相关疑难解决方法(0)

我的WPF应用程序有一个UserControl应该看起来和行为像弹出窗口,但它不是一个窗口.控件不会从Window类中下降的原因是因为它包含第三方虚拟屏幕键盘,并且该控件必须与TextBox它在单击其按钮时将输入字符发送到的控件位于同一窗口中.如果键盘控件不在同一窗口中,则甚至看不到TextBox控件.

拖动对话框时,我遇到的问题是表现糟糕.它足够慢,鼠标离开拖动区域,它停止跟随鼠标.我需要一个更好的方法.

以下是控件的xaml摘录:

<Grid Name="LayoutRoot">
    <Grid.RowDefinitions>
        <RowDefinition Height="Auto" />
        <RowDefinition Height="*" />
    </Grid.RowDefinitions>
    <Border Background="{DynamicResource PopupBackground}"
            BorderBrush="{DynamicResource PopupBorder}"
            BorderThickness="5,5,5,0"
            MouseLeftButtonDown="Grid_MouseLeftButtonDown"
            MouseLeftButtonUp="Grid_MouseLeftButtonUp"
            MouseMove="Grid_MouseMove">
    . . .
    </Border>
</Grid>

这是鼠标事件处理程序:

    private void Grid_MouseLeftButtonDown( object sender, MouseButtonEventArgs e ) {
        Canvas canvas = Parent as Canvas;
        if ( canvas == null ) {
            throw new InvalidCastException( "The parent of a KeyboardPopup control must be a Canvas." );
        }
        DraggingControl = true;
        CurrentMousePosition = e.GetPosition( canvas …