How to register message handler prior to ShowDialog() blocking call?

问题

I'm using a Messenger class in order to send data between view models. There is an AppView that hosts two main views in a content control, and up until now have had no issue with sending/receiving data this way.

Issue:

Now I added a ProductView that shows a separate dialog to the AppView. But when I call Messenger.Default.Send<ProductModel>(SelectedProduct); after calling .ShowDetailDialog() this blocks the Send code call, until the dialog is closed.

I tried the other way around, calling the Send code first, then opening the dialog. But this means that the message handler in the receiving VM doesn't register in time before the message is sent.

Does anyone know of a solution, to prevent the dialog from blocking the send call? Or alternatively register the ProductVM message handler prior to sending message and showing dialog?

Below is a summary of the related classes:

CustomerOrdersVM (sending code):

    private void EditOrder(object obj)
    {
        _dialogService.ShowDetailDialog();    
        Messenger.Default.Send<ProductModel>(SelectedProduct);            
    }

ProductVM (receiving code):

    public ProductViewModel()
    {
        Messenger.Default.Register<ProductModel>(this, OnSelectedProductReceived);              
    }

DialogService:

class DialogService : IDialogService
{

    Window productView = null;

    public DialogService()
    {

    }


    public void ShowDetailDialog()
    {
         productView = new ProductView();
        productView.ShowDialog();
    }
}

AppVM (Main VM's are registered, ProductVM is independent of this VM):

    public ApplicationViewModel()
    {
        // Add available pages
        PageViewModels.Add(new CustomerDetailsViewModel(customerDataService, countryDataService, dialogService));
        PageViewModels.Add(new CustomerOrdersViewModel(orderDataService, dialogService));
        PageViewModels.Add(new OrderStatisticsViewModel());

        // Set starting page
        CurrentPageViewModel = PageViewModels[0];  
    }

AppView: (holds the AppVM views):

<Window x:Class="MongoDBApp.Views.ApplicationView"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        xmlns:views="clr-namespace:MongoDBApp.Views"
        xmlns:vm="clr-namespace:MongoDBApp.ViewModels">


    <Window.Resources>
        <DataTemplate DataType="{x:Type vm:CustomerDetailsViewModel}">
            <views:CustomerDetailsView />
        </DataTemplate>
        <DataTemplate DataType="{x:Type vm:CustomerOrdersViewModel}">
            <views:CustomerOrdersView />
        </DataTemplate>
        <DataTemplate DataType="{x:Type vm:OrderStatisticsViewModel}">
            <views:OrderStatisticsView />
        </DataTemplate>
    </Window.Resources>

    <Window.DataContext>
        <vm:ApplicationViewModel />
    </Window.DataContext>

    <Grid>
        <Grid.RowDefinitions>
            <RowDefinition Height=".07*" />
            <RowDefinition Height="1*" />
        </Grid.RowDefinitions>


        <TabControl Grid.Row="1"
                    ItemsSource="{Binding PageViewModels}"
                    SelectedItem="{Binding CurrentPageViewModel}"
                    TabStripPlacement="Top">
            <TabControl.ItemTemplate>
                <DataTemplate>
                    <TextBlock Text="{Binding Name}" />
                </DataTemplate>
            </TabControl.ItemTemplate>
            <TabControl.ItemContainerStyle>
                <Style TargetType="{x:Type TabItem}">
                    <Setter Property="IsEnabled" Value="{Binding IsEnabled}" />
                </Style>
            </TabControl.ItemContainerStyle>
        </TabControl>
    </Grid>
</Window>

回答1:

You could solve the problem a few ways:

1. Don't use ShowDialog(). Use Show(), and make the dialog window TopMost and parented to the main window.

2. Register the ProductView in the constructor of the DialogService (do you really need a new ProductView each time anyway?)

3. Make the DialogService (or a utility class inside of it) register at construction time for the message, and then pass the message on to any displayed ProductViews

Personally I like #2- since you are using ShowDialog, it implies that only one ProductView is ever needed at a time. For example:

class DialogService : IDialogService
{
    Window productView = null;
    ProductView _productView;

    public DialogService()
    {
         _productView = new ProductView();
    }

    public void ShowDetailDialog()
    {
        _productView.ShowDialog();
    }
}

来源：https://stackoverflow.com/questions/34347091/how-to-register-message-handler-prior-to-showdialog-blocking-call