Correct way to update property in ViewModel from Model - wpf

Question

I'm fairly novice with WPF. It's my understanding that data changes in the model, and it should notify the viewmodel, and the view will bind to properties and things alike in the viewmodel. Is this correct? If so, I've been reading that the model should implement INotifyPropertyChanged, and look something like this
public class LoginModel : INotifyPropertyChanged
{
public event PropertyChangedEventHandler PropertyChanged;
public void NotifyPropertyChanged(string propName)
{
PropertyChanged?.Invoke(this, new PropertyChangedEventArgs(propName));
}
public bool Authenticated { get; set; }
}
and in my ViewModel, I have a property "AuthResult", that should get the update from the Model property "Authenticated"
public partial class view1 : UserControl, INotifyPropertyChanged{
public bool AuthResult
{
get
{
return _authVal;
}
set
{
_authVal = value;
NotifyPropertyChanged("AuthResult");
}
}
public event PropertyChangedEventHandler PropertyChanged;
public void NotifyPropertyChanged(string propName)
{
PropertyChanged?.Invoke(this, new PropertyChangedEventArgs(propName));
}
}
I know this current implementation is incorrect. I've found that I should be subscribing to the PropertyChanged notification from my model like so:
LoginModel.PropertyChanged += new System.ComponentModel.PropertyChangedEventHandler(LoginModel_PropertyChanged);
void LoginModel_PropertyChanged(object sender, System.ComponentModel.PropertyChangedEventArgs e)
{
if(e.PropertyName == "Authenticated")
{
//do something
}
}
I don't see where the "AuthResult" property should be updated. Would I do something in the If statement like AuthResult = _model.Authenticated;?
EDITED:
and in my constructor?
LoginModel _model;
public view1(LoginModel model)
{
_model = model;
InitializeComponent();
}

Answer 1

If the model implements the INotifyPropertyChanged interface you can bind directly to it from the view:
<Button Content="Button" IsEnabled="{Binding Authenticated}" />
Note that the LoginModel class must raise the PropertyChanged event whenever the Authenticated property is set to a new value.
You could also expose the entire model entity through the view model class:
public class ViewModel
{
public ViewModel(LoginModel model)
{
Model = model;
}
public LoginModel Model { get; }
}
...and bind to it like this:
<Button Content="Button" IsEnabled="{Binding Model.Authenticated}" />
It is still the model class that must implement the INotifyPropertyChanged interface and raise change notifications.
Another option is for the view model to wrap any property of the model class that you want to be able to bind to from the view. Then you bind to a property of the view model class that in turn wraps a property of the model class something like this:
public class ViewModel
{
private readonly LoginModel _model;
public ViewModel(LoginModel model)
{
_model = model;
}
public bool AuthResult
{
get
{
return _model.Authenticated;
}
set
{
_model.Authenticated = value;
NotifyPropertyChanged("AuthResult");
}
}
}
<Button Content="Button" IsEnabled="{Binding AuthResult}" />
The benefit of using this latter approach is that view has no dependency upon the model class. It binds to the view model class only and this is how the MVVM design pattern typically is meant to be implemented.
But if you do bind to a (wrapper) property of the view model and want the view to be updated whenever a property of the model class is set, the model has to notify the view model that it has changed one way or another, i.e. it has to raise some kind of event or similar. And this typically means implementing the INotifyPropertyChanged interface. The view model can then subscribe to the PropertyChanged event of the model and raise its own PropertyChanged event for the data bound property whenever the model is updated, e.g.:
public class ViewModel
{
private readonly LoginModel _model;
public ViewModel(LoginModel model)
{
if (model == null)
throw new ArgumentNullException("model");
_model = model;
_model.PropertyChanged += OnModelChanged;
}
private void OnModelChanged(object sender, PropertyChangedEventArgs e)
{
if (e.PropertyName == "Authenticated")
NotifyPropertyChanged("AuthResult");
}
public bool AuthResult
{
get
{
return _model.Authenticated;
}
set
{
_model.Authenticated = value;
NotifyPropertyChanged("AuthResult");
}
}
}

Answer 2

Just use Model as member in the ViewModel
public class ViewModel
{
private Model _myModel;
public Model MyModel
{
get { return _myModel; }
set
{
if (Equals(_myModel, value)) return;
_myModel = value;
NotifyPropertyChanged(nameof(MyModel));
}
}
}
Then in xaml you can bind properties of Model
<CheckBox IsChecked="{Binding MyModel.Authenticated}" />
With this approach your ViewModel will be "build" around your Model.
In case you don't want that models implement INotifyPropertyChanged than create a "Facade" class of model in use it in same way as previous example.
public class ModelFacade : INotifyPropertyChanged
{
private Model _myModel;
public bool Authenticated
{
get { return _myModel.Authenticated; }
set
{
_myModel.Authenticated = value;
NotifyPropertyChanged(nameof(Authenticated));
}
}
}
public class ViewModel
{
private ModelFacade _myModel;
public ModelFacade MyModel
{
get { return _myModel; }
set
{
if (Equals(_myModel, value)) return;
_myModel = value;
NotifyPropertyChanged(nameof(MyModel));
}
}
}