问题
I was trying to register for a message outside of a ViewModel in a static constructor but apparently that registration didn't take: the registered action never ran when messages were sent. I tried passing in null or a new object for the recipient parameter when registering but that didn't work.
I have a feeling the specifying the recipient must be important somehow, but I don't know why. I thought that all recipients were supposed to get broadcasted messages anyway. Is there a way to make this work or is this simply not supported?
回答1:
The recipient is important when you use Messenger.Register for Messenger.Send it is not important. However, there is a known bug in the WeakReferenceWeakAction implementation in MVVM that holds a reference to the recipient, although, it should release it.
In short, if you are inside a ViewModel, make sure that you call Cleanup. In a view, call Messenger.Unregister(this) in the Unloaded event e.g.
public MyView() {
this.Unloaded += (o, e) => { Messenger.Unregister(this); }
}
In other classes your will have to either implement IDisposable or use some other mechanism to unregister the message recipient.
See also:
- MVVM Light Listener not releasing / deterministic finalization for registered object?
- Does mvvm light v3 unregister work properly?
- When to dispose ViewModel in MVVM Light
回答2:
Nevermind, I went into the source code and figured out what was happening. It's adding the recipient as a WeakReference to figure out if it's alive or not. I was just passing in new object() as the recipient and it wasn't firing because it thought my recipient was dead.
来源：https://stackoverflow.com/questions/7695045/registering-for-messages-outside-of-a-viewmodel-in-mvvm-light