ViewModel not activated from ActivateItem()

Question

When starting the application I'd like to open by default a user control, but only the empty shell view displayed.

Following the usual Caliburn pattern and using SimpleContainer in the ShellViewModel constructor everything is executed fine, but can't see the activated vm.

The ShellViewModel

public class ShellViewModel : Conductor<object>, IHandle<NewLayoutCreatedEvent>
    {
        private readonly SimpleContainer _container;
        private readonly IEventAggregator _events;
        private readonly IWindowManager _manager;

        public ShellViewModel(IEventAggregator events, IWindowManager manager, SimpleContainer container)
        {
            _events = events;
            _events.Subscribe(this);

            _manager = manager;
            _container = container;

            ActivateItem(_container.GetInstance<WorkLayoutViewModel>());
        }
}

The WorkLayoutViewModel is inheriting from Screen. In the Bootstrapper class what happens is after running the constructor of WorkLayoutViewModel:

protected override object GetInstance(Type service, string key)
{
    return _container.GetInstance(service, key);
}

called once with null key and service = ShellViewModel then continue with:

protected override IEnumerable<object> GetAllInstances(Type service)
{
    return _container.GetAllInstances(service);
}

where service again ShellViewModel, which runs the InitializeComponent() in the ShellView class. After this jumping to:

protected override void OnStartup(object sender, StartupEventArgs e)
{
    DisplayRootViewFor<ShellViewModel>();
}

where the sender is the app itself and no args. Following this the App class OnStartup() override called where I'm setting up language and calling base.OnStartup()

I believe the GetAllInstances should call the WorkLayoutViewModel too not just the ShellViewModel, this is not happening and the view is not displayed.

As requested the ShellView:

<Window x:Class="MyApp.Views.ShellView"
        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"
        xmlns:local="clr-namespace:MyApp.Views"
        xmlns:lang="clr-namespace:MyApp.Language;assembly=MyApp.Language"
        xmlns:interact="clr-namespace:System.Windows.Interactivity;assembly=System.Windows.Interactivity"
        xmlns:cal="http://www.caliburnproject.org"
        mc:Ignorable="d"
        Title="{x:Static lang:Resources.AppName}" Height="800" Width="1000" WindowStartupLocation="CenterScreen">
    <DockPanel>
        <StackPanel DockPanel.Dock="Top">
            <Menu FontSize="14" Height="27">
                <Menu.ItemsPanel>
                    <ItemsPanelTemplate>
                        <DockPanel HorizontalAlignment="Stretch" />
                    </ItemsPanelTemplate>
                </Menu.ItemsPanel>
                <MenuItem Header="{x:Static lang:Resources.Mnu_File}" HorizontalAlignment="Left">
                    <MenuItem Header="{x:Static lang:Resources.Mnu_New}">
                        <MenuItem.Icon>
                            <Image Source="/Images/new.png" />
                        </MenuItem.Icon>
                        <interact:Interaction.Triggers>
                            <interact:EventTrigger EventName="Click">
                                <cal:ActionMessage MethodName="NewLayout"/>
                            </interact:EventTrigger>
                        </interact:Interaction.Triggers>
                    </MenuItem>
                </MenuItem>
            </Menu>

            <ToolBar Height="36">
                <Button ToolTip="{x:Static lang:Resources.Tip_New}">
                    <Image Source="/Images/new.png" />
                    <interact:Interaction.Triggers>
                        <interact:EventTrigger EventName="Click">
                            <cal:ActionMessage MethodName="NewLayout"/>
                        </interact:EventTrigger>
                    </interact:Interaction.Triggers>
                </Button>
            </ToolBar>
        </StackPanel>
        <Grid>
            <ContentControl x:Name="ActivateItem" Margin="10 5 10 10" />
        </Grid>
    </DockPanel>
</Window>

and the WorkLayoutView is just a standard UserControl with red back ground.

Are you trying to open a window by default? Display a control inside a window by default? — David Bentley
The windowmanager is for a popup dialog to add a new workitem, but initially I'd like to display a usercontrol. Also the ShellView contains a Dockpanel for the menu and toolbar. — Paxsentry
When you say you would like to display a usercontrol, do you mean a new window with a usercontrol or the mainwindow with a usercontrol. Just saying you want to display a user control tells us nothing. — David Bentley
I'd like to open the WorkLayoutView (the usercontrol) inside the ShellView (main window). — Paxsentry
Ok. Well there are actually lots of ways to do this and all are dependent on how you have your window code (XAML and C#). Post some code so we can maybe direct you into the right direction. — David Bentley

Jack Hughes Jack Hughes · Accepted Answer · 2019-07-26T14:10:46

Your ContentControl inside the window should probably be called ActiveItem not ActivateItem. ActiveateItem is the method that sets the ActiveItem property.

ViewModel not activated from ActivateItem()

1 Answers