Setting Renderer on feature layer by code

[MarcoRosa]

Hi to all, i've this case:

if i set simple renderer on xaml this work fine ,

    <esri:SimpleRenderer x:Key="DefaultFillRenderer">
    <esri:SimpleRenderer.Symbol>
                    <esriSymbols:SimpleFillSymbol BorderBrush="Blue" Fill="#330000FF" />
                </esri:SimpleRenderer.Symbol>
            </esri:SimpleRenderer>

if i set renderer throught code does't work ....


FeatureLayer dyn_ftlyr = new FeatureLayer();
dyn_ftlyr.ID = "MyFeatureLayer";
dyn_ftlyr.ID = "MyUrl";

dyn_ftlyr.Renderer = LayoutRoot.Resources["DefaultFillRenderer"] as SimpleRenderer ;


Where is my error ? thankkkkssss

[DominiqueBroux]

dyn_ftlyr.ID = "MyUrl";

I guess it's just a typo and that you are initializing the Url method and not the ID

dyn_ftlyr.Renderer = LayoutRoot.Resources["DefaultFillRenderer"] as SimpleRenderer ;

Did you verify in debug that the renderer is not null?

[MarcoRosa]

hi dominique , that's true .... renderer its null.
it's wrong the way to set the renderer ?

[DominiqueBroux]

that's true .... renderer its null.
it's wrong the way to set the renderer ?

The Renderer can be set this way but it needs to be defined in the resources.
In your case, it looks like the "DefaultFillRenderer" is not defined in the resources of the Layout grid.