How do I set WPF xaml form's Design DataContext to class that uses generic type parameters

Tod picture Tod · Nov 23, 2011 · Viewed 7.4k times · Source

Originally my .xaml form used the following line to set the Designer's DataContext where the view model was a non-generic type (note I'm talking about the Design time DataContext not the actual DataContext that will be used at runtime).

<Window ...
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"    
d:DataContext="{d:DesignInstance Dialogs:CustomerSearchDlogViewModel}"
...>

Now instead of CustomerSearchDlogViewModel I have a generic SearchDialogViewModel but I can't figure out what syntax will work in the <Window> tag to let me specify that view model.

Answer

H.B. picture H.B. · Nov 23, 2011

That is not possible unless the markup extension (DesignInstance) provides properties to pass type arguments, which i doubt. So you might want to subclass as suggested or write your own markup extension which creates generic instances (in fact that is what i am doing right now).

Edit: This extension should do it:

public class GenericObjectFactoryExtension : MarkupExtension
{
    public Type Type { get; set; }
    public Type T { get; set; }

    public override object ProvideValue(IServiceProvider serviceProvider)
    {
        var genericType = Type.MakeGenericType(T);
        return Activator.CreateInstance(genericType);
    }
}

Initially i had some problems getting the object type from a type-name but you can let the XAML parser resolve the type for you which is neat:

DataContext="{me:GenericObjectFactory Type={x:Type Dialogs:CustomerSearchDlogViewModel`1},
                                      T=Data:Customer}"

(Note the `1 at the end to reference a generic type. If you drop the x:Type wrapping the backtick will cause an error.)