Linq to
JSON是故来操作JSON对象的.可以用于快速查询,修改及创造JSON对象.当JSON对象内容比较复杂,而我们只是要中的一模一样稍微有数据常常,可以设想下Linq
to JSON来读取和改部分的数额如果未反序列化全部.

二.创建JSON数组和目标

在进展Linq to JSON之前,首先使打听一下用来操作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

json 1

            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());

json 2

结果:

json 3

除去,还好透过瞬间智来获取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

json 4

            JArray arr = new JArray();
            arr.Add(new JValue(1));
            arr.Add(new JValue(2));
            arr.Add(new JValue(3));
            Console.WriteLine(arr.ToString());

json 5

结果:

json 6

三.使用Linq to JSON

1.查询
先是准备Json字符串,是一个分包员工基本信息的Json

string json = "{\"Name\" : \"Jack\", \"Age\" : 34, \"Colleagues\" : [{\"Name\" : \"Tom\" , \"Age\":44},{\"Name\" : \"Abel\",\"Age\":29}] }";

①落该员工的真名

json 7

            //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());

json 8

结果:

json 9

②沾取该员工同事的装有人名

json 10

            //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);

json 11

“Children()”可以回来所有数组中之靶子

结果:

json 12

2.修改

①现行咱们发现沾之json字符串中Jack的年华应当为35

            //将json转换为JObject
            JObject jObj = JObject.Parse(json);
            jObj["Age"] = 35;
            Console.WriteLine(jObj.ToString());

结果:

json 13

在意不要通过以下措施来修改:

            JObject jObj = JObject.Parse(json);
            JToken age = jObj["Age"];
            age = 35;

②现行咱们发现Jack的同事Tom的年龄错了,应该吗45

json 14

            //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());

json 15

结果:

json 16

3.删除
①现我们怀念抹Jack的同事

            JObject jObj = JObject.Parse(json);
            jObj.Remove("Colleagues");//跟的是属性名称
            Console.WriteLine(jObj.ToString());

结果:

json 17

②现行咱们发现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());

结果:

json 18

4.添加
①咱们发现Jack的消息遭到遗失了部门信息,要求我们必须上加在Age的后

            //将json转换为JObject
            JObject jObj = JObject.Parse(json);
            jObj["Age"].Parent.AddAfterSelf(new JProperty("Department", "Personnel Department"));
            Console.WriteLine(jObj.ToString());

结果:

json 19

②本我们以发现,Jack公司来了一个初同事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());

结果:

json 20

四.简化查询语句

运用函数SelectToken可以简化查询语句,具体:
①行使SelectToken来查询称

            JObject jObj = JObject.Parse(json);
            JToken name = jObj.SelectToken("Name");
            Console.WriteLine(name.ToString());

结果:

json 21

②运SelectToken来询问所有同事的讳

            JObject jObj = JObject.Parse(json);
            var names = jObj.SelectToken("Colleagues").Select(p => p["Name"]).ToList();
            foreach (var name in names)
                Console.WriteLine(name.ToString());

结果:

json 22

③查询最后一叫做同事的年纪

            //将json转换为JObject
            JObject jObj = JObject.Parse(json);
            var age = jObj.SelectToken("Colleagues[1].Age");
            Console.WriteLine(age.ToString());

结果:

json 23

 

FAQ

1.而Json中的Key是转之不过结构不移,如何获取所假设的情节?

例如:

json 24

 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 }

json 25

其中的”2013-05-31
14:31″是变的key,如何获得其中的”name”,”query”,”amount”,”delta”等消息为?
由此Linq可以好粗略地形成:

json 26

 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; }
}

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