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

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}"
    />