当popup中的datagrid捕获鼠标时,wpf弹出窗口不会自动关闭

Question

我有一个弹出窗口,StaysOpen=False所以我想通过点击弹出窗口外的任何地方来关闭它.在弹出窗口我有一个DataGrid.如果我打开弹出窗口然后单击其他位置,弹出窗口将被关闭.但是如果在点击弹出窗口之前我将点击列标题,就不会发生这种情况DataGrid.测试XAML:

<Window x:Class="Test.MainWindow"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    Title="MainWindow" Height="350" Width="525" VerticalAlignment="Center" HorizontalAlignment="Center" Foreground="Black">
<Grid>
    <ToggleButton x:Name="btn" VerticalAlignment="Top">Open</ToggleButton>
    <Popup StaysOpen="False" IsOpen="{Binding IsChecked, ElementName=btn}" > 
        <DataGrid Width="150" Height="150">
            <DataGrid.Columns>
                <DataGridTextColumn Header="Column" />
            </DataGrid.Columns>
        </DataGrid>
    </Popup>
</Grid>
</Window>

我认为这是因为列标题在点击时捕获鼠标并且弹出窗口不再接收鼠标事件.我试图在LostMouseCapture事件上添加一个处理程序,以便通过弹出窗口捕获鼠标,但它似乎并不那么容易.有任何想法吗？

Answer 1

也许它会有所帮助。附加行为：

public class DataGridColumnHeaderReleaseMouseCaptureBehavior {
    public static DataGrid GetReleaseDGCHeaderBehavior(DependencyObject obj) {
        return (DataGrid)obj.GetValue(ReleaseDGCHeaderBehaviorProperty);
    }

    public static void SetReleaseDGCHeaderBehavior(DependencyObject obj, Boolean value) {
        obj.SetValue(ReleaseDGCHeaderBehaviorProperty, value);
    }

    public static readonly DependencyProperty ReleaseDGCHeaderBehaviorProperty =
        DependencyProperty.RegisterAttached("ReleaseDGCHeaderBehavior",
            typeof(DataGrid),
            typeof(DataGridColumnHeaderReleaseMouseCaptureBehavior),
            new UIPropertyMetadata(default(DataGrid), OnReleaseDGCHeaderBehaviorPropertyChanged));

    private static Popup _popup;

    private static void OnReleaseDGCHeaderBehaviorPropertyChanged(DependencyObject d, DependencyPropertyChangedEventArgs e) {
        var oldGrid = (DataGrid)e.OldValue;
        if (oldGrid != null)
            oldGrid.MouseLeave -= OnMouseLeave;
        var refSender = d as Popup;
        _popup = refSender;
        if (refSender != null) {
            var refGrid = e.NewValue as DataGrid;
            if (refGrid != null) {
                refGrid.MouseLeave += OnMouseLeave;
            }
        }
    }
    static void OnMouseLeave(object sender, MouseEventArgs args) {
        if (_popup != null)
            typeof(Popup).GetMethod("EstablishPopupCapture", System.Reflection.BindingFlags.NonPublic | System.Reflection.BindingFlags.Instance).Invoke(_popup, null);
    }
}

XAML：

<Popup x:Name="popup"
bhvrs:DataGridColumnHeaderReleaseMouseCaptureBehavior.ReleaseDGCHeaderBehavior="{Binding ElementName=dataGrid}">
  <DataGrid x:Name="dataGrid"/>
</Popup>

Answer 2

我想你只是偶然发现了一个普通的老错误。我已经复制了这个，但找不到合理的方法来让它工作。我认为你应该向 Microsoft 提交错误。它似乎是一个捕获鼠标并取消捕获的组件，它不会将捕获恢复到原始捕获组件。