Pass ViewModel to UserControl

Question

MySpecialView is a complex image control, I would like to reuse it from different views and pass its ViewModel as in this example.

MainWindow.x

Answer 1

You are too used to "View-First-Approach" (VFA). Your situation is better off using "ViewModel-First-Approach" (VMFA). In VFA, you place your child views from the main View, and each subview is linked to the respective ViewModel via DataContext.

In VMFA, your ViewModel holds references of sub-ViewModels. You expose these ViewModel references through property binding, and the View display them via DataTemplate.

MainWindow.xaml

<Window x:Class="YouBug.MainWindow"
    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:YouBug"
    mc:Ignorable="d"
    DataContext="{Binding MainViewModel}"
    Title="MainWindow" Height="350" Width="525">
<Grid>
    <Grid.Resources>
        <DataTemplate DataType="{x:Type vm:MySpecialViewModel}">
            <local:MySpecialViewModel />
        </DataTemplate>
    </Grid.Resources>
    <ContentControl Content={Binding MySpecialView}" />
</Grid>

MainViewModel

public class MainViewModel : ViewModelBase // Not sure why you didn't subclass ViewModelBase in your question
{
    private MySpecialViewModel _mySpecialViewModel;
    public MySpecialViewModel MySpecialViewModel
    {
        get
        {
            return _mySpecialViewModel;
        }
        set
        {
            if (value != _mySpecialViewModel)
            {
                _mySpecialViewModel = value;
                RaisePropertyChanged(); // The property changed method call
            }
        }
    }

    public MainViewModel()
    {
        MySpecialViewModel = new MySpecialViewModel();
        //gets not displayed!
        Task.Run(() => MySpecialViewModel.changeImage(5000, "C:\\Users\\user\\Pictures\\Capture.PNG"));
    }
}

MySpecialView does not need that DependencyProperty, nor set the DataContext. The DataContext is set automatically by the DataTemplate part. Your MySpecialViewModel can stay as it is now.

Edit

I just realized your MainWindow is not doing the DataContext correctly either.

MainWindow.xaml.cs

public partial class MainWindow: Window
{
    public MainWindow()
    {
        InitializeComponents();
        this.DataContext = new MainViewModel();
    }
}

Answer 2

Do not specify viewmodel property in you view, use DataContext. See the following code.

public partial class MySpecialView : UserControl
{

    public MySpecialView()
    {
        InitializeComponent();
    }
}

ViewModel for special:

public class MySpecialViewModel : ViewModelBase
{
    public BitmapSource imageSource { get; set; }
    public BitmapSource ImageSource { get { return imageSource; }
        set { if (value != imageSource)
            {
                imageSource = value;
 RaisePropertyChanged("ImageSource");
            }
        } }

    public MySpecialViewModel()
    {
        //gets displayed
        ImageSource = new BitmapImage(new Uri("C:\\Users\\user\\Pictures\\test.jpg"));

        //gets displayed aswell
        Task.Run(() => changeImage(10000, "C:\\Users\\user\\Pictures\\clickMe.png"));
    }

    public async void changeImage(int sleep, string uri)
    {
        await Task.Delay(sleep);
        BitmapSource source = new BitmapImage(new Uri(uri));
        source.Freeze();
        ImageSource = source;
    }

}

In XAML special:

    <UserControl x:Class="YouBug.MySpecialView"
             xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
             xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
             xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006" 
             xmlns:d="http://schemas.microsoft.com/expression/blend/2008" 
             xmlns:local="clr-namespace:YouBug"
             mc:Ignorable="d" 
             d:DesignHeight="300" d:DesignWidth="300">
    <Grid>
        <Image Source="{Binding ImageSource}" />
    </Grid>

For main:

public class MainViewModel : ViewModelBase
{
    public MySpecialViewModel SpecialViewModel
    { 
        get { return _specialViewModel; }
        set
        {
          if (value != _specialViewModel)
          {
             _specialViewModel= value;
             RaisePropertyChanged("SpecialViewModel");
          }
        } 
     }

    private MySpecialViewModel _specialViewModel;

        public MainViewModel()
        {
            MySpecialViewModel = new MySpecialViewModel();
            //gets not displayed!
            Task.Run(() => MySpecialViewModel.changeImage(5000, "C:\\Users\\user\\Pictures\\Capture.PNG"));
        }
    }

And in XAML:

<Window x:Class="YouBug.MainWindow"
    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:YouBug"
    mc:Ignorable="d"
    Title="MainWindow" Height="350" Width="525">
<Window.DataContext>
    <MainWindowViewModel/
</Window.DataContext>
<Grid>
    <local:MySpecialView DataContext="{Binding Path=SpecialViewModel}"></local:MySpecialView>
</Grid>