JSON TO JOBJECT转换的使用方法

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Json

string 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":1369981898
22 }

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<Trend>(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; }
}