Question

以下类派生自System.Windows.Controls.UserControl。在所述类中，我调用OpenFileDialog来打开XAML文件（工作流文件）。接下来，我在右键单击鼠标时实现动态菜单。菜单没有显示。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;
}

Answer 1

我自己遇到过这个问题，我用过这个：

ContextMenu.IsOpen = true;

MSDN Documentation on ContextMenu

Answer 2

你必须调用ContextMenu的Show(Control, Point) 方法。此外，我不会在每次单击控件时实例化新的上下文菜单，而是我会做类似的事情：

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);
    }
}

Answer 3

我认为您需要致电contextMenu.Show

为什么ContextMenu（弹出菜单）不显示？

3 个答案: