C# 关于JArray和JObject封装JSON对象

1、JObject：基本的json对象

    /// <summary>
    /// Gets the j object.
    /// </summary>
    /// <returns></returns>
    public JObject GetJObject() {
        var obj = new JObject {{"Name", "Mark" } };
        return obj;
    }

2、JObject：嵌套子对象（JObject嵌JObject）

    /// <summary>
    /// Gets the j object.
    /// </summary>
    /// <returns></returns>
    public JObject GetJObject() {
        var obj = new JObject {{"Name", "Mark"}, {"Age", 8 }};
        var info = new JObject {{"Phone", "132****7777"}, {"Gender", "男"}};
        obj.Add("Info", info);
        return obj;
    }

3、JArray：基本json对象中的数组

    /// <summary>
    /// Gets the j array.
    /// </summary>
    /// <returns></returns>
    public JArray GetJArray() {
        var jarray = new JArray();
        var mark = new JObject { { "Name", "Mark" }, { "Age", 8 } };
        var jack = new JObject { { "Name", "Jack" }, { "Age", 9 } };
        jarray.Add(mark);
        jarray.Add(jack);
        return jarray;
    }

4、JArray： 多个json对象数组

    /// <summary>
    /// Gets the j array.
    /// </summary>
    /// <returns></returns>
    public JObject GetJArray() {
        var obj = new JObject();
        var student = new JArray
        {
            new JObject {{ "Name", "Mark" }, { "Age", 8 } },
            new JObject {{ "Name", "Jack" }, { "Age", 9 } }
        };
        var results = new JArray
        {
            new JObject {{ "Subject", "语文"}, { "Score", 100}},
            new JObject {{ "Subject", "数学" }, { "Score", 88}}
        };
        obj.Add("Student", student);
        obj.Add("Results", results);
        return obj;
    }

5、JArray：json数组嵌套数组（一个学生对应多个课程分数）

    /// <summary>
    /// Gets the results.
    /// </summary>
    /// <returns></returns>
    public JObject GetResults() {
        var mark = new JObject { { "Name", "Mark" }, { "Age", "8" } };
        var results = new JArray
        {
            new JObject {{ "Subject", "语文"}, { "Score", 100}},
            new JObject {{ "Subject", "数学" }, { "Score", 88}}
        };            
        mark.Add("Results", results);
        return mark;
    }

//--------------------------------------------------------------------------

Linq to JSON是用来操作JSON对象的.可以用于快速查询,修改和创建JSON对象.当JSON对象内容比较复杂,而我们仅仅需要其中的一小部分数据时,可以考虑使用Linq to JSON来读取和修改部分的数据而非反序列化全部.

二.创建JSON数组和对象

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());

结果:

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());

结果:

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());

结果:

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

结果：

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());

结果：

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());

结果：

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());

结果：

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());

结果：

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());

结果：

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());

结果：

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());

结果：

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());

结果：

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());

结果：

 

FAQ

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?

例如:

 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; }
}