CODEWITHSUNDEEP
c#wpfmvvm-light
Arh Hokagi
Arh Hokagi

Reputation: 296

Mvvm light messenger, communication between viewmodels


In my ViewmodelA I want to open a new window in ViewModelB, so i used the messenger class, but the probleme is that i need to reference ViewModelB inside the ViewModelA, so that ViewModelB can listen to the messages.
here is my implementation
in ViewModelA:

 private void btnAddExecute()
 {  
    // I need to instanciate ViewModelB otherwise it wont work/listen
    ViewModelB vb= new ViewModelB();

    Messenger.Default.Send(new NotificationMessage("ShowWindow"));       
 }

in ViewModelB i listen to the broadcasted messages in it's constructor.

is there anyway to decouple ViewmodelA from ViewmodelB ?

Upvotes: 2

Views: 1661

Answers (1)

MrZander
MrZander

Reputation: 3120

I don't exactly see where the coupling is occurring if you are using the messenger properly. There is no need to reference ViewModelB

Edit

Here is a way to do it without reference to an instance ViewModelB. It uses a singleton to register for messages and create ViewModelBs when it receives the notification message. I haven't tested this, it is just an idea. Make sure ViewModelBCreator is used at some point so that the static constructor is called.

public class ViewModelBCreator()
{
    private static ViewModelBCreator instance;
    static ViewModelBCreator() { instance = new ViewModelBCreator(); }
    private ViewModelBCreator()
    {
        Messaging.Messenger.Default.Register<NotificationMessage>(this, true, NotificationMessageReceived);
    }
    private static void NotificationMessageReceived(NotificationMessage notification)
    {
        var vm = ViewModelB();
        //Do stuff with the new ViewModelB
    }
}  

public class ViewModelB
{ 
    public ViewModelB()
    {
        //etc . . .
    }
}

public class ViewModelA
{
    public void OpenTheWindow()
    {
        Messenger.Default.Send(new NotificationMessage("ShowWindow"));  
    }
}

Upvotes: 2

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