面，因此事件之后弹出的任何窗口当然都会保持在顶部。

public partial class MainWindow : Window
{
    protected override void OnContentRendered(EventArgs e)
    {
        base.OnContentRendered(e);

        Topmost = true;
        Topmost = false;
    }
    protected override void OnInitialized(EventArgs e)
    {
        base.OnInitialized(e);

        Topmost = true;
        Topmost = false;
    }

    ....
}

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;

为了便于重用，我构建了一个扩展方法。

using System.Windows.Forms;
    namespace YourNamespace{
        public static class WindowsFormExtensions {
            public static void PutOnTop(this Form form) {
                form.Show();
                form.Activate();
            }// END PutOnTop()       
        }// END class
    }// END namespace

调用表单构造函数

namespace YourNamespace{
       public partial class FormName : Form {
       public FormName(){
            this.PutOnTop();
            InitalizeComponents();
        }// END Constructor
    } // END Form            
}// END namespace

记住不要把显示该窗口的代码放在PreviewMouseDoubleClick处理程序中，因为活动窗口将切换回处理该事件的窗口。 只要把它放在MouseDoubleClick事件处理程序中，或者通过设置e.Handled为True来停止冒泡。

在我的情况下，我正在处理一个列表视图上的PreviewMouseDoubleClick，并没有设置e.Handled = true，然后它会引发MouseDoubleClick事件，将焦点集中到原始窗口。