如何使用 DataContext 属性在 XAML 窗口上设置 ViewModel？

这个问题几乎说明了一切。

我有一个窗口，并试图使用完整的名称空间将 DataContext 设置为 ViewModel，但我似乎做错了什么。

<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">
166297 次浏览
小开

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>
小开

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>
小开
最佳答案

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}"
/>
小开

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.

小开

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>


提交答案