2

以下类派生自System.Windows.Controls.UserControl. 在所述类中，我调用OpenFileDialog以打开 XAML 文件（工作流文件）。接下来，我在鼠标右键单击时实现了一个动态菜单。菜单不显示。这是线程问题还是 UI 问题？在我的研究中，我一直无法找到解决方案。

提前致谢。

private void File_Open_Click(object sender, RoutedEventArgs e)
{
    var fileDialog = new OpenFileDialog();

    fileDialog.Title  = "Open Workflow";
    fileDialog.Filter = "Workflow| *.xaml";

    if (fileDialog.ShowDialog() == DialogResult.OK)
    {
        LoadWorkflow(fileDialog.FileName);
        MouseDown += new System.Windows.Input.MouseButtonEventHandler(mouseClickedResponse);
     }
}

private void mouseClickedResponse(object sender, System.Windows.Input.MouseEventArgs e)
{
    if (e.RightButton == MouseButtonState.Pressed)
    {
         LoadMenuItems();
    }
}

private void LoadMenuItems()
{
    System.Windows.Controls.ContextMenu contextmenu = new System.Windows.Controls.ContextMenu();   
    System.Windows.Controls.MenuItem item1 = new System.Windows.Controls.MenuItem();
    item1.Header = "A new Test";
    contextmenu.Items.Add(item1);
    this.ContextMenu = contextmenu;
    this.ContextMenu.Visibility = Visibility.Visible;
}
4

3 回答 3

7

我自己遇到了这个问题，我用这个：

ContextMenu.IsOpen = true;

ContextMenu 上的 MSDN 文档

于 2012-11-19T03:25:29.430 回答
0

您必须调用 ContextMenu 的Show(Control, Point)方法。此外，每次单击控件时，我都不会实例化新的上下文菜单，而是会执行以下操作：

MyClass()
{
     // create the context menu in the constructor:

     this.ContextMenu = new System.Windows.Forms.ContextMenu();   
     System.Windows.Forms.MenuItem item1 = new System.Windows.Forms.MenuItem();
     item1.Text = "A new Test";
     this.ContextMenu.Items.Add(item1);
}


private void mouseClickedResponse(object sender, System.Windows.Input.MouseEventArgs e)
{
    if (e.RightButton == MouseButtonState.Pressed)
    {
          // show the context menu as soon as the right mouse button is pressed
          this.ContextMenu.Show(this, e.Location);
    }
}
于 2012-08-23T20:41:40.920 回答
-1

我想你需要打电话contextMenu.Show

于 2012-08-23T20:27:18.407 回答