Question

我有一个listview

 <ListView Name="listView2">
        <ListView.View>
            <GridView>
                <!-- First Column -->
                <GridViewColumn Width="105" Header="ID">
                    <GridViewColumn.CellTemplate>
                        <DataTemplate>              
                            <StackPanel>    
                                      <!-- here is the event xID_Loaded -->    
                                <TextBox Loaded="xID_Loaded" Text="{Binding ID}"></TextBox>
                                <Popup Height="Auto"  Width="100" IsOpen="True" >
                                    <ListView Margin="2">
                                    </ListView>
                                </Popup>
                            </StackPanel>
                        </DataTemplate>
                    </GridViewColumn.CellTemplate>
                </GridViewColumn>

                <!-- Second Column -->
                <GridViewColumn Width="185" Header="Description">
                    <GridViewColumn.CellTemplate>
                        <DataTemplate>
                            <TextBox Text="{Binding Description}" ></TextBox>
                        </DataTemplate>
                    </GridViewColumn.CellTemplate>
                </GridViewColumn>
               etc...

无论如何，listview被绑定到一个可观察的集合。我在列上使用 Loaded =＆＃34; xID_Loaded＆＃34; 事件来每次向listview添加新行时初始化弹出控件。

因此在我的代码背后我有：

private void xID_Loaded(object sender, RoutedEventArgs e)
{
    // sender is the textbox

    // I can get the popup relative to the sender
    var stackPanel = (StackPanel)(((Control)sender).Parent);
    var popup = return (Popup)sp.Children[1];

    // every time window moves I want to reset the location
    // of the popu by doing:
    this.LocationChanged += (a, b) =>
    {
        popup.IsOpen = false;   // this will ensure that the popup moves with the control
        popup.IsOpen = true;
    };

   // some more code to initialice the popup
   // ...

现在我想要一个名为ResetAllPopups()的方法，其中该方法将重置所有弹出窗口。万一有人滚动或为了watever原因我想重置所有popus。

我知道我可以通过以下方式解决这个问题：

遍历listview中的listviewitems，使用VisualTreeHelper查找弹出控件，然后重置找到的每个弹出窗口。

我认为如果能找到订阅this.LocationChanged事件处理程序的所有事件会更好。如何执行this.LocationChanged事件所包含的所有事件，就像我在哪里更改窗口的位置一样？

Answer 1

您认为以UI为中心，只需将IsOpen绑定到您的项目，然后对其进行迭代，无需获取Popup实例。

Answer 2

    public event EventHandler reset_Popus;

    private void xID_Loaded(object sender, RoutedEventArgs e)
    {
        // I can get the popup relative to the sender
       var stackPanel = (StackPanel)(((Control)sender).Parent);
       var popup = return (Popup)sp.Children[1];


        Action<object,EventArgs> tempEvent = (object a, EventArgs b) =>
        {
            popup.IsOpen = false;
            popup.IsOpen = true;
        };

        reset_Popus += new EventHandler(tempEvent);
        this.LocationChanged += new EventHandler(tempEvent);
  }


  // finally the method that I needed
  public void ResetAllPopups()
  {
      foreach (var d in reset_Popus.GetInvocationList())
      {
          d.DynamicInvoke(null, null);
      }
  }

触发订阅特定事件处理程序的所有事件

2 个答案: