Two dynamically assigned ContentControls in single view in Caliburn.Micro - silverlight

Question

I have a UserControl that contains two ContentControls that need to have different UserControl Views bound to them at runtime. The attached-Property solution noted here does not seem to work in Silverlight. Or, I am doing something wrong. I also, found this, but it did not bring any joy either.
I had a single ContentControl working by naming it 'ActiveItem'. But, of course, I cannot have two ContentControls with the same name.
Thanks in advance for any help,
Jim

Answer 1

Just expose two public properties on your main view model, each one being an instance of the type of view model you wish to display. Then, in your view have a ContentControl with the corresponding name. E.g:
public class MyMainViewModel
{
private NavigationViewModel navigation;
private MyContentViewModel main;
public MyMainViewModel()
{
// better to inject factories using constructor injection here
this.Navigation = new NavigationViewModel();
this.Main = new MyContentViewModel();
}
public NavigationViewModel Navigation
{
get { return navigation; }
set { navigation= value; NotifyOfPropertyChanged(() => this.Navigation); }
}
public MyContentViewModel Main
{
get { return main; }
set { main= value; NotifyOfPropertyChanged(() => this.Main); }
}
...
}
<ContentControl x:Name="Navigation" />
...
<ContentControl x:Name="Main" />

Answer 2

This is an old question, but in case anyone is having the same issue, I post here my way of handling it from the beginning and in a more thorough manner:
Your main window that contain both (or even more than two) of your User Controls must be inherited from Caliburn.Micro.Conductor<Screen>.Collection.AllActive;
Your User Controls must be inherited from Caliburn.Micro.Screen;
You must also keep naming conventions in mind. If you use MenuUC as the name of a ContentControl in your View, also create a property named MenuUC in your ViewModel;
Initialize your UserControl as I do in Constructor;
Now you can use ActivateItem(MenuUC) and DeactivateItem(MenuUC) everywhere in your code. Caliburn.Micro automatically detects which one you want to work with.
Example XAML View code:
<Window x:Class="YourProject.Views.YourView"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
mc:Ignorable="d"
Title="YourViewTitle" Width="900" Height="480">
<Grid>
<Grid.ColumnDefinitions>
<ColumnDefinition Width="*"/>
<ColumnDefinition Width="4*"/>
</Grid.ColumnDefinitions>
<Grid.RowDefinitions>
<RowDefinition Height="auto"/>
<RowDefinition Height="*"/>
</Grid.RowDefinitions>
<!-- Menu Side Bar -->
<ContentControl Grid.Row="0" Grid.Column="0" x:Name="MenuUC" />
<!-- Panel -->
<Border Grid.Column="1" Grid.RowSpan="2" BorderThickness="1,0,0,0" BorderBrush="#FF707070" >
<ContentControl x:Name="PanelUC" />
</Border>
</Grid>
</Window>
Example C# ViewModel code:
class YourViewModel : Conductor<Screen>.Collection.AllActive
{
// Menu Side Bar
private MenuUCViewModel _menuUC;
public MenuUCViewModel MenuUC
{
get { return _menuUC; }
set { _menuUC = value; NotifyOfPropertyChange(() => MenuUC); }
}
// Panel
private Screen _panelUC;
public Screen PanelUC
{
get { return _panelUC; }
set { _panelUC = value; NotifyOfPropertyChange(() => PanelUC); }
}
// Constructor
public YourViewModel()
{
MenuUC = new MenuUCViewModel();
ActivateItem(MenuUC);
PanelUC = new FirstPanelUCViewModel();
ActivateItem(PanelUC);
}
// Some method that changes PanelUC (previously FirstPanelUCViewModel) to SecondPanelUCViewModel
public void ChangePanels()
{
DeactivateItem(PanelUC);
PanelUC = new SecondPanelUCViewModel();
ActivateItem(PanelUC);
}
}
In the above example, ChangePanels() acts as a method to load new User Control into your ContentControl.
Also read this question, it might be help you further.