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Home/ Questions/Q 1039501
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T15:06:28+00:00 2026-05-16T15:06:28+00:00

I wanted to do a quick user control for my app, but to keep

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I wanted to do a quick user control for my app, but to keep things in a MVVM style I thought I’d set the DataContext of the XAML to the code behind in my UserControl.

i.e.

DataContext="{Binding RelativeSource={RelativeSource Self}}"

This allows me to bind the XAML to properties in my code behind.

Everything went well until I came to bind the Visibility of an instance of the control to a Visibility property on a ViewModel.

<Controls:RoundProgress Visibility="{Binding ProgressVisibility}" Width="100" Height="100"></Controls:RoundProgress>

The Visibility no longer works – if I remove my tinkerings with the DataContext from the User Control – the visibility works!

Can someone set me right please? Thanks

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  1. Editorial Team
    Editorial Team
    2026-05-16T15:06:28+00:00Added an answer on May 16, 2026 at 3:06 pm

    Don’t set the DataContext of the UserControl itself from the internal XAML. By doing that you override the inherited DataContext and make your Binding look for a ProgressVisibility property on the UC instead of your ViewModel. Instead set the DataContext on an element inside the UC:

    <UserControl x:Class...>
      <Grid DataContext="{Binding RelativeSource={RelativeSource Mode=FindAncestor, AncestorType={x:Type UserControl}}}">
        ...
      </Grid>
    </UserControl>
    
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