How to show MenuFlyout in code?

Hong 1,206 Reputation points
2020-02-25T21:53:02.807+00:00

The MenuFlyout shows up perfectly upon right-clicking the ListView

 <ListView Name="lvDeviceTiles"  
       ItemsSource="{x:Bind listDeviceItems}"  
       DataContext="{x:Bind}"  
       Margin="0">  
     <!--ContextFlyout shows up automatically after right-click-->  
     <ListView.ContextFlyout>  
         <MenuFlyout   
             x:Name="menuFlyoutContext"  
             Opening="menuFlyoutContext_Opening">  
           ...    
         </MenuFlyout>  
     </ListView.ContextFlyout>  
 </ListView>  

How can I show it in C# code?
I tried:

FlyoutBase.ShowAttachedFlyout(lvDeviceTiles);  
lvDeviceTiles.ContextFlyout.ShowAt(lvDeviceTiles);  
lvDeviceTiles.ContextFlyout.ShowAt(myButton);  

None of them work.

Universal Windows Platform (UWP)
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Accepted answer
  1. Roy Li - MSFT 32,721 Reputation points Microsoft Vendor
    2020-02-26T06:50:06.477+00:00

    Hello,

    Welcome to Microsoft Q&A!

    I have to say that you are in the right direction. To show the MenuFlyout control, you could just call FlyoutBase.ShowAt Method to do that. But please note that MenuFlyout control is derived from FlyoutBase. So you just need to call MenuFlyout.ShowAt method in your scenario.

    Like this:

            private void Button_Click(object sender, RoutedEventArgs e)  
            {  
                menuFlyoutContext.ShowAt(lvDeviceTiles);  
            }  
    

    Thank you!

    1 person found this answer helpful.

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