you can do it with a dynamic object
dynamic obj = JsonConvert.DeserializeObject<ExpandoObject>(jsonString);
obj.Values.valueName4 = "value4";
System.Console.WriteLine(JsonConvert.SerializeObject(obj));
标签：OLE div stack tty class name bsp expand line
you can do it with a dynamic object
dynamic obj = JsonConvert.DeserializeObject<ExpandoObject>(jsonString);
obj.Values.valueName4 = "value4";
System.Console.WriteLine(JsonConvert.SerializeObject(obj));
Adding property to a json object in C#
标签：OLE div stack tty class name bsp expand line
原文地址：https://www.cnblogs.com/chucklu/p/11171928.html
