我有一个C＃/。NET应用程序，我想实现以下行为：

I have a C#/.NET app and I want to implement the following behavior:

我有一个弹出式菜单。每当在用户点击任何东西的 的是没有的弹出式菜单，我想在弹出菜单中关闭该应用程序。

I have a popup menu. Whenever the user clicks on anything within the application that is not the popup menu, I want the popup menu to close.

不过，只要用户不在应用程序，我不希望任何事情发生。

However, whenever a user is not in the application I don't want anything to happen.

我想通过LostFocus事件来管理这一点，但我无法确定我的应用程序是否是活动窗口。在code看起来是这样的。

I'm trying to manage this through the LostFocus event, but I'm having trouble determining whether my application is the active window. The code looks something like this.

    private void Button_LostFocus(object sender, System.EventArgs e)
    {
        if (InActiveWindow()) {
           CloseMenu()
        }
        else {
           // not in active window, do nothing
        }
    }

我需要知道的是如何实现的InActiveWindow（）方法。

What I need to know is how to implement the InActiveWindow() method.

推荐答案

您可以P /调用到的 GetForegroundWindow（）和比较HWND返回给应用程序的form.Handle属性。

You could P/Invoke into GetForegroundWindow(), and compare the HWND returned to the application's form.Handle property.

一旦你的把柄，你也可以的P / Invoke的 GetAncestor（），以获取根所有者窗口。这应该是您的应用程序的主，启动窗口的句柄，如果这是你的应用程序。

Once you have the handle, you can also P/Invoke GetAncestor() to get the root owner window. This should be the handle of your application's main, startup window, if this is in your application.