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    lt 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  gt

User · Answer

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>

User · Answer

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

User · Answer

Try this instead    lt 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  gt       lt Window DataContext gt           lt VM MainViewModel   gt       lt  Window DataContext gt   lt  Window gt

User · Answer

You need to instantiate the MainViewModel and set it as datacontext  In your statement it just consider it as string value         lt 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  gt         lt Window DataContext gt           lt local MainViewModel  gt         lt  Window DataContext gt

User · Answer

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    lt 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       gt       lt Application Resources gt           lt local MainViewModel x Key  MainViewModel    gt       lt  Application Resources gt   lt  Application gt    In MainWindow xaml    lt 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         gt

[.net] How do I set a ViewModel on a window in XAML using DataContext property?

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