如何从复杂的JObject中获取价值？c#

using Newtonsoft.Json;
using Newtonsoft.Json.Linq;

// 复杂的JSON字符串
string jsonString = "{\"name\":\"John\",\"age\":30,\"address\":{\"street\":\"123 Main St\",\"city\":\"New York\"}}";

// 将JSON字符串转换为JObject对象
JObject jObject = JObject.Parse(jsonString);

// 获取name属性的值
string name = (string)jObject["name"];
Console.WriteLine("Name: " + name);

// 获取address对象中的city属性的值
string city = (string)jObject["address"]["city"];
Console.WriteLine("City: " + city);

// 使用SelectToken()方法获取特定属性的值
JToken ageToken = jObject.SelectToken("age");
int age = (int)ageToken;
Console.WriteLine("Age: " + age);