CODEWITHSUNDEEP
c#xamlmvvmdata-bindingdatacontext
Brian Var
Brian Var

Reputation: 6227

How to re-wire View data context to XAML from C#?

I've set my View's data context to it's associated ViewModel in the View's code behind. But after reading a question on the MVVM pattern, it's suggested to move this glue code to the View's XAML mark up.

Googling has shown me examples of setting the context below in XAML, by setting a namespace to the VM and setting data context.

Although in my case, the MainViewModel takes a parameter of a CustomerRepository instance, which I'm not sure how to set up in XAML, base on the previous example.

Does anyone know how to move the data context to the View's XAML?

This is how I've currently set the View's code behind in C#:

    public partial class MainView : Window
    {
        private MainViewModel ViewModel { get; set; }

        public MainView()
        {
            InitializeComponent();
            ViewModel = new MainViewModel(CustomerRepository.Instance);
            this.DataContext = ViewModel;

        }

    }

Upvotes: 0

Views: 103

Answers (1)

Martin
Martin

Reputation: 5623

You can instantiate your view model in your xaml like this:

<Window x:Class="MyWindow"
            xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
            xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
            Title="Test" 
            xmlns:viewModel="clr-namespace:ViewModels"> 
        <Window.DataContext>
            <viewModel:ExampleViewModel/>
        </Window.DataContext>
</Window>


You explained that your view model constructor has a parameter for your repository. Do you need this constructor for unit testing purposes?

Usually you can just new up your repository instance in a parameterless constructor of the view model.

public class MainViewModel : ObservableObject, INotifyPropertyChanged
{

    private static CustomerRepository _customerRepository;

    // existing constructor
    public MainViewModel(CustomerRepository customerRepository)
    {
        _customerRepository = customerRepository;
    }

    // new parameterless constructor
    public MainViewModel(CustomerRepository customerRepository)
    {
        if (DesignerProperties.GetIsInDesignMode(this))
        {
            _customerRepository = new CustomerRepository();         
        }
    }
}


Check if is in design mode

There is one more thing you need to think of when creating a view model in xaml: The view models constructor is called at design time when you open the view. So you need to wrap any code in the constructor which does not make any sense at design time into a "only when not in design time" condition. If you do not do this, your view will open with an error in the designer.

This is explained in this response: Check if in design mode

Upvotes: 1

Related Questions