Question

我有一个GridControl，其ItemSource绑定到SearchResultViewModel.SearchResults。在这个GridControl中，我有一个DetailDescriptor，它包含另一个GridControl，其ItemSource我也想设置为SearchResultViewModel.SearchResults。

这是我的代码：

<dxg:GridControl ItemsSource="{Binding Path=SearchResultViewModel.SearchResults, Mode=OneWay}">
        <dxg:GridControl.DetailDescriptor>
            <dxg:ContentDetailDescriptor>
                <dxg:ContentDetailDescriptor.ContentTemplate>
                    <DataTemplate>
                        <dxg:GridControl ItemsSource="{Binding Path=SearchResultViewModel.SearchResults, Mode=OneWay}" MaxHeight="200">
                            <dxg:GridControl.Columns>
                                <dxg:GridColumn Header="Heating Deviation" FieldName="HeatingDeviation"/>
                            </dxg:GridControl.Columns>
                        </dxg:GridControl>
                    </DataTemplate>
                </dxg:ContentDetailDescriptor.ContentTemplate>
            </dxg:ContentDetailDescriptor>
        </dxg:GridControl.DetailDescriptor>

SearchResults看起来像这样

public struct SearchResult
{
    public SearchResult(SepoContainer sepo, int heatingDeviation)
    {
        HeatingDeviation = heatingDeviation;
        SepoContainer = sepo;
    }

    public int HeatingDeviation { get; }

    public SepoContainer { get; }
}

我想我理解错误，因为我不知道如何继续

Answer 1

ContentDetailDescriptor.DataContext会自动设置为行对象，您可以使用RelativeSource中的Binding来检索排名GridControl，从而达到您想要的效果。

<dxg:ContentDetailDescriptor.ContentTemplate>
    <DataTemplate>
        <dxg:GridControl ItemsSource="{Binding ItemsSource, RelativeSource={RelativeSource AncestorType=dxg:GridControl}, Mode=OneWay}" MaxHeight="200">
            <dxg:GridControl.Columns>
                <dxg:GridColumn Header="Heating Deviation" FieldName="HeatingDeviation"/>
            </dxg:GridControl.Columns>
        </dxg:GridControl>
    </DataTemplate>
</dxg:ContentDetailDescriptor.ContentTemplate>

DevExpress DataGrid：如何在DetailDescriptor中设置GridControl的ItemSource

1 个答案: