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

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}"
    />

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>

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>

Tags:

.Net

Wpf

Xaml

Mvvm