在进行Linq to JSON之前,首先要了解一下用于操作Linq to JSON的类.
| 类名 | 说明 |
JObject | 用于操作JSON对象 |
JArray | 用语操作JSON数组 |
JValue | 表示数组中的值 |
JProperty | 表示对象中的属性,以"key/value"形式 |
JToken | 用于存放Linq to JSON查询后的结果 |
1.创建JSON对象
JObject staff = new JObject(); staff.Add(new JProperty("Name", "Jack")); staff.Add(new JProperty("Age", 33)); staff.Add(new JProperty("Department", "Personnel Department")); staff.Add(new JProperty("Leader", new JObject(new JProperty("Name", "Tom"), new JProperty("Age", 44), new JProperty("Department", "Personnel Department")))); Console.WriteLine(staff.ToString()); 除此之外，还可以通过一下方式来获取JObject.JArray类似。
| 方法 | 说明 |
JObject.Parse(string json) | json含有JSON对象的字符串，返回为JObject对象 |
JObject.FromObject(object o) | o为要转化的对象，返回一个JObject对象 |
JObject.Load(JsonReader reader) | reader包含着JSON对象的内容，返回一个JObject对象 |
2.创建JSON数组
JArray arr = new JArray();arr.Add(new JValue(1));arr.Add(new JValue(2));arr.Add(new JValue(3));Console.WriteLine(arr.ToString()); 三.使用Linq to JSON
1.查询
首先准备Json字符串，是一个包含员工基本信息的Jsonstring json = "{\"Name\" : \"Jack\", \"Age\" : 34, \"Colleagues\" : [{\"Name\" : \"Tom\" , \"Age\":44},{\"Name\" : \"Abel\",\"Age\":29}] }"; ①获取该员工的姓名
//将json转换为JObject JObject jObj = JObject.Parse(json);//通过属性名或者索引来访问，仅仅是自己的属性名，而不是所有的JToken ageToken = jObj["Age"];Console.WriteLine(ageToken.ToString()); ②获取该员工同事的所有姓名
//将json转换为JObject JObject jObj = JObject.Parse(json);var names=from staff in jObj["Colleagues"].Children() select (string)staff["Name"];foreach (var name in names) Console.WriteLine(name); "Children()"可以返回所有数组中的对象 2.修改 ①现在我们发现获取的json字符串中Jack的年龄应该为35
//将json转换为JObject JObject jObj = JObject.Parse(json); jObj["Age"] = 35; Console.WriteLine(jObj.ToString()); 注意不要通过以下方式来修改：
JObject jObj = JObject.Parse(json); JToken age = jObj["Age"]; age = 35; ②现在我们发现Jack的同事Tom的年龄错了，应该为45
//将json转换为JObject JObject jObj = JObject.Parse(json); JToken colleagues = jObj["Colleagues"]; colleagues[0]["Age"] = 45; jObj["Colleagues"] = colleagues;//修改后，再赋给对象 Console.WriteLine(jObj.ToString()); 3.删除 ①现在我们想删除Jack的同事
JObject jObj = JObject.Parse(json); jObj.Remove("Colleagues");//跟的是属性名称 Console.WriteLine(jObj.ToString()); ②现在我们发现Abel不是Jack的同事，要求从中删除 JObject jObj = JObject.Parse(json); jObj["Colleagues"][1].Remove(); Console.WriteLine(jObj.ToString()); 4.添加 ①我们发现Jack的信息中少了部门信息，要求我们必须添加在Age的后面
//将json转换为JObject JObject jObj = JObject.Parse(json); jObj["Age"].Parent.AddAfterSelf(new JProperty("Department", "Personnel Department")); Console.WriteLine(jObj.ToString()); ②现在我们又发现，Jack公司来了一个新同事Linda //将json转换为JObject JObject jObj = JObject.Parse(json); JObject linda = new JObject(new JProperty("Name", "Linda"), new JProperty("Age", "23")); jObj["Colleagues"].Last.AddAfterSelf(linda); Console.WriteLine(jObj.ToString()); 四.简化查询语句
使用函数SelectToken可以简化查询语句，具体：
①利用SelectToken来查询名称JObject jObj = JObject.Parse(json); JToken name = jObj.SelectToken("Name"); Console.WriteLine(name.ToString()); ②利用SelectToken来查询所有同事的名字 JObject jObj = JObject.Parse(json); var names = jObj.SelectToken("Colleagues").Select(p => p["Name"]).ToList(); foreach (var name in names) Console.WriteLine(name.ToString()); ③查询最后一名同事的年龄 //将json转换为JObject JObject jObj = JObject.Parse(json); var age = jObj.SelectToken("Colleagues[1].Age"); Console.WriteLine(age.ToString()); 1.如果Json中的Key是变化的但是结构不变，如何获取所要的内容? 1 { 2 "trends": 3 { 4 "2013-05-31 14:31": 5 [ 6 {"name":"我不是谁的偶像", 7 "query":"我不是谁的偶像", 8 "amount":"65172", 9 "delta":"1596"},10 {"name":"世界无烟日","query":"世界无烟日","amount":"33548","delta":"1105"},11 {"name":"最萌身高差","query":"最萌身高差","amount":"32089","delta":"1069"},12 {"name":"中国合伙人","query":"中国合伙人","amount":"25634","delta":"2"},13 {"name":"exo回归","query":"exo回归","amount":"23275","delta":"321"},14 {"name":"新一吻定情","query":"新一吻定情","amount":"21506","delta":"283"},15 {"name":"进击的巨人","query":"进击的巨人","amount":"20358","delta":"46"},16 {"name":"谁的青春没缺失","query":"谁的青春没缺失","amount":"17441","delta":"581"},17 {"name":"我爱幸运七","query":"我爱幸运七","amount":"15051","delta":"255"},18 {"name":"母爱10平方","query":"母爱10平方","amount":"14027","delta":"453"}19 ]20 },21 "as_of":136998189822 } 其中的"2013-05-31 14:31"是变化的key，如何获取其中的"name","query","amount","delta"等信息呢? 通过Linq可以很简单地做到: var jObj = JObject.Parse(jsonString); var tends = from c in jObj.First.First.First.First.Children() select JsonConvert.DeserializeObject(c.ToString());public class Trend{ public string Name { get; set; } public string Query { get; set; } public string Amount { get; set; } public string Delta { get; set; }}