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

101
votes

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">
5
.netwpfxamlmvvm
Following up on Mike Nakis, I was trying to create the ViewModel manually and subscribe to events in it, only to find that the framework was creating another ViewModel. Hence the viewModel I was subscribed to was not the one attached to the view. – jlady
Does this mean that besides instantiating the viewmodel yourself, you were also specifying the type of the viewmodel in some other way ? A secondary advantage of viewmodels requiring constructor parameters is that the framework either cannot instantiate them, or it has to pass default values for those parameters, in which case you can easily detect instantiation by the framework. – Mike Nakis
The XAML designer might also need to be able to instantiate viewmodels, but this designer was never of any usefulness to me, (it just causes problems,) so I don't use it, so I personally do not care about that usage case. – Mike Nakis
DataContext="{Binding Source={x:Type BuildAssistantUI.ViewModels.MainViewModel}}" can we use like this {x:Type}?.. but, it is not working. – Vintage Coder

5 Answers

157
votes

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>
120
votes

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}"
    />
11
votes

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>
4
votes

There is also this 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 ) //you need to add this.
    {
        base.OnStartup( e );
        MainWindow = new MainView();
        MainWindow.DataContext = new MainViewModel( e.Args );
        MainWindow.Show();
    }
}

<Rant>

All of the solutions previously proposed require that MainViewModel must have a parameterless constructor.

Microsoft is under the impression that systems can be built using parameterless constructors. If you are also under that impression, go ahead and use some of the other solutions.

For those that know 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>

3
votes

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.