Question

I’ve developed an application with WPF and MVVM. In it I have a Window with a DataGrid. It’s ViewModel contains some properties for the window and one property for the DataGrid (an ObservableCollection<DataGridItemViewModel>).

In the window xaml I set the design DataContext in this way:

<Window x:Class="XXX"

    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
    xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"

            mc:Ignorable="d"

            d:DataContext="{d:DesignInstance TheTypeOfTheWindowViewModelHere}">

Then I want to set the design DataContext onto the DataGrid this way:

<DataGrid ItemsSource="{Binding Path=PropertyOfTheDataGrid}" d:DataContext="{d:DesignInstance DataGridItemViewModel}" >

But then I get a warning telling me that cannot find PropertyOfTheDataGrid inside DataGridItemViewModel.

I thought I was setting only the DataContext of the ItemsSource but not I’m not sure if I’m doing it wrong or if its an issue of some kind.

Thanks in advance.

Answer 1

I’m not quite sure what you were expecting? From your naming standard, you have a DataGridItemViewModel which suggests you were expecting to apply a view model context to each data grid item?

Normally, you would apply one view model to your entire view, and then have a property on that view model which is for example an ObservableCollection, which is your collection of items for your grid. Then you would set the ItemsSource of your DataGrid to bind to that collection property.

ItemsSource="{Binding MyItems}"

You wouldn’t normally need to set the data context of the grid directly, it would use the data context of the view (a Window in this case).