SuperMap.Web.iServerJava6R.Data.GetFeaturesResult.FromJson C# (CSharp) Метод

        public static GetFeaturesResult FromJson(JsonObject json)
        {
            if (json == null)
                return null;

            GetFeaturesResult result = new GetFeaturesResult();
            result.FeatureCount = (int)json["featureCount"];
            if (result.FeatureCount < 1)
            {
                return null;
            }

            JsonArray features = (JsonArray)json["features"];
            if (features != null && features.Count > 0)
            {
                result.Features = new FeatureCollection();

                for (int i = 0; i < features.Count; i++)
                {
                    ServerFeature f = ServerFeature.FromJson((JsonObject)features[i]);
                    result.Features.Add(f.ToFeature());
                }
            }

            return result;
        }