Adding property to a json object in C#
you can do it with a dynamic object
dynamic obj = JsonConvert.DeserializeObject<ExpandoObject>(jsonString); obj.Values.valueName4 = "value4"; System.Console.WriteLine(JsonConvert.SerializeObject(obj));
you can do it with a dynamic object
dynamic obj = JsonConvert.DeserializeObject<ExpandoObject>(jsonString); obj.Values.valueName4 = "value4"; System.Console.WriteLine(JsonConvert.SerializeObject(obj));