Question

我为LineSymbol创建了一个ControlTemplate：

            <esri:SimpleLineSymbol 
        x:Key="PolylineSymbol" 
        Width="3"
        >
        <esri:SimpleLineSymbol.ControlTemplate>
            <ControlTemplate>
                <Grid
                    >
                    <VisualStateManager.VisualStateGroups>
                        <VisualStateGroup x:Name="CommonStates">
    ...
                        </VisualStateGroup>
                        <VisualStateGroup x:Name="SelectionStates">
    ...
                        </VisualStateGroup>
                    </VisualStateManager.VisualStateGroups>
                    <Path 
                        x:Name="Element" Fill="{x:Null}"
                        Stroke="Navy" StrokeThickness="3"
                                StrokeLineJoin="Round" StrokeStartLineCap="Round" StrokeEndLineCap="Round">
                        <Path.ContextMenu>
                            <ContextMenu 
                                x:Name="popUpMenu"
                                DataContext="{Binding RelativeSource={RelativeSource FindAncestor, AncestorType={x:Type esri:Graphic}}}">
                                <MenuItem 
                                    x:Name="miSelect"
                                    Header="Select" 
                                    IsCheckable="True"
                                    IsChecked="{Binding Selected, FallbackValue=False}"
                                    />
                                ...
                            </ContextMenu>
                        </Path.ContextMenu>
                    </Path>
                </Grid>
            </ControlTemplate>
        </esri:SimpleLineSymbol.ControlTemplate>
    </esri:SimpleLineSymbol>

除了IsChecked在“选择”菜单项上的绑定外，一切都运行良好：由于SymbolElement和Symbol都没有从FrameworkElement继承，因此没有任何绑定表达式会将我从此处带到此SimpleLineSymbol的包含图形。

我还尝试了一个Click事件（它给出了一个发送者，一个命令（支持一个参数）或一个MouseRightButtonDown事件（在图形上） - 没有方法从我路径上的右键单击点开始符号包含图形...

菜单的DataContext在VS2012的设计器中看起来没问题，但是在运行时它不起作用，因为Menu位于Symbol的ControlTemplate中定义的路径内，而不是FrameworkElement！

我已经为ContextMenu添加了一个名称，但我无法从ViewModel（我创建图形和符号的位置）中检索它;如果我能够这样做，我将能够添加所需的datacontext在代码中：

      var graphic = new Graphic { Symbol = Resources["PolylineSymbol"] as SimpleLineSymbol;
      var menu = graphic.Symbol.ControlTemplate.FindName("popUpMenu", graphic.Symbol); // ???
      menu.DataContext = graphic;

）

有什么想法吗？

Answer 1

如果我正确理解您的问题，似乎您在WPF中遇到了常见问题。解决方案是利用Tag的{{1}}属性，使用ContextMenu.PlacementTarget property将Path“传递”到DataContext。此获取或设置ContextMenu在打开时所处的UIElement 。试试这个：

ContextMenu

Answer 2

解决方案是实现上下文菜单的Opened。在后面的代码中，我在其DataContext上分配了视图模型的实例。

    <Path.ContextMenu>
            <ContextMenu 
                    Opened="PopUpMenu_OnOpened"
            >

代码背后的代码：

    private void PopUpMenu_OnOpened(object sender, RoutedEventArgs e)
    {
        var menu = sender as ContextMenu;
        if (menu != null)
        {
            menu.DataContext = ViewModel;
        }
    }

另一个挑战是获取点击的图形和点击的点。解决方案是在View模型中创建属性，并在LeftMouseDown和RightMouseDown上分配两者。

    private Graphic GetPolylineGraphic(ESRI.ArcGIS.Client.Geometry.Geometry geometry = null)
    {
        var drawLayer = Model.GetDrawLayer(MyMap, "Polyline");
        var graphic = new Graphic
        {
            // clone the resourced PolylineSymbol (from Model)
            Symbol = new SimpleLineSymbol 
                {
                    Color = PolylineSymbol.Color,
                    Width = PolylineSymbol.Width,
                    ControlTemplate = PolylineSymbol.ControlTemplate
                }
        };

        if (geometry != null) graphic.Geometry = geometry;

        graphic.MouseLeftButtonDown += GraphicOnMouseLeftButtonDown;
        graphic.MouseRightButtonDown += GraphicOnMouseRightButtonDown;
        drawLayer.Graphics.Add(graphic);

        return graphic;
    }

    private Graphic m_clickedGraphic;
    public Graphic ClickedGraphic
    {
        get { return m_clickedGraphic; }
        set
        {
            if (!Equals(m_clickedGraphic, value))
            {
                m_clickedGraphic = value;
                OnPropertyChanged(value);
            }
        }
    }

    private MapPoint m_clickedPoint;
    public MapPoint ClickedPoint
    {
        get { return m_clickedPoint; }
        set
        {
            if (m_clickedPoint != value)
            {
                m_clickedPoint = value;
                OnPropertyChanged(value);
            }
        }
    }

    private void GraphicOnMouseRightButtonDown(object sender, MouseButtonEventArgs args)
    {
        //// This does not work because GraphicElement is internal!!!
        //var s = args.Source; 
        //ClickedGraphic = ((GraphicElement)(e.Source)).Graphic;
        //ClickedPoint = ((GraphicElement)(e.Source)).Origin;
        ClickedGraphic = sender as Graphic;
        ClickedPoint = MyMap.ScreenToMap(args.GetPosition(MyMap));

        //// not here - else context menu won't pop!
        //args.Handled = true;
    }

    private void GraphicOnMouseLeftButtonDown(object sender, MouseButtonEventArgs args)
    {
        var g = sender as Graphic;
        if (g != null)
        {
            ClickedGraphic = g;
            ClickedPoint = MyMap.ScreenToMap(args.GetPosition(MyMap));

            // select/unselect the graphic on left click
            if (g.Selected) g.UnSelect();
            else g.Select();
            args.Handled = true;
        }
    }

为了使一切正常，我必须克隆符号。

如何在ESRI中的LineSymbol上绑定上下文菜单？

2 个答案: