功能点击侦听器问题android GeoJsonLayer

Question

[您好，我正在使用Google Maps android api，并且我正在使用Google Maps android geojson实用程序。我将3层geojson添加到地图中，只有road_layer（LineString）具有功能click事件。在添加samsad_boundary_layer（多边形）之前，单击事件工作正常。问题是，如果单击了[[samsad_boundary_layer功能，尽管该图层（samsad_boundary_layer）没有功能单击事件，但它触发了[[road_layer功能单击事件。

private void showBoundery(){
    try {
        samsad_boundary_layer=new GeoJsonLayer(mMap,R.raw.samsad_boundary,MapActivity.this); //1st Layer
    } catch (IOException e) {
        e.printStackTrace();
    } catch (JSONException e) {
        e.printStackTrace();
    }
    samsad_boundary_layer.getDefaultPolygonStyle().setZIndex(-12f);
    GeoJsonPolygonStyle polygonStyle=samsad_boundary_layer.getDefaultPolygonStyle();
    polygonStyle.setStrokeWidth(5f);
    polygonStyle.setStrokeColor(getResources().getColor(R.color.purple));
    samsad_boundary_layer.addLayerToMap();
}

private void retrieveFileFromResource() {
    try {
        temp_layer=new GeoJsonLayer(mMap,R.raw.area_survey,this); //2nd Layer
        temp_layer.addLayerToMap();
        road_layer=new GeoJsonLayer(mMap,R.raw.road_survey,this); //3rd Layer
        road_layer.setOnFeatureClickListener(new Layer.OnFeatureClickListener() {
            @Override
            public void onFeatureClick(Feature feature) {
                if (feature!=null){
                    Iterator<HashMap> flavoursIter = feature.getProperties().iterator();
                   while (flavoursIter.hasNext()){
                        Map.Entry entry = (Map.Entry)flavoursIter.next();
                        //Log.v("SURVEYOR_FEATURE",entry.getKey().toString()+":"+entry.getValue());

                    }

                }
            }
        });
        GeoJsonLineStringStyle lineStyle=road_layer.getDefaultLineStringStyle();
        lineStyle.setZIndex(10f);
        lineStyle.setWidth(2f);
        lineStyle.setColor(getResources().getColor(R.color.light_green));
        road_layer.addLayerToMap();
    } catch (IOException e) {
        e.printStackTrace();
    } catch (JSONException e) {
        e.printStackTrace();
    }
}

Answer 1