Question

我需要在点击按钮时显示视图。这是事件Click

的代码

private void Button_Click(object sender, RoutedEventArgs e)
{
    Button s = sender as Button;
    //some stuff needed to recognise which button was pressed
    myPopup.PlacementTarget = s;
    myPopup.Placement = System.Windows.Controls.Primitives.PlacementMode.Top;
    myPopup.IsOpen = true;
}

这是xaml

中定义的元素

<Popup IsOpen="False" StaysOpen="True" Name="myPopup">
       <view:myCustomView />
</Popup>

我有2个问题。

1-我的观点背景是黑色的，即使我没有设置它（我认为它应该是透明的？）

2-当我点击弹出窗口内的任何地方时，这只会消失

显示我的观点的正确方法是什么？我不想将拥有该按钮的ViewModel的任何内容绑定到这个具有自己的viewmodel的新视图

请注意，在事件Button_Click上，我会向ViewModel myCustomView发送一些参数，这些参数会改变其部分功能（这就是为什么我需要创建一个新的实例每次触发Button_Click事件时查看

编辑1：感谢EdPlunkett的回答，我能够用后台解决问题。我只需要设置AllowTransparency =“True”

EDIT2：

我按照建议通过Code Behind定义了我的Popup，所以我的代码现在是：

  private void Button_Click(object sender, RoutedEventArgs e)
        {
            Button s = sender as Button;
            System.Windows.Controls.Primitives.Popup popup = new System.Windows.Controls.Primitives.Popup();
            popup.AllowsTransparency = true;
            popup.Child = new myCustomView();
            popup.PlacementTarget = s;
            popup.Placement = System.Windows.Controls.Primitives.PlacementMode.Top;
            popup.IsOpen = true;
            popup.StaysOpen = true;
        }

问题是，当我在myCustomView中定义的任何控件内部单击时，Popup会失去焦点并关闭。

Answer 1

我怀疑Popup是在另一个Popup中定义的，例如ComboBox的下拉列表，对吧？

您可以尝试在click事件处理程序中以编程方式创建Popup元素：

private void Button_Click(object sender, RoutedEventArgs e)
{
    Button s = sender as Button;
    System.Windows.Controls.Primitives.Popup popup = new System.Windows.Controls.Primitives.Popup();
    popup.AllowsTransparency = true;
    popup.Child = new myCustomView();
    //some stuff needed to recognise which button was pressed
    popup.PlacementTarget = s;
    popup.Placement = System.Windows.Controls.Primitives.PlacementMode.Top;
    popup.IsOpen = true;
    popup.StaysOpen = true;
}

这可以让您在弹出视图中单击而不关闭Popup。确保视图设置为Background：

popup.Child = new myCustomView() { Background = Brushes.White };

我尝试了你的解决方案，虽然情况稍微好一点，但我仍然有结束问题。我可以在弹出窗口内单击而不关闭它，但是当我在我的视图中定义的任何文本框内单击时，弹出窗口关闭

因此，将PlacementTarget设置为父ItemsControl，然后设置VerticalOffset的{{1}}和HorizontalOffset属性，以在屏幕上指定其确切位置，例如：

Popup

您应该调整private void btn_Click(object sender, RoutedEventArgs e) { Button s = sender as Button; System.Windows.Controls.Primitives.Popup popup = new System.Windows.Controls.Primitives.Popup(); popup.AllowsTransparency = true; popup.Child = new myCustomView(); //some stuff needed to recognise which button was pressed popup.PlacementTarget = ic; //<-- "ic" is the name of the parent ItemsControl Point p = s.TranslatePoint(new Point(0, 0), ic); popup.VerticalOffset = p.Y; popup.HorizontalOffset = p.X; popup.IsOpen = true; popup.StaysOpen = true; }和VerticalOffset的值以满足您的要求。

如何使用弹出控件来显示视图？

1 个答案: