How to update a view on LinearLayout for a fragment

Question

I have a fragment , GraphFragment which plots all the data I need using GraphView on a graph. I want to give the user the option to change the gridcolor of the GraphView but when I try to update the view it doesn't seem to work (nothing changes). The setGridColor function relised on the specific listview index the user presses.

public class GraphFragment extends Fragment {

    private GraphView graphView;    
    LinearLayout layout;

    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container,
            Bundle savedInstanceState) {

        ViewGroup rootView = (ViewGroup) inflater.inflate(R.layout.graph_fragment, container, false);

        //Initialise the glucose level graph
        graphView = new LineGraphView(this.getActivity(), "Blood Glucose (mmol/l)");  

        graphView.getGraphViewStyle().setVerticalLabelsWidth(40);

        //Add the test data
        graphView.addSeries(testSeries);

        layout = (LinearLayout) rootView.findViewById(R.id.graph1);  
        layout.addView(graphView);
    }

    public void setGridColor (int position){
        if(position==0){
            //graphView = new LineGraphView(this.getActivity(), "Blood Glucose (mmol/l)");
            layout.removeView(graphView);
            graphView.getGraphViewStyle().setGridColor(Color.RED);
            layout.addView(graphView);              
        }
        if(position==1){
            //graphView = new LineGraphView(this.getActivity(), "Blood Glucose (mmol/l)");
            layout.removeView(graphView);
            graphView.getGraphViewStyle().setGridColor(Color.BLUE);
            layout.addView(graphView);
        }
        if(position==2){
            //graphView = new LineGraphView(this.getActivity(), "Blood Glucose (mmol/l)");
            layout.removeView(graphView);
            graphView.getGraphViewStyle().setGridColor(Color.GREEN);
            layout.addView(graphView);
        }
    }
}

Answer 1

更改后，尝试运行graphView.invalidate() 。