CODEWITHSUNDEEP
wpflayoutgriditemscontrol
Ben
Ben

Reputation: 1032

Dynamically Display Grid Content

I have 3 different layouts (similar to I guess what you would call Skins but the layouts are hugely different, not just changes to colors and fonts) which I have developed for my application. The layouts are used for displaying the same data, but in a completely different format. Each of these layouts have been constructed within their own Grid.

I want my application to decide which layout to display dynamically based on a string value available at runtime.

What's the best way to get a parent Grid to display a Child Grid dynamically?

I'm trying to find some sort of magical DataTemplate / DataBinding / Templating method but just can't seem to find the best way. Alternatively, should I be looking at a different method of displaying these different layouts? Like an ItemsControl or similar?

Ben

Upvotes: 1

Views: 436

Answers (1)

Rachel
Rachel

Reputation: 132618

I usually use a ContentControl and DataTrigger to determine what ContentTemplate to use.

For example,

<ContentControl Content="{Binding MyViewModel}">

    <ContentControl.Resources>
        <DataTemplate x:Key="DefaultTemplate">
            <TextBlock Text="DefaultTemplate" />
        </DataTemplate>

        <DataTemplate x:Key="TemplateA">
            <TextBlock Text="Template A" />
        </DataTemplate> 

        <DataTemplate x:Key="TemplateB">
            <TextBlock Text="Template B" />
        </DataTemplate>
    </ContentControl.Resources>

     <ContentControl.Style>
         <Style TargetType="{x:Type ContentControl}">
             <Setter Property="ContentTemplate" Value="{StaticResource DefaultTemplate}" />
             <Style.Triggers>
                 <DataTrigger Binding="{Binding SelectedView}" Value="ViewA">
                     <Setter Property="ContentTemplate" Value="{StaticResource TemplateA}" />
                 </DataTrigger>
                 <DataTrigger Binding="{Binding SelectedView}" Value="ViewB">
                     <Setter Property="ContentTemplate" Value="{StaticResource TemplateB}" />
                 </DataTrigger>
             </Style.Triggers>
         </Style>
     </ContentControl.Style>

 </ContentControl>

Upvotes: 1

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