CODEWITHSUNDEEP
c#wpfuser-interfacecontextmenupopupmenu
user1298925
user1298925

Reputation: 2424

Why does not the ContextMenu(popup menu) show up?

The following class derives from System.Windows.Controls.UserControl. In said class I call OpenFileDialog to open a XAML file (workflow file). Next, I implement a dynamic menu when right clicking the mouse. The menu does not show up. Is this a threading problem or a UI problem? In my research I've been unable to discover a solution.

Thanks in advance.

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;
}

Upvotes: 2

Views: 13031

Answers (3)

Dunstad
Dunstad

Reputation: 71

Came across this problem myself, I used this:

ContextMenu.IsOpen = true;

MSDN Documentation on ContextMenu

Upvotes: 7

flix
flix

Reputation: 609

you have to call the ContextMenu's Show(Control, Point) method. furthermore i wouldn't instantiate a new context menu each time the control is clicked, instead i would do something like that:

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);
    }
}

Upvotes: 0

DGH
DGH

Reputation: 11549

I think you need to call contextMenu.Show

Upvotes: -2

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