如何使用户控件在窗口上像窗口一样可拖动

Question

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 );
        e.Handled = true;
    }

    private void Grid_MouseLeftButtonUp( 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." );
        }

        if ( DraggingControl ) {
            Point mousePosition = e.GetPosition( canvas );

            // Correct the mouse coordinates in case they go off the edges of the control
            if ( mousePosition.X < 0.0 ) mousePosition.X = 0.0; else if ( mousePosition.X > canvas.ActualWidth ) mousePosition.X = canvas.ActualWidth;
            if ( mousePosition.Y < 0.0 ) mousePosition.Y = 0.0; else if ( mousePosition.Y > canvas.ActualHeight ) mousePosition.Y = canvas.ActualHeight;

            // Compute the new Left & Top coordinates of the control
            Canvas.SetLeft( this, Left += mousePosition.X - CurrentMousePosition.X );
            Canvas.SetTop( this, Top += mousePosition.Y - CurrentMousePosition.Y );
        }
        e.Handled = true;
    }

    private void Grid_MouseMove( object sender, MouseEventArgs e ) {
        Canvas canvas = Parent as Canvas;
        if ( canvas == null ) {
            // It is not.  Throw an exception
            throw new InvalidCastException( "The parent of a KeyboardPopup control must be a Canvas." );
        }

        if ( DraggingControl && e.LeftButton == MouseButtonState.Pressed ) {
            Point mousePosition = e.GetPosition( canvas );

            // Correct the mouse coordinates in case they go off the edges of the control
            if ( mousePosition.X < 0.0 ) mousePosition.X = 0.0; else if ( mousePosition.X > canvas.ActualWidth  ) mousePosition.X = canvas.ActualWidth;
            if ( mousePosition.Y < 0.0 ) mousePosition.Y = 0.0; else if ( mousePosition.Y > canvas.ActualHeight ) mousePosition.Y = canvas.ActualHeight;

            // Compute the new Left & Top coordinates of the control
            Canvas.SetLeft( this, Left += mousePosition.X - CurrentMousePosition.X );
            Canvas.SetTop ( this, Top  += mousePosition.Y - CurrentMousePosition.Y );

            CurrentMousePosition = mousePosition;
        }
        e.Handled = true;
    }

请注意,控件必须放在Canvas使用它的窗口中.

我不能使用DragMove它,因为它是Window类的一个方法,这个类来自UserControl.如何提高此控件拖动的性能？我是否必须使用Win32 API？

Answer 1

您只需使用MouseDragElementBehavior即可.

UPD 关于MouseDragElementBehavior行为的重要事项:

MouseDragElementBehavior行为不适用于处理MouseClick事件的任何控件(例如,Button,TextBox和ListBox控件).如果您需要能够拖动其中一种类型的控件,请将该控件设置为可以拖动的控件的子项(例如,边框).然后,您可以将MouseDragElementBehavior行为应用于父元素.

您还可以像这样实现自己的拖动行为:

public class DragBehavior : Behavior<UIElement>
{
    private Point elementStartPosition;
    private Point mouseStartPosition;
    private TranslateTransform transform = new TranslateTransform();

    protected override void OnAttached()
    {
        Window parent = Application.Current.MainWindow;
        AssociatedObject.RenderTransform = transform;

        AssociatedObject.MouseLeftButtonDown += (sender, e) => 
        {
            elementStartPosition = AssociatedObject.TranslatePoint( new Point(), parent );
            mouseStartPosition = e.GetPosition(parent);
            AssociatedObject.CaptureMouse();
        };

        AssociatedObject.MouseLeftButtonUp += (sender, e) =>
        {
            AssociatedObject.ReleaseMouseCapture();
        };

        AssociatedObject.MouseMove += (sender, e) =>
        {
            Vector diff = e.GetPosition( parent ) - mouseStartPosition;
            if (AssociatedObject.IsMouseCaptured)
            {
                transform.X = diff.X;
                transform.Y = diff.Y;
            }
        };
    }
}

Answer 2

根据@DmitryMartovoi 的回答中的信息，我想出了一种方法来完成这项工作。我仍然给 Dmitry 一个 +1，因为如果没有他的贡献，我将无法弄清楚这一点。

我所做的是TranslateTransform在UserControl's构造函数中创建了一个并将其分配给它的RenderTransform属性：

RenderTransform = new TranslateTransform();

在 XAML 中，我为Border用户单击以拖动整个控件的控件命名：

<Border Background="{DynamicResource PopupBackground}"
        BorderBrush="{DynamicResource PopupBorder}"
        BorderThickness="5,5,5,0"
        MouseLeftButtonDown="Grid_MouseLeftButtonDown"
        MouseLeftButtonUp="Grid_MouseLeftButtonUp"
        MouseMove="Grid_MouseMove"
        Name="TitleBorder">

    . . .
</Border>

最后，我修改了各种鼠标事件处理程序，如下所示：

private void Grid_MouseLeftButtonDown( object sender, MouseButtonEventArgs e ) {
    CurrentMousePosition = e.GetPosition( Parent as Window );
    TitleBorder.CaptureMouse();
}

private void Grid_MouseLeftButtonUp( object sender, MouseButtonEventArgs e ) {
    if ( TitleBorder.IsMouseCaptured ) {
        TitleBorder.ReleaseMouseCapture();
    }
}

private void Grid_MouseMove( object sender, MouseEventArgs e ) {
    Vector diff = e.GetPosition( Parent as Window ) - CurrentMousePosition;
    if ( TitleBorder.IsMouseCaptured ) {
        ( RenderTransform as TranslateTransform ).X = diff.X;
        ( RenderTransform as TranslateTransform ).Y = diff.Y;
    }
}

这很好用。UserControl当您拖动 时，整个及其所有内容都会平滑移动Border，并与鼠标保持一致。UserControl如果您单击其表面上的任何其他位置，则整体不会移动。

再次感谢@DmitryMartovoi 提供的代码。

编辑：我正在编辑这个答案，因为上面的代码虽然有效，但并不完美。它的缺陷是当您单击标题栏区域和开始拖动之前，控件会弹回到屏幕上的原始位置。这很烦人，而且完全错误。

我想出的实际上完美无缺的方法涉及首先将控件放入Canvas. 控件的父级为 a 很重要，Canvas否则以下代码将不起作用。我也停止使用RenderTransform. 我添加了一个名为canvastype的私有属性Canvas。我Loaded在弹出控件中添加了一个事件处理程序来进行一些重要的初始化：

private void KeyboardPopup_Loaded( object sender, RoutedEventArgs e ) {
    canvas = Parent as Canvas;
    if ( canvas == null ) {
        throw new InvalidCastException( "The parent of a KeyboardPopup control must be a Canvas." );
    }    
}

完成所有这些后，以下是修改后的鼠标事件处理程序：

private void TitleBorder_MouseLeftButtonDown( object sender, MouseButtonEventArgs e ) {
    StartMousePosition = e.GetPosition( canvas );
    TitleBorder.CaptureMouse();
}

private void TitleBorder_MouseLeftButtonUp( object sender, MouseButtonEventArgs e ) {
    if ( TitleBorder.IsMouseCaptured ) {
        Point mousePosition = e.GetPosition( canvas );
        Canvas.SetLeft( this, Canvas.GetLeft( this ) + mousePosition.X - StartMousePosition.X );
        Canvas.SetTop ( this, Canvas.GetTop ( this ) + mousePosition.Y - StartMousePosition.Y );
        canvas.ReleaseMouseCapture();
    }
}

private void TitleBorder_MouseMove( object sender, MouseEventArgs e ) {
    if ( TitleBorder.IsMouseCaptured && e.LeftButton == MouseButtonState.Pressed ) {
        Point mousePosition = e.GetPosition( canvas );

        // Compute the new Left & Top coordinates of the control
        Canvas.SetLeft( this, Canvas.GetLeft( this ) + mousePosition.X - StartMousePosition.X );
        Canvas.SetTop ( this, Canvas.GetTop ( this ) + mousePosition.Y - StartMousePosition.Y );
        StartMousePosition = mousePosition;
    }
}

当您再次单击标题栏以移动控件时，控件会停留在您放置它的位置，并且仅在您单击标题栏时才会移动。单击控件中的任何其他位置什么都不做，拖动是平滑且响应迅速的。