.Net 4 WPF DataGrid's ItemsSource - works programmatically, but not declaratively

Question

I am writing a WPF program using C# and targeting .Net 4.0. I am trying to follow the MVVM pattern, where my view (code-behind) has minimal-to-no code.

I have a List<MyRecord> that I want to display in the GUI using a DataGrid. In my XAML, I have the following:

<DataGrid x:Name="RecordGrid" ...>
   <DataGrid.Columns>
      <DataGridTextColumn Binding="{Binding RecId}" Header="Record ID"/>
      <DataGridTextColumn Binding="{Binding Name}" Header="Name"/>
   </DataGrid.Columns>
</DataGrid>

All that's left is to bind this grid to a data collection.

When I bind it in the code-behind file, it works fine:

RecordGrid.ItemsSource = MyRecordList;

However, I would prefer to bind it declaratively in XAML. So I tried this:

<DataGrid x:Name="RecordGrid" ItemsSource="{Binding MyRecordList}" ...>

but it silently doesn't work. There is no XAML binding error message when the datagrid loads. I set a breakpoint on MyRecordList's get method, and it's never invoked as long as ItemsSource is defined declaratively.

How can I get my datagrid to pull from MyRecordList via XAML?

@default.kramer, I haven't set or changed my DataContext. My entire test program consists of an Application and MainWindow. — Philip, Feb 05 '13 at 22:24
You should probably ready [this](http://stackoverflow.com/a/7262322/644812) and [this](http://msdn.microsoft.com/en-us/library/ms750612.aspx) — default.kramer, Feb 05 '13 at 22:25

Blachshma · Accepted Answer · 2013-02-05T22:37:31.893

0

You need to set the DataContext if you want the binding to work... The MVVM pattern has a View and a ViewModel. The View is your UI - e.g. the Window (Let's call it MainWindow) and your ViewModel is where you have the RecordGrid property and all theother commands/properties (let's call the class MainWindowViewModel)

You need to connect them both, This is done by specifying in your View, who the DataContext is (in our case the MainWindowViewModel class).

So you'll want to do something like this in your View's constructor:

    public MainWindow()
    {
        this.DataContext = new MainWindowViewModel();
        InitializeComponent();

edited Feb 05 '13 at 22:37

answered Feb 05 '13 at 22:31

Blachshma

17,097
4
58
72

Thanks. That's what I normally do, but while playing with DataGrid I wanted to have everything in one file for simplicity. I thought if I don't set DataContext on MainWindow, WPF will use MainWindow as the default context. Seems like that's not the case. So what's the default DataContext? Is there one? – Philip Feb 05 '13 at 22:37
So if everything is in the same class, e.g. your Property is in the Window's code behind - use `this.DataContext = this;` – Blachshma Feb 05 '13 at 22:38
And if I don't -- what's the default? – Philip Feb 05 '13 at 22:38
There is no default, DataContext is an object and therefore null by default. Bindings won't work without setting a DataContext. – Blachshma Feb 05 '13 at 22:39
You wrote you have a `List` -> Where did you declare it? – Blachshma Feb 05 '13 at 22:40
In the test program, I declared it as a member of the MainWindow class. – Philip Feb 05 '13 at 22:41
Then 1. Make sure it's a *Property* 2. Set the DataContext to `this` like I stated in my previous comment – Blachshma Feb 05 '13 at 22:42

.Net 4 WPF DataGrid's ItemsSource - works programmatically, but not declaratively

1 Answers1