How do I set a ViewModel on a window in XAML using DataContext property?

Question

The question pretty much says it all.

I have a window, and have tried to set the DataContext using the full namespace to the ViewModel, but I seem to be doing something wrong.

<Window x:Class="BuildAssistantUI.BuildAssistantWindow"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    DataContext="BuildAssistantUI.ViewModels.MainViewModel">

Answer 1

Try this instead.

<Window x:Class="BuildAssistantUI.BuildAssistantWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        xmlns:VM="clr-namespace:BuildAssistantUI.ViewModels">
    <Window.DataContext>
        <VM:MainViewModel />
    </Window.DataContext>
</Window>

Answer 2

In addition to the solution that other people provided (which are good, and correct), there is a way to specify the ViewModel in XAML, yet still separate the specific ViewModel from the View. Separating them is useful for when you want to write isolated test cases.

In App.xaml:

<Application
    x:Class="BuildAssistantUI.App"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    xmlns:local="clr-namespace:BuildAssistantUI.ViewModels"
    StartupUri="MainWindow.xaml"
    >
    <Application.Resources>
        <local:MainViewModel x:Key="MainViewModel" />
    </Application.Resources>
</Application>

In MainWindow.xaml:

<Window x:Class="BuildAssistantUI.MainWindow"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    DataContext="{StaticResource MainViewModel}"
    />

Answer 3

You need to instantiate the MainViewModel and set it as datacontext. In your statement it just consider it as string value.

     <Window x:Class="BuildAssistantUI.BuildAssistantWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        xmlns:local="clr-namespace:BuildAssistantUI.ViewModels">
      <Window.DataContext>
        <local:MainViewModel/>
      </Window.DataContext>

Answer 4

There is also this much better way of specifying the viewmodel:

    using Wpf = System.Windows;

    public partial class App : Wpf.Application //your skeleton app already has this.
    {
        protected override void OnStartup( Wpf.StartupEventArgs e ) //add this.
        {
            base.OnStartup( e );
            MainWindow = new MainView();
            MainWindow.DataContext = new MainViewModel( e.Args );
            MainWindow.Show();
        }
    }

The above mechanism applies to the main viewmodel only. To address a comment below which is asking what about user controls, the mechanism translates as follows for child viewmodels:

    public class ParentViewModel
    {
        public MyChildViewModel ChildViewModel { get; }
    
        public ParentViewModel()
        {
            ChildViewModel = new MyChildViewModel( ... );
        }
    }

ParentView.xaml:

    [...]
        xmlns:local="clr-namespace:the-namespace-of-my-wpf-stuff"
    [...]
        <local:MyChildView DataContext="{Binding ChildViewModel}" />
    [...]

^{<Rant>All of the solutions previously proposed require viewmodels to have parameterless constructors. Microsoft is under the impression that systems can be built using parameterless constructors. If you are also under that impression, the please, by all means, do go ahead and use some of the other solutions. For those who understand that constructors must have parameters, and therefore the instantiation of objects cannot be left in the hands of magic frameworks, the proper way of specifying the viewmodel is the one I showed above.</Rant>}

Answer 5

You might want to try Catel. It allows you to define a DataWindow class (instead of Window), and that class automatically creates the view model for you. This way, you can use the declaration of the ViewModel as you did in your original post, and the view model will still be created and set as DataContext.

See this article for an example.

Answer 6

Setting it in the xaml did not work for me. I reviewed my code over and over and everthing looked fine. What solved my problem was setting the DataContext in the WindowIsLoaded event handler:

PrintDrsAnywayViewModel TheWindowsViewModel = new PrintDrsAnywayViewModel();

private void MyWindowIsLoaded(object sender, RoutedEventArgs e)
{
    DataContext = TheWindowsViewModel;
    TheWindowsViewModel.WindowHeaderMessage = "snot";
}

The xaml, which was the same before and after it worked is this:

<TextBlock Text="{Binding WindowHeaderMessage, Mode=OneWay}"
          Margin="20,20,0,20" FontSize="20" />

WindowHeaderMessage is a string property in the viewmodel file which implements INotifyPropertyChanged

hth