How to receive DialogResult using mvvm-light Messenger - wpf

Question

I'm trying to use the mvvm-light messenger capability to open a custom confirm password dialog in my view, triggered by a command in my viewmodel.
I think I understand the usage of Messenger.Default.Register and Messenger.Default.Send.
But how do I get the dialog results back in my viewmodel?
To me the sending seems to be a one way street...
Could someone help a beginner with a small C#/WPF code sample?
Thanks for any help

Answer 1

IMHO it is better to use the NotificationMessageAction<T> as it is cut out for this task.
On the sender side:
var msg = new NotificationMessageAction<MessageBoxResult>(this, "GetPassword", (r) =>
{
if (r == MessageBoxResult.OK)
{
// do stuff
}
});
Messenger.Default.Send(msg);
And on the receiver side:
Messenger.Default.Register<NotificationMessageAction<MessageBoxResult>>(this, (m) =>
{
if (m.Notification == "GetPassword") {
var dlg = new PasswordDialog();
var result = dlg.ShowDialog();
m.Execute(result);
}
});
I believe that this approach is cleaner as it does not create an unnecessary dependency from the View to the ViewModel (although this way round is not so bad). For better readability consider sub-classing the NodificationMessageAction<MessageResult>. I.e.
public class ShowPasswordMessage : NotificationMessageAction<MessageBoxResult>
{
public ShowPasswordMessage(object Sender, Action<MessageBoxResult> callback)
: base(sender, "GetPassword", callback)
{
}
}
Then the sender
var msg = new ShowPasswordMessage(this, (r) =>
{
if (r == MessageBoxResult.OK)
{
// do stuff
}
});
Messenger.Default.Send(msg);
and receiver side
Messenger.Default.Register<ShowPasswordMessage>(this, (m) =>
{
var dlg = new PasswordDialog();
var result = dlg.ShowDialog();
m.Execute(result);
});
becomes a lot clearer.
And verry important unregister the recipient as else you might create a memory leak.

Answer 2

In Register method you can show a dialog and pass the YourViewModel reference.
Messenger.Default.Register<YourViewModel>(this, "showDialog", viewModel=>
{
var dlg = new Dialog();
viewModel.Result = dlg.ShowDialog();
});
somewhere in your code you can throw Send() message with a reference to YourViewModel like this:
Messenger.Default.Send(viewModel, "showDialog");

Answer 3

In order to achieve the above using DialogMessage as the title suggests,
one may use the following:
sender side:
void SendMessage(String msgText)
{
DialogMessage messege = new DialogMessage(msgText, res =>
{
callback(res);
})
//set more dialog properties using the Initializer
{ Button = MessageBoxButton.OKCancel, Caption = "" };
Messenger.Default.Send(messege, "mb1");
}
public void callback(MessageBoxResult res)
{
if (res == MessageBoxResult.OK)
{ /*do something*/ }
}
receiver side:
void Rec()
{
Messenger.Default.Register<DialogMessage>(
this, "mb1", msg => ShowDialog(msg));
Unloaded += Unreg;
}
void ShowDialog(DialogMessage msg)
{
var result = MessageBox.Show(
msg.Content,
msg.Caption,
msg.Button);
msg.Callback(result);
}
note the explicit call to the Callback method in the last line of the receiver.
msg.Callback(result);