CODEWITHSUNDEEP
wpfxamlmvvmcaliburn.micro
superjos
superjos

Reputation: 12705

How to show ContentControl in designer when using Calibrun.Micro viewModel-first approach?

I'm using Caliburn.Micro (CM) in a WPF application with ViewModel-first approach. I'm composing the main view with a command bar and an active item. Main viewModel sets the property for the command bar viewModel, and navigates to active item correctly.

Everything looks fine at runtime, the issue is only related to design-time: the main view shows empty in designer and I cannot find how to set it correctly. I managed to having this working in other scenarios, e.g. when setting the datacontext at design time for a whole Window or UserControl, i.e. when that's the root UI element in XAML. But now I'm not able to to this for child ContentPresenter UI elements within a Window.

This is an excerpt of the main view I'm composing:

<Window x:Class="...MainView" ...>
  <DockPanel ...>
    <!-- this one binds to a property of type CommandBarViewModel -->
    <ContentControl x:Name="CommandBar" ... />
    <ContentControl x:Name="ActiveItem" ... />
  </DockPanel>
</Window>

I've checked a number of related reads, but none of them seems to fit/solve my issue. This question is basically the same as mine, but has no answers. That has a reference to this other question which it seems to me is going for a View-first approach, judging by the cal:View.Model bindings.

I tried adding a design-time context like the following (fake namespace not shown for brevity):

<ContentControl x:Name="CommandBar" ... 
   d:DataContext="{d:DesignInstance Type=fake:DesignTimeCommandBarViewModel, IsDesignTimeCreatable=True}" 
   cal:Bind.AtDesignTime="True"/>

but then I incur in one of two cases:

Have you got any idea about this? Maybe this should be solved with a design-time version of the hosting MainViewModel?

Other references I checked: this answer, from Rob Eisenberg himself, this CM thread, this other SO

Edit
Following my last (auto-)hint, I'm trying also creating and instantiating a DesignTimeMainViewModel, not inheriting from MainViewModel, which exposes the same properties and sets a DesignTimeCommandBarViewModel in its default constructor. In this case, in place of the command bar the designer shows the classic CM complaint: cannot find view for the DesignTimeCommandBarViewModel. What's next?

Upvotes: 3

Views: 1822

Answers (1)

superjos
superjos

Reputation: 12705

Well, here's the solution I found: I'd be glad to hear about better ways or other suggestions.

Host MainView XAML specifies a design-time data-context pointing to a design-time version of the Main view-model which, by the way, does not inherit from the runtime version MainViewModel. ContentControl items are left untouched.

<Window x:Class="...MainView" ...
  d:DataContext="{d:DesignInstance Type=fake:DesignTimeMainPanelViewModel, IsDesignTimeCreatable=True}"
  cal:Bind.AtDesignTime="True">

  <DockPanel ...>
    <ContentControl x:Name="CommandBar" ... />
    <ContentControl x:Name="ActiveItem" ... />
  </DockPanel>

</Window>

DesignTimeMainPanelViewModel has the same public properties as MainPanelViewModel, has a default c'tor without dependencies and its c'tor sets the CommandBar property to a new instance of DesignTimeCommandBarViewModel:

public class DesignTimeMainPanelViewModel
{
    public DesignTimeMainPanelViewModel()
    {
        CommandBar = new DesignTimeCommandBarViewModel();
        ActiveItem = ...some instance here as well...;
    }

    public DesignTimeCommandBarViewModel CommandBar { get; private set; }
    public IScreen ActiveItem { get; private set; }
}

DesignTimeCommandBarViewModel class is decorated with a custom Attribute having only one required parameter, the System.Type of the view associated with that view-model.

During bootstrap the code adds a new ViewLocator strategy to get the view Type from the view-model Type, by setting a new ViewLocator.LocateTypeForModelType.

The new locator function will try to find a view Type if the standard locator function cannot find one. Granted, it will look for the custom attribute on view-model Type, and if found that would be the returned view Type. Here's the gist of that:

Type viewType = _previousLocate(viewModelType, displayLocation, context);

if (viewType == null)
{
    FakeViewAttribute fakeViewAttr = Attribute.GetCustomAttribute(viewModelType, typeof(FakeViewAttribute)) as FakeViewAttribute;

    if (fakeViewAttr != null) viewType = fakeViewAttr.ViewType;
}

return viewType;

Upvotes: 2

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