C#关于JArray和JObject封装JSON对象

￡神魔★判官ぃ 2021-06-26 16:06 446阅读 0赞

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;
        }

![在这里插入图片描述][20181018132801420]

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;
        }

![在这里插入图片描述][20181018132910782]

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;
        }

![在这里插入图片描述][20181018132937503]

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;
        }

![在这里插入图片描述][20181018133006151]

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;
        }

![在这里插入图片描述][201810181330270]

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

## Json.Net系列教程 4.Linq To JSON ##

### 一.Linq to JSON是用来干什么的? ###

`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.

<table> 
 <thead> 
  <tr> 
   <th align="left">类名</th> 
   <th align="left">说明</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td align="left">JObject</td> 
   <td align="left">用于操作JSON对象</td> 
  </tr> 
  <tr> 
   <td align="left">JArray</td> 
   <td align="left">用语操作JSON数组</td> 
  </tr> 
  <tr> 
   <td align="left">JValue</td> 
   <td align="left">表示数组中的值</td> 
  </tr> 
  <tr> 
   <td align="left">JProperty</td> 
   <td align="left">表示对象中的属性,以"key/value"形式</td> 
  </tr> 
  <tr> 
   <td align="left">JToken</td> 
   <td align="left">用于存放Linq to JSON查询后的结果</td> 
  </tr> 
 </tbody> 
</table>

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:  
![在这里插入图片描述][20181018134007302]

除此之外，还可以通过一下方式来获取`JObject.JArray`类似。

<table> 
 <thead> 
  <tr> 
   <th align="left">方法</th> 
   <th align="left">说明</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td align="left"><code>JObject.Parse(string json)</code></td> 
   <td align="left">json含有JSON对象的字符串，返回为JObject对象</td> 
  </tr> 
  <tr> 
   <td align="left"><code>JObject.FromObject(object o)</code></td> 
   <td align="left">o为要转化的对象，返回一个JObject对象</td> 
  </tr> 
  <tr> 
   <td align="left"><code>JObject.Load(JsonReader reader)</code></td> 
   <td align="left">reader包含着JSON对象的内容，返回一个JObject对象</td> 
  </tr> 
 </tbody> 
</table>

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:  
![在这里插入图片描述][20181018134335743]

### 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:

{ 
    "trends":
    { 
    "2013-05-31 14:31":
    [
    { "name":"我不是谁的偶像",
    "query":"我不是谁的偶像",
    "amount":"65172",
    "delta":"1596"},
    { "name":"世界无烟日","query":"世界无烟日","amount":"33548","delta":"1105"},
    { "name":"最萌身高差","query":"最萌身高差","amount":"32089","delta":"1069"},
    { "name":"中国合伙人","query":"中国合伙人","amount":"25634","delta":"2"},
    { "name":"exo回归","query":"exo回归","amount":"23275","delta":"321"},
    { "name":"新一吻定情","query":"新一吻定情","amount":"21506","delta":"283"},
    { "name":"进击的巨人","query":"进击的巨人","amount":"20358","delta":"46"},
    { "name":"谁的青春没缺失","query":"谁的青春没缺失","amount":"17441","delta":"581"},
    { "name":"我爱幸运七","query":"我爱幸运七","amount":"15051","delta":"255"},
    { "name":"母爱10平方","query":"母爱10平方","amount":"14027","delta":"453"}
    ]
    },
    "as_of":1369981898
    }

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:

[20181018132801420]: /images/20210626/a4f4a023c0fb48308a946b78cd7d3b52.png
[20181018132910782]: /images/20210626/ca91a2aaf20540ecabaf098efac37d12.png
[20181018132937503]: /images/20210626/48c566ef16a9477db6c1a264fe42e33b.png
[20181018133006151]: /images/20210626/78f10327748347b28bf5c7d7b56fb201.png
[201810181330270]: /images/20210626/087e370267354204a08d238c6852961a.png
[20181018134007302]: /images/20210626/91214a16163f4f3087387a8177a0e516.png
[20181018134335743]: /images/20210626/3dd99d68333541da9778bdac579feec6.png