对于Json和对象转换的学习
学习这个的用处有非常多的:
在传输数据过程中比較查看数据比較清晰,代码也较清晰。也能够避免split函数带来的隐藏bug
当然也有不足:
准备工具较繁琐,须要准备对象(当然一般项目中每张表都会有相应的Model类,少量数据不建议适用)
------请教各位大侠。能不能直接将Json字符串转换为对象(对象格式无需我们自定义,由于Json中已经非常明白对象的格式了)
介绍两种方式:
第一种:
适用.NET Framework3.5以上版本号的
命名空间为using System.Runtime.Serialization.Json;
//将一个对象转换为Json字符串 public static string ObjectToJson_(object obj) { DataContractJsonSerializer serializer = new DataContractJsonSerializer(obj.GetType()); MemoryStream stream = new MemoryStream(); serializer.WriteObject(stream, obj); byte[] dataBytes = new byte[stream.Length]; stream.Position = 0; stream.Read(dataBytes, 0,(int)stream.Length); return Encoding.UTF8.GetString(dataBytes); } //将一个Json字符串转换为对象 public static object JsonToObject_(string jsonString, Type type) { DataContractJsonSerializer serilizer = new DataContractJsonSerializer(type); MemoryStream stream = new MemoryStream(Encoding.UTF8.GetBytes(jsonString)); return serilizer.ReadObject(stream); }
另外一种:
须要引用Newtonsoft.Json.dll 下载地址
// 从一个对象信息生成Json串 public static string ObjectToJson(object obj) { return JsonConvert.SerializeObject(obj); } // 从一个Json串生成对象信息 public static object JsonToObject(string jsonString, Type obj) { return JsonConvert.DeserializeObject(jsonString, obj); }
//将提交的数据Json转换为Model string sendMessage = "[{\"Type\":\"精美小炒\", \"MyLunch\":[{\"Name\":\"番茄炒蛋\", \"Price\":\"10\"}, {\"Name\":\"耗油牛肉\", \"Price\":\"14\"}, {\"Name\":\"金针菇肥牛\", \"Price\":\"16\"}, {\"Name\":\"虾仁炒蛋\", \"Price\":\"15\"}]}, {\"Type\":\"精美套餐\", \"MyLunch\":[{\"Name\":\"商务套餐\", \"Price\":\"15\"}, {\"Name\":\"红烧猪排套餐\", \"Price\":\"12\"}, {\"Name\":\"椒盐排条套餐\", \"Price\":\"10\"}, {\"Name\":\"茄汁牛排套餐\", \"Price\":\"10\"}]}, {\"Type\":\"特色盖浇饭\", \"MyLunch\":[{\"Name\":\"回锅肉盖浇饭\", \"Price\":\"12\"}, {\"Name\":\"尖椒牛柳盖浇饭\", \"Price\":\"13\"}, {\"Name\":\"蒜苗肉丝盖浇饭\", \"Price\":\"15\"}, {\"Name\":\"辣子鸡盖浇饭\", \"Price\":\"12\"}]}]"; List<Meal> getModel = new List<Meal>(); getModel = (List<Meal>)JsonToObject_(sendMessage, typeof(List<Meal>)); //将Model转换为Json List<Meal> MealList = new List<Meal>(); Meal modelMeal = new Meal(); List<Lunch> myLunch = new List<Lunch>(); Lunch modelLunch = new Lunch(); modelLunch.Name = "台式卤肉"; modelLunch.Price = "15"; myLunch.Add(modelLunch); modelMeal.MyLunch = myLunch; modelMeal.Type = "星期五套餐"; MealList.Add(modelMeal); string jsonString = ObjectToJson_(MealList); result = jsonString;
大概就是这样子了,错误求指正!
QAQ
posted on 2017-04-28 19:41 gavanwanggw 阅读(122) 评论(0) 编辑 收藏 举报