如何使用MEF初始化viewModel？

Question

我有一个名为ModuleMenu的模块.在这个模块中,我有一个名为MenuView的UserControl和一个名为UserControlViewModel的Corresponding ViewModel.我还有一个名为Module的类.所有代码如下:

MenuView.xmal

<UserControl ..............>

    <ListBox ItemsSource="{Binding MenuItems, Converter={StaticResource dummy}}" DisplayMemberPath="MenuItemName" SelectedItem="{Binding SelectedMenuItem}" >

        <ListBox.ItemsPanel>
            <ItemsPanelTemplate>
                <StackPanel IsItemsHost="True" Orientation="Horizontal"/>
            </ItemsPanelTemplate>
        </ListBox.ItemsPanel>

        <ListBox.Resources>
            <Style TargetType="ListBoxItem">
                <Setter Property="Margin" Value="10,0" />
            </Style>
        </ListBox.Resources>

    </ListBox>

</UserControl>

MenuView.xaml.cs

[Export]
[PartCreationPolicy(CreationPolicy.NonShared)]
public partial class MenuView : UserControl
{
    public MenuView()
    {
        InitializeComponent();
    }
}

UserControlViewModel.cs

[Export]
[PartCreationPolicy(CreationPolicy.NonShared)]
public class MenuViewModel : ViewModelBase
{
    IServiceFactory _ServiceFactory;

    [ImportingConstructor]
    public MenuViewModel(IServiceFactory serviceFactory)
    {
        _ServiceFactory = serviceFactory;
    }

    protected override void OnViewLoaded()
    {
        _MenuItems = new ObservableCollection<MenuItem>();

        WithClient<IMenuItemService>(_ServiceFactory.CreateClient<IMenuItemService>(), menuItemClient =>
            {
                MenuItem[] menuItems = menuItemClient.GetAllParentMenuItemsWithChildren();
                if (menuItems != null)
                {
                    foreach (MenuItem menuItem in menuItems)
                    {
                        _MenuItems.Add(menuItem);
                    }

                    _SelectedMenuItem = _MenuItems[2];
                }

            });
    }

    private ObservableCollection<MenuItem> _MenuItems;

    public ObservableCollection<MenuItem> MenuItems
    {
        get
        {
            return _MenuItems;
        }
        set
        {
            if (_MenuItems != value)
            {
                _MenuItems = value;
                OnPropertyChanged(() => MenuItems, false);
            }
        }
    }

    private MenuItem _SelectedMenuItem;

    public MenuItem SelectedMenuItem
    {
        get
        {
            return _SelectedMenuItem;
        }
        set
        {
            if (_SelectedMenuItem != value)
            {
                _SelectedMenuItem = value;
                OnPropertyChanged(() => SelectedMenuItem);
            }
        }
    }

}

Module.cs

[ModuleExport(typeof(Module), InitializationMode=InitializationMode.WhenAvailable)]
public class Module : IModule
{
    IRegionManager _regionManager;

    [ImportingConstructor]
    public Module(IRegionManager regionManager)
    {
        _regionManager = regionManager;
    }

    public void Initialize()
    {
        _regionManager.Regions[RegionNames.MenubarRegion].Add(ServiceLocator.Current.GetInstance<MenuView>());
    }
}

现在在我的主项目中,我有一个名为BootStrapper.cs的类,如下所示:

public class Bootstrapper : MefBootstrapper
{
    protected override DependencyObject CreateShell()
    {
        return Container.GetExportedValue<Shell>();
    }

    protected override void InitializeShell()
    {
        base.InitializeShell();
        App.Current.MainWindow = (Window)Shell;
        App.Current.MainWindow.Show();
    }

    protected override void ConfigureAggregateCatalog()
    {
        base.ConfigureAggregateCatalog();
        AggregateCatalog.Catalogs.Add(new AssemblyCatalog(typeof(Bootstrapper).Assembly));
        AggregateCatalog.Catalogs.Add(new AssemblyCatalog(typeof(ModuleMenu.Module).Assembly));
    }
}

在我的App.xaml中:

<Application ..............>
    <Application.Resources>
        <DataTemplate DataType="{x:Type modMenu:MenuViewModel}">
            <modMenu:MenuView />
        </DataTemplate>
    </Application.Resources>
</Application>

最后在App.xaml.cs中:

public partial class App : Application
{
    protected override void OnStartup(StartupEventArgs e)
    {
        base.OnStartup(e);

        Bootstrapper bootstrapper = new Bootstrapper();
        bootstrapper.Run();

    }
}

当我运行应用程序时,我按预期获得shell.它显示了我的MenuView,但未加载MenuView中的数据.我尝试使用虚拟转换器调试它,并显示viewModel永远不会被初始化.

那么,现在我的问题是如何初始化viewModel？

更新:

在尝试您的代码后,我得到如下异常:

尝试将视图添加到区域'MenubarRegion'时发生异常.

- The most likely causing exception was was: 
    'Microsoft.Practices.ServiceLocation.ActivationException: Activation  
    error occured while trying to get instance of type MenuView, key "" ---> 
    Microsoft.Practices.ServiceLocation.ActivationException: Activation 
    error occured while trying to get instance of type MenuView, key ""

at Microsoft.Practices.Prism.MefExtensions.MefServiceLocatorAdapter
                  .DoGetInstance(Type serviceType, String key)

at Microsoft.Practices.ServiceLocation.ServiceLocatorImplBase
                  .GetInstance(Type serviceType, String key)

--- End of inner exception stack trace ---

at Microsoft.Practices.ServiceLocation.ServiceLocatorImplBase
                  .GetInstance(Type serviceType, String key)

at Microsoft.Practices.ServiceLocation.ServiceLocatorImplBase
                  .GetInstance(Type serviceType)

at Microsoft.Practices.Prism.Regions.RegionViewRegistry
                  .CreateInstance(Type type)

at Microsoft.Practices.Prism.Regions.RegionViewRegistry
                  .<>c__DisplayClass1.<RegisterViewWithRegion>b__0()

at Microsoft.Practices.Prism.Regions.Behaviors
                  .AutoPopulateRegionBehavior
                  .OnViewRegistered(Object sender, ViewRegisteredEventArgs e)'.

 But also check the InnerExceptions for more detail or call 
                  .GetRootException().

当我看一下内部异常时,我收到以下错误消息:

{"Activation error occured while trying to get instance of type MenuView, key \"\""}

UPDATE2:

这是ServiceFactory类型的Export:

[Export(typeof(IServiceFactory))]
[PartCreationPolicy(CreationPolicy.NonShared)]
public class ServiceFactory : IServiceFactory
{
    public T CreateClient<T>() where T : IServiceContract
    {
        return ObjectBase.Container.GetExportedValue<T>();
    }
}

Answer 1

您将DataTemplateview 定义为viewmodel的视图,但实际上并未使用它.

使用Prism有很多方法可以解决您的问题,请参阅此主题.

您可以DataContext在XAML中设置视图的属性:

<UserControl.DataContext>
    <my:MyViewModel/>
</UserControl.DataContext>

您可以在视图的构造函数中创建一个viewmodel:

public MyView()
{
    InitializeComponent();
    this.DataContext = new MyViewModel();
}

更好的方法是通过依赖注入导入viewmodel:

[ImportingConstructor]
public MyView(MyViewModel viewModel)
{
    InitializeComponent();
    this.DataContext = viewModel;
}

您可以使用Prism的viewmodel位置服务:

<MyView prism:ViewModelLocator.AutoWireViewModel="True"/>

最后但并非最不重要:您也可以使用DataTemplate,但您应该设置一个DataContext对象让WPF为您创建视图,而不是在代码中创建视图.

更新: 所以,你想使用该DataTemplate功能.嗯,这不是"棱镜"做事的方式,因为在这种情况下,视图将由WPF创建,而不是由Prism创建IRegion.但无论如何这都是可能的.

我将解释其中的区别:在所有其他(Prism)方法中,视图是'master',它将首先创建,然后将创建一个适当的viewmodel并附加到视图.在Prism中,您可以定义应创建的视图,何时(导航)和位置(区域).在该DataTemplate方法中,viewmodel(data)是'master',它将首先创建,WPF确定如何显示它们创建视图.所以在这种情况下,你不能使用Prism的区域和导航,因为WPF处理所有的事情.

所以我真的建议你使用上面的任何方法,但不是DataTemplate一个.

您可以按如下方式创建视图:

_regionManager.Regions[RegionNames.MenubarRegion].Add(ServiceLocator.Current.GetInstance<MenuView>());

我建议你把它改成:

_regionManager.RegisterViewWithRegion(RegionNames.MenubarRegion, typeof(MenuView));

ServiceLocator直接使用不是一个好的模式(除非你无法避免),所以让Prism为你实例化视图.

在视图的构造函数中,只需添加viewmodel的依赖项注入,并将视图设置DataContext为它.瞧!你有它.

[Export]
[PartCreationPolicy(CreationPolicy.NonShared)]
public partial class MenuView : UserControl
{
    [ImportingConstructor]
    public MenuView(MenuViewModel viewModel)
    {
        this.InitializeComponent();
        this.DataContext = viewModel;
    }
}

使用DataTemplate,你只能用'旧的普通WPF方式'.您必须手动创建viewmodel的实例并将其作为父(shell的)viewmodel的属性公开(或使其静态,但这是一种糟糕的方法).

<Window>
  <ContentControl Content="{Binding MenuViewModelInstace}"/>
</Window>

然后,WPF将为您创建视图以显示视图模型,但您应该直接在XAML标记中定义它,如前所述.棱镜无法帮助你.