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Home/ Questions/Q 8649225
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T13:30:30+00:00 2026-06-12T13:30:30+00:00

I have a FlipView with some controls in the DataTemplate. Since it’s in a

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I have a FlipView with some controls in the DataTemplate. Since it’s in a FlipView, it will generate multiple copies of those controls. If I want to find the control that is in the current FlipView view (i.e. on screen / visible), how can I do it?

I can check the loaded event on the control inside but it will be called multiple times and I won’t know which one is being shown.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-12T13:30:31+00:00Added an answer on June 12, 2026 at 1:30 pm

    Create an Attached Dependency Property that you can bind to the RichTextBlock from the ViewModel, such as:

    public static class MyStaticClass
    {
        public static readonly DependencyProperty IsVisible = DependencyProperty.RegisterAttached("IsVisible", typeof(bool), typeof(MyStaticClass), new PropertyMetadata(false, OnVisibilityChanged));
    
        private static void OnVisibilityChanged(DependencyObject d, DependencyPropertyChangedEventArgs e)
        {
            var rtb = (RichTextBlock)d;
            var isVisible = (bool)e.NewValue;
    
            // Do something to rtb.Inlines  
        }
    }
    

    With this property, you can bind it to the ViewModel’s IsSelected property:

    <FlipView ItemsSource="{Binding SomeList}" SelectedItem="{Binding SelectedVM, Mode=TwoWay}">
        <FlipView.ItemTemplate>
            <DataTemplate>
                <RichTextBlock ns:MyStaticClass.IsVisible="{Binding IsSelected}" />
            </DataTemplate>
        </FlipView.ItemTemplate>
    </FlipView>
    

    When the SelectedItem is changed, you can set IsSelected on the child View Model to true, and that triggers the MyStaticClass.OnVisibilityChanged event.

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