使用 C# TouchMove 进行跳跃和不完整的 XAML 窗口拖动

Question

我有一个用于触摸屏的未装饰的小 xaml 窗口。用户必须能够使用触摸和拖动来移动窗口。当前在触摸和拖动中，窗口向拖动的方向移动，但只是部分移动；并且似乎有两个窗口而不是一个，使触摸和拖动看起来很跳跃。

此行为表现在开发系统（使用 Visual Studio Professional 2015 的 Surface Pro 3）以及生产系统（Windows 7，无键盘或鼠标）上。

我基于 Microsoft 的 example 这个 C#。

using System.Windows;
using System.Windows.Input;

namespace XAMLApp
{
    public partial class MainWindow : Window
    {
        public MainWindow()
        {
            InitializeComponent();
        }

        private TouchDevice windowTouchDevice;
        private Point lastPoint;

        private void Circle_TouchUp(object sender, TouchEventArgs e)
        {
            // Do stuff.
        }

        private void Window_TouchDown(object sender, TouchEventArgs e)
        {
            e.TouchDevice.Capture(this);

            if (windowTouchDevice == null)
            {
                windowTouchDevice = e.TouchDevice;
                lastPoint = windowTouchDevice.GetTouchPoint(null).Position;
            }

            e.Handled = true;
        }

        private void Window_TouchMove(object sender, TouchEventArgs e)
        {
            if (e.TouchDevice == windowTouchDevice)
            {
                var currentTouchPoint = windowTouchDevice.GetTouchPoint(null);

                var deltaX = currentTouchPoint.Position.X - lastPoint.X;
                var deltaY = currentTouchPoint.Position.Y - lastPoint.Y;

                Top += deltaY;
                Left += deltaX;

                lastPoint = currentTouchPoint.Position;

                e.Handled = true;
            }
        }

        private void Window_TouchLeave(object sender, TouchEventArgs e)
        {
            if (e.TouchDevice == windowTouchDevice)
                windowTouchDevice = null;

            e.Handled = true;
        }
    }
}

这里有一些用于窗口的 xaml。

<Window x:Name="AppWindow" x:Class="XAMLApp.MainWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
        xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
        xmlns:local="clr-namespace:XAMLApp"
        mc:Ignorable="d"
        Title="XAML Application"
        Height="25" Width="25"
        AllowsTransparency="True" WindowStyle="None" ResizeMode="NoResize"
        ScrollViewer.VerticalScrollBarVisibility="Hidden" ScrollViewer.HorizontalScrollBarVisibility="Hidden"
        ShowInTaskbar="False" ToolTip="XAML Application" Topmost="True" UseLayoutRounding="True"
        MaxHeight="25" MaxWidth="25" MinHeight="25" MinWidth="25"
        Left="0" Top="0" Background="Transparent"
        TouchDown="Window_TouchDown"
        TouchMove="Window_TouchMove"
        TouchLeave="Window_TouchLeave">
    <Grid>
        <Ellipse x:Name="Circle" Fill="Black" HorizontalAlignment="Left"
                 Height="24" Margin="0" Stroke="Black" VerticalAlignment="Top"
                 Width="24" ScrollViewer.HorizontalScrollBarVisibility="Hidden" ScrollViewer.VerticalScrollBarVisibility="Hidden"
                 TouchUp="Circle_TouchUp" />
    </Grid>
</Window>

我尝试将Grid 替换为Canvas。那没什么区别。我还尝试将 Manipulation 用作 Microsoft 的 demonstrated。尝试拖动窗口时，被告知转换对窗口无效。

如何使用鼠标左键单击并拖动使触摸和拖动的行为与DragMove() 相同？

Answer 1

通过TouchDevice.GetTouchPoint方法检索到的TouchPoint与窗口的Top和Left属性不共享同一个坐标系。您需要做的就是将检索到的 X 和 Y 值转换为屏幕坐标：

private void Window_TouchMove(object sender, TouchEventArgs e)
{
    if (e.TouchDevice == windowTouchDevice)
    {
        var currentTouchPoint = windowTouchDevice.GetTouchPoint(null);

        var locationOnScreen = this.PointToScreen(new Point(currentTouchPoint.Position.X, currentTouchPoint.Position.Y));

        var deltaX = locationOnScreen.X - lastPoint.X;
        var deltaY = locationOnScreen.Y - lastPoint.Y;

        Top += deltaY;
        Left += deltaX;

        lastPoint = locationOnScreen;

        e.Handled = true;
    }
}

编辑：

当然，不同坐标系的事情适用于整个项目，所以Window_TouchDown事件处理方法需要以类似的方式进行适配：

private void Window_TouchDown(object sender, TouchEventArgs e)
{
    e.TouchDevice.Capture(this);

    if (windowTouchDevice == null)
    {
        windowTouchDevice = e.TouchDevice;
        var currentTouchPoint = windowTouchDevice.GetTouchPoint(null);
        var locationOnScreen = this.PointToScreen(new Point(currentTouchPoint.Position.X, currentTouchPoint.Position.Y));
        lastPoint = locationOnScreen;
    }

    e.Handled = true;
}

Answer 2

代替

Top += deltaY;
Left += deltaX;

我只是需要

Top += currentTouchPoint.Position.Y;
Left += currentTouchPoint.Position.X;

呃。