I've followed your help sample and created a custom style selector like this:-
I have also created the necessary styles in my view XAML:-
The problem is, I'm programmatically creating the RadGridView columns in my view model. I can't simply assign an instance of my custom style selector to the column's CellStyleSelector property as there is nothing linking that instance to the styles in the XAML. Am I going about this the wrong way?
public class BlockTypeStyleSelector : StyleSelector{ public override Style SelectStyle(object item, DependencyObject container) { if (item is SampleRun) { var sampleRun = item as SampleRun; switch (sampleRun.BlockType) { case BlockType.NormalBlock: return NormalBlockStyle; case BlockType.CalibrationBlock: return CalibrationBlockStyle; } } return null; } public Style NormalBlockStyle { get; set; } public Style CalibrationBlockStyle { get; set; }}I have also created the necessary styles in my view XAML:-
<GridStyleSelectors:BlockTypeStyleSelector x:Key="blockTypeStyle"> <GridStyleSelectors:BlockTypeStyleSelector.NormalBlockStyle> <Style TargetType="telerik:GridViewCell"> <Setter Property="Background" Value="White"/> </Style> </GridStyleSelectors:BlockTypeStyleSelector.NormalBlockStyle> <GridStyleSelectors:BlockTypeStyleSelector.CalibrationBlockStyle> <Style TargetType="telerik:GridViewCell"> <Setter Property="Background" Value="PaleGoldenrod"/> </Style> </GridStyleSelectors:BlockTypeStyleSelector.CalibrationBlockStyle></GridStyleSelectors:BlockTypeStyleSelector>The problem is, I'm programmatically creating the RadGridView columns in my view model. I can't simply assign an instance of my custom style selector to the column's CellStyleSelector property as there is nothing linking that instance to the styles in the XAML. Am I going about this the wrong way?