I've created a new WPF project, and threw in a DataGrid. Now I'm trying to figure out the easiest way to bind a collection of data to it.
The example I downloaded seems to do it in the window c'tor:
public partial class MainWindow : Window
{
public MainWindow()
{
InitializeComponent();
DataContext = new ViewModel();
}
}
But then the bindings don't seem to appear in the Visual Studio's Properties window. I'm pretty sure there's a way to set the data context in XAML too... it would make me even happier if I could do it directly through the properties window, but all the Binding options are empty. What's the typical approach?
Edit: At 14 minutes, he starts to talk about other methods of setting the data context, such as static resources, and some "injection" method. I want to learn more about those!
What I typically do is use MVVM. You can implement a simplified version by setting the data context in your code behind and having a model type class that holds your data.
Example: In your code behind
DataContext = Model; // where Model is an instance of your model
then in your view
<DataGrid .... ItemsSource="{Binding SomeProperty}">....
Where SomeProperty is an enumerable property on your view model
You can also set a data context in XAML by using the DataContext property
<uc:SomeUserControl DataContext="{Binding AnotherProperty}"....
This will run your user control within the DataContext of the AnotherProperty on your model.
Note that this is grosely simplified but it'll get you on your way.
Have a look at the MVVM design pattern. This pattern is very suitable for wpf applications.
There is described where to store your data and how to bind your ui to the data.
Related
I am pretty new to WPF and right now I am trying to get used to the MVVM pattern. Right now I have a simple application in which I have a collection of ViewModels that I display in a grid. When I doubleclick on the row in the grid I want to show a details View of the ViewModel.
The problem I am having right now is that I already have a fully instanced ViewModel, but I can't seem to pass it into the view. When I try to load that View it turns up empty. I already found out that this is due to the fact that when a View gets loaded it creates it's own instance of the backing ViewModel. So obviously I need to get around this behaviour and somehow pass the instanced ViewModel into the View when it is created. I could use a constructor in the View that takes a ViewModel and set the datasource in there. However, taking this approach but would mean that I need to construct the View in the ViewModel and thus making the ViewModel aware of the View. This I something I would like to avoid since I am trying to uphold the MVVM pattern.
So what should I do in this case? Should I just break the MVVM pattern or are there some nice and clean sollutions for this that fit in the MVVM pattern?
There are many ways of passing a view model to a view, as you call it, or setting a view model as the DataContext of either a Window or UserControl, as others may call it. The simplest is just this:
In a view constructor:
public partial class SomeView
{
InitializeComponent();
DataContext = new SomeViewModel();
}
A more MVVM way might be to define DataTemplates in App.xaml for each view model that defines which view each will use:
<DataTemplate DataType="{x:Type YourViewModelsPrefix:YourViewModel">
<YourViewsPrefix:YourView />
</DataTemplate>
...
<DataTemplate DataType="{x:Type YourViewModelsPrefix:AnotherViewModel">
<YourViewsPrefix:AnotherView />
</DataTemplate>
Now whenever the Framework comes across an instance of these view model classes, it will render the associated view. You can display them by having a property of the type of your view model using a ContentControl like this:
<ContentControl Content="{Binding YourViewModelProperty}" />
Or even in a collection like this:
<ListBox ItemsSource="{Binding YourViewModelCollectionProperty}" />
"Should I just break the MVVM pattern?"
Well, please consider to learn more about the pattern, to know what it is to "break it". The main purpose of this pattern is to keep responsability clear, thus to obtain testable and maintainable code. There are a lot of ressource for that as show in this question:
MVVM: Tutorial from start to finish?
Anyway to be more specific about your question, what you are looking for is how to set the DataContext.
"somehow pass the instanced ViewModel into the View when it is created"
Yes, you get it, if you assign the dataContext with a viewModel in the constructor of your view, it could work but it it is acceptable only if the viewModel has the responsability to create the view (which could be acceptable in really few situation). You could even write something like that to directly set DataContext from outside your view:
var l_window = new MyView { DataContext = new MyViewModel() };
l_window.Show();
Of course the main drawback is that this code is not testable. If you would like to test it you should use a mockable service to manage the view creation.
A more common solution is to inject the dataContext with an IOC container (like prism). You create all required ViewModel when the software started and you store them in this IOC container. Then, when the view is created, you ask this container to get you an instance of your viewModel.
An example could be: export your viewModel in PRISM:
[Export]
public class MyViewModel {...}
And then Import it in your view:
[Import]
private MyViewModel ViewModel
{
set { this.DataContext = value; }
get { return this.DataContext as MyViewModel; }
}
Hope it helps.
I agree with #Sheridan's answer and would only like to add another way to instantiate a view with a view model: you could use the Factory Pattern, maybe like this:
public class ViewFactory
{
public UIElement Create(object context)
{
// Create the view model
// You can pass in various information by parameters
// as I do with context (Constructor Injection)
var viewModel = new ViewModel(context);
// Create the view and set the view model as data context
var view = new View { DataContext = viewModel };
return view;
}
}
You can call this factory from within a method of your view model and then assign it to e.g. a property that is data bound to the UI. This allows for a bit more flexibility - but #Sheridan's solution is also fine.
Hi guys I am very new to WPF. I have two datacontexts in two different classes which are being binded by the elements in the View producing datatriggers, and one or the other wouldn't work as they cannot bind both datacontexts together. How do I bind xaml from two different classes using datacontext? Is there any alternative way could make it easier?
Class A
public Window1()
{
InitializeComponent();
Appointments = new Appointments();
DataContext = Appointments;
}
Class B
private void FilterAppointments()
{
this.DataContext = this;
...
Firstly, you should never use DataContext = this; in any UserControl in a serious WPF Application. Secondly, you should look up the MVVM design pattern, which provides the idea of a view model for each view. Your Window or UserControl are the 'Views' and your view models are simply classes that contain all of the data properties that you need to display in your view.
Therefore, you should declare a view model class (that implements the INotifyPropertyChanged interface) and put whatever you wanted to data bind into that. Finally, you should set that object as the DataContext property value. In that way, you'll have access to all the data that you need.
Looking again at your question, it just occurred to me that you may have set the DataContext to this so that you could use properties that you declared in your Window or UserControl. If this is the case, then you should not set the DataContext to this, instead using a RelativeSource Binding to access the properties. That would free up the actual DataContext to be set however you like. Try this Binding within the Window or UserControl:
<TextBlock Text="{Binding PropertyName, RelativeSource={RelativeSource
AncestorType={x:Type YourPrefix:YourWindowOrControl}}}" />
I'm new to WPF and I'm writing a simple test app to familiarize myself with it. My test app will detect all joysticks I have attached to my computer and display information about it. So far, I have this ViewModel:
public class JoystickViewModel
{
public ObservableCollection<Joystick> Joysticks { get; set; }
public JoystickViewModel()
{
GetAttachedJoysticks();
}
private void GetAttachedJoysticks()
{
// populate Joysticks collection by using SlimDX
}
}
And this is my codebehind for my MainWindow.xaml:
public partial class MainWindow : Window
{
public MainWindow()
{
InitializeComponent();
DataContext = new JoystickViewModel();
}
}
And my XAML for MainWindow:
<Window ...>
<Grid>
<ComboBox ItemsSource="{Binding Joysticks}"
DisplayMemberPath="Information.ProductName"/>
</Grid>
</Window>
I followed a tutorial that also populated the ViewModel in its constructor.
My question is, how should I populate the ViewModel? It seems sort of weird to me that I'm population the collection in the ViewModel constructor. Should this logic be in MainWindow's codebehind instead? Or somewhere else altogether? The end goal is to not only have this collection populated, but also updated periodically to reflect the current state (user plugged in new joystick, unplugged existing one, etc...).
The MainWindow code behind is definitively not the place where "business" logic should occur, as the View should be kept as simple as possible.
Keep your fetch/update logic inside of your viewmodel, this way you can test it easily and independently.
From a learning perspective, it's important to keep concerns separated :
the View is bound to the ViewModel, and has no intelligence
the ViewModel has knowledge on how to get the Model
the Model represents the data
In your case, the VM knowledge is at the moment a call inside it's constructor. Later you can change this to call some IJoystickDataService interface, and wire everything using a MVVM framework.
I would have your JoySticks observable collection property (and the code that populates it) in a Model class. The viewmodel simply exposes this same property to the view for binding. The vm should be as thin as possible - ideally just exposing properties that are in the model for binding and not doing any kind of 'business' logic (i.e. populating joystick info as in your case).
I'm writting a form in WPF/c# with the MVVM pattern and trying to share data with a user control. (Well, the User Controls View Model)
I either need to:
Create a View model in the parents and bind it to the User Control
Bind certain classes with the View Model in the Xaml
Be told that User Controls arn't the way to go with MVVM and be pushed in the correct direction. (I've seen data templates but they didn't seem ideal)
The usercontrol is only being used to make large forms more manageable so I'm not sure if this is the way to go with MVVM, it's just how I would of done it in the past.
I would like to pass a class the VM contruct in the Xaml.
<TabItem Header="Applicants">
<Views:ApplicantTabView>
<UserControl.DataContext>
<ViewModels:ApplicantTabViewModel Client="{Binding Client} />
</UserControl.DataContext>
</Views:ApplicantTabView>
</TabItem>
public ClientComp Client
{
get { return (ClientComp)GetValue(ClientProperty); }
set { SetValue(ClientProperty, value); }
}
public static readonly DependencyProperty ClientProperty = DependencyProperty.Register("Client", typeof(ClientComp),
typeof(ApplicantTabViewModel),
new FrameworkPropertyMetadata
(null));
But I can't seem to get a dependancy property to accept non static content.
This has been an issue for me for a while but assumed I'd find out but have failed so here I am here.
Thanks in advance,
Oli
Oli - it is OK (actually - recommended) to split portions of the View into UserControl, if UI became too big - and independently you can split the view models to sub view models, if VM became too big.
It appears though that you are doing double-instantiations of your sub VM. There is also no need to create Dependency Property in your VM (actually, I think it is wrong).
In your outer VM, just have the ClientComp a regular property. If you don't intend to change it - the setter doesn't even have to fire a property changed event, although it is recommended.
public class OuterVm
{
public ClientComp Client { get; private set; }
// instantiate ClientComp in constructor:
public OuterVm( ) {
Client = new ClientComp( );
}
}
Then, in the XAML, put the ApplicantTabView, and bind its data context:
...
<TabItem Header="Applicants">
<Views:ApplicantTabView DataContext="{Binding Client}" />
</TabItem>
I answered a similar question as yours recently: passing a gridview selected item value to a different ViewModel of different Usercontrol
Essentially setting up a dependency property which allows data from your parent view to persist to your child user control. Abstracting your view into specific user controls and hooking them using dependency properties along with the MVVM pattern is actually quite powerful and recommended for Silverlight/WPF development, especially when unit testing comes into play. Let me know if you'd like any more clarification, hope this helps.
I'm just getting started with the MVVM pattern in WPF and I decided that the most elegant way to structure my code was injecting the view-model in to the view's constructor.
This is all well and good, but ReSharper gives a warning in the XAML that my view doesn't have a default constructor. I'm assuming that this is so that I can construct my view in XAML if required, but that's only a guess.
What am I giving up by requiring my view to take a view-model in the constructor?
Edit: My view constructor looks like this:
public ExampleView(ExampleViewModel viewModel)
{
if (viewModel == null) throw new ArgumentNullException("viewModel");
DataContext = viewModel;
}
Answer: I settled on the following set up, where the DesignTime namespace contains mocked up versions of the ViewModel for testing and design time support.
ExampleView.xaml.cs
public ExampleView()
{
InitializeComponent();
}
public ExampleView(IExampleViewModel viewModel)
: this()
{
DataContext = viewModel;
}
ExampleView.xaml
<UserControl
x:Class="Wpf.Examples.ExampleView"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:DesignTime="clr-namespace:Wpf.Examples.DesignTime">
<UserControl.DataContext>
<DesignTime:ExampleViewModel/>
</UserControl.DataContext>
</UserControl>
As you correctly recognized, requiring a non-default constructor will deny you using that control from XAML. That also means no more design-support and your designers will probably hate you. Finally you break all sorts of nice data binding scenarios. Like using the control as an ItemTemplate.
As a remedy for the missing design support, I would suggest implementing a default constructor which creates a mocked view-model which doesn't need any infrastructure. That way you can support design mode very elegantly and putting the view in a XAML file (e.g. for testing) will do something sensible.
As a remedy for the missing data binding support, you should ponder whether it might be better to consume the view model via the DataContext of your WPF control. This is common in WPF and---as far as I can tell---the intended way to pass the model to the view in WPF.
Assuming that you don't need designer support then I see no reasons.
To keep designer support you need a default constructor. When you define your own constructor you basically loose the autogenerated default constructor. Just create an explicit default constructor and you should be fine.