wpf中的可拖动弹出控件

Question

我在wpf中需要一个可拖动的弹出控件,并且想知道你们中的任何人是否可以帮助我......我确实看到了以下帖子:

拖动WPF弹出控件

但这不应该如何运作......？当我点击并拖动它总是重置到一个特定的点,而且评论者说这不是一个有效的方法......？有没有人有其他选择？

谢谢!

Answer 1

我们可以写一个行为来制作任何Popup可拖动的行为.下面是一个与文本框关联的弹出窗口的示例XAML,该文本框在文本框聚焦时打开并保持打开状态:

<Grid>
    <StackPanel>
        <TextBox x:Name="textBox1" Width="200" Height="20"/>
    </StackPanel>
    <Popup PlacementTarget="{Binding ElementName=textBox1}" IsOpen="{Binding IsKeyboardFocused, ElementName=textBox1, Mode=OneWay}">
        <i:Interaction.Behaviors>
            <local:MouseDragPopupBehavior/>
        </i:Interaction.Behaviors>
        <TextBlock Background="White">
            <TextBlock.Text>Sample Popup content.</TextBlock.Text>
        </TextBlock>
    </Popup>
</Grid>

以下是允许我们拖动的行为Popup:

public class MouseDragPopupBehavior : Behavior<Popup>
{
    private bool mouseDown;
    private Point oldMousePosition;

    protected override void OnAttached()
    {
        AssociatedObject.MouseLeftButtonDown += (s, e) =>
        {
            mouseDown = true;
            oldMousePosition = AssociatedObject.PointToScreen(e.GetPosition(AssociatedObject));
            AssociatedObject.Child.CaptureMouse();
        };
        AssociatedObject.MouseMove += (s, e) =>
        {
            if (!mouseDown) return;
            var newMousePosition = AssociatedObject.PointToScreen(e.GetPosition(AssociatedObject));
            var offset = newMousePosition - oldMousePosition;
            oldMousePosition = newMousePosition;
            AssociatedObject.HorizontalOffset += offset.X;
            AssociatedObject.VerticalOffset += offset.Y;
        };
        AssociatedObject.MouseLeftButtonUp += (s, e) =>
        {
            mouseDown = false;
            AssociatedObject.Child.ReleaseMouseCapture();
        };
    }
}

如果您不熟悉行为,请安装Expression Blend 4 SDK并添加以下命名空间:

xmlns:i="http://schemas.microsoft.com/expression/2010/interactivity"

并添加System.Windows.Interactivity到您的项目.