MVVM How to pass object in Itemscontrol-CodePudding

Is there any way to bind whole object while generating list by itemscontrol? MainWindowViewModel.cs

   private List<TrackModel> _trackModels=new List<TrackModel>();

    public List<TrackModel> TrackModels
    {
        get { return _trackModels; }
        set { _trackModels = value; }
    }

MainWindow.Xaml

<ItemsControl ItemsSource="{Binding TrackModels}">
        <ItemsControl.ItemTemplate>
            <DataTemplate>
                <Label Content="{Binding TrackModel}"/> <!--Here should be my UserControl-->
            </DataTemplate>
        </ItemsControl.ItemTemplate>
    </ItemsControl>

The thing is that I want to create my own UserControl that will receive TrackModel instance because I will need it to get FileName of track to play music. Is there any way to pass TRACKMODEL to my UserControl while using ItemsControl?

CodePudding user response：

You do not need more than this:

<ItemsControl ItemsSource="{Binding TrackModels}">
    <ItemsControl.ItemTemplate>
        <DataTemplate>
            <local:YourUserControl/>
        </DataTemplate>
    </ItemsControl.ItemTemplate>
</ItemsControl>

It requires that YourUserControl does not explicitly set its own DataContext property, but instead simply binds its child elements to the automatically provided DataContext object, which is a TrackModel element from the TrackModels collection.

CodePudding user response：

There is another option if you wanted to explore it though it maps to the same property.

If you give your ItemsControl an x:Name property such as myitems you will be able to access the ItemsSource property from the code behind and set the content there.