1 string jsonStr = "{"data": {"ssoToken": "70abd3d8a6654ff189c482fc4842468c","account":"admin","userType":"platformAdmin","realName": "超级管理员","sex": 0,"sexName":"男","email":"alina_dong@163.com","mobile":"15120757948","createdDt": "2013-08-16 00:00:00","updatedDt": "2014-12-10 00:00:00" },"isSuccess": true}";
当 .Net 程序接收到了这段JSON字符串数据的时候，大家肯定会想到使用 Newtonsoft.Json 去序列化（SerializeObject）和反序列化（DeserializeObject）一个对象。
使用 SerializeObject 的示例：
1 A a = new A(); 2 a.age = 11; 3 a.name = "Jack"; 4 B b = new B(); 5 b.sex = "Man"; 6 //b.money = 12; 7 a.B = b; 8 string str = Newtonsoft.Json.JsonConvert.SerializeObject(a); 9 10 输出结果：{"age": 11, "name": "Jack", "B": {"sex": "Man", "money": ""}}
使用 DeserializeObject 的示例：
1 string jsonStr = @"{"age": 11, "name": "Jack", "B": {"sex": "Man", "money": ""}}"; 2 var a = Newtonsoft.Json.JsonConvert.DeserializeObject<A>(jsonStr); 3 4 结果：a.age = 11;.......
好了，言归正传，如何使用 dynamic 去解析一个Json字符串呢？
1 string jsonStr = "{"data": {"ssoToken": "70abd3d8a6654ff189c482fc4842468c","account":"admin","userType":"platformAdmin","realName": "超级管理员","sex": 0,"sexName":"男","email":"alina_dong@163.com","mobile":"15120757948","createdDt": "2013-08-16 00:00:00","updatedDt": "2014-12-10 00:00:00" },"isSuccess": true}"; 2 var loginInfo = JsonConvert.DeserializeObject<dynamic>(jsonStr); 3 var user = loginInfo.data; 4 string ssoToken = user.ssoToken; 5 string account = user.account;
这样，不用创建loginInfo，user照样能解析JSON，而且不会因为那边增加字段报错啦。