C#Newtonsoft.Json用法详解

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());

// 创建数组
jarray array = new JArray();
array.Add(new JValue("吃饭"));
array.Add(new JValue("睡觉"));
obj.Add("Favorites", array);
obj.Add("Remark", null);
 
Console.WriteLine(array.ToString());

JArray array = new JArray("吃饭", "睡觉");

string json = "{\"Name\" : \"Jack\", \"Age\" : 34, \"Colleagues\" : [{\"Name\" : \"Tom\" , \"Age\":44},{\"Name\" : \"Abel\",\"Age\":29}] }";
//将json转换为JObject
JObject jObj = JObject.Parse(json);
JToken ageToken =  jObj["Age"];
Console.WriteLine(ageToken.ToString());

public class Student
{
    public int ID { get; set; }
    public string Name { get; set; }
}
 
Student student = new Student();
student.ID = "1";
student.Name = "jack";
 
//序列化为JSON
string json = JsonConvert.SerializeObject(student);

public class Student
{
    public int ID { get; set; }
    public string Name { get; set; }
}
 
string json = ".....";
Student student = JsonConvert.DeserializeObject<Student>(json);

string json = "{\"ID\":1,\"Name\":\"张三\",\"Favorites\":[\"吃饭\",\"睡觉\"]}";
 
JObject obj = JObject.Parse(json);