CODEWITHSUNDEEP
c#windowswindows-mobile
gmtek
gmtek

Reputation: 811

sOpen Menu popup programmatically

I am developing a Windows Mobile 6.5.3 application. My question is how can I open the Menu popup programmatically.

What I did is I simulate the touch event. by using

[DllImport("coredll")] private static extern void mouse_event(MOUSEEVENTF dwFlags, int dx, int dy, int dwData, int dwExtraInfo);

but here the problem is I need to specify the location of menu button and also it is showing some glance of mouse pointer, which I don't like...

Is there any message which I can send and the popup for the menu will open?

As requested I am adding some more information. I want to open the popup which opens after pressing the menu button programmatically.

enter image description here

Upvotes: 0

Views: 289

Answers (1)

Andrew Mack
Andrew Mack

Reputation: 312

Get the context menu from the control then use the Show() method of the context menu.

EDIT: answer modified and code changed

This is what I've done to have it work for me.

ContextMenu ctxMenu = BTN_TheButtonControlThatYouPressed.ContextMenu; ctxMenu.Placement = System.Windows.Controls.Primitives.PlacementMode.MousePoint; ctxMenu.IsOpen = true;

Now, because your context menu has been assigned to your form instead of your button you simply swap out the "BTN_TheButtonControl..." for a simple "this" which should "target" the form.

SO - add the following code to the button's click event and you should be okay.

ContextMenu ctxMenu = this.ContextMenu; ctxMenu.Placement = System.Windows.Controls.Primitives.PlacementMode.MousePoint; ctxMenu.IsOpen = true;

** not sure what's going on with the carriage returns... can't get 'em to go for just the second block of code...

Upvotes: 0

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