Newtonsoft.Json之JArray, JObject, JProperty,JValue-白红宇

Newtonsoft.Json之JArray, JObject, JProperty,JValue

阅读量：4589 次

发布时间：2019-06-09

本文共 4555 字，大约阅读时间需要 15 分钟。

https://www.cnblogs.com/shy1766IT/p/5304741.html

http://www.cnblogs.com/usharei/archive/2012/04/24/2467578.html

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());

JArray arr = new JArray();

arr.Add(new JValue(1));

arr.Add(new JValue(2));

arr.Add(new JValue(3));

Console.WriteLine(arr.ToString());

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

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());

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

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());

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());

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());

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());

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());

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());

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());

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());

定义一个错误提示：

JObject errors = new JObject();

if (productName.Length <= 0)

{

errors.Add("ProductName", new JValue("该输入项为必输项"));

}

//获取json里的值string jsonStr = "";//Json Str字符串            JToken json = JToken.Parse(jsonStr);//转化为JToken（JObject基类）            string xx = json.Value
       
        ("xx");//获取Json里xx键的值            JToken arr = json["arr"];//获取Json里的数组  {arr:[{yy:1,zz:2},{yy:3,zz:4}]}            foreach (JToken baseJ in arr)//遍历数组            {                int yy = baseJ.Value
        
         ("yy");            }            string yy1 = json["arr"][0].Value
         
          ("yy");//也可以酱紫，多层的获取            string yy2 = json["arr"][0]["yy"] != null ? json["arr"][0]["yy"].ToString() : "";//这个和上面句等价,不要直接ToString，容易报错