C#中Jobject转string方法实现

using Newtonsoft.Json.Linq;

JObject jObject = new JObject
{
    ["name"] = "张三",
    ["age"] = 25,
    ["city"] = "北京"
};

string jsonString = jObject.ToString();
Console.WriteLine(jsonString);
// 输出: {"name":"张三","age":25,"city":"北京"}

using Newtonsoft.Json;

string jsonString = jObject.ToString(Formatting.None);
// 紧凑格式，无缩进

string formattedJson = jObject.ToString(Formatting.Indented);
Console.WriteLine(formattedJson);
// 输出:
// {
//   "name": "张三",
//   "age": 25,
//   "city": "北京"
// }

using Newtonsoft.Json;

string jsonString = JsonConvert.SerializeObject(jObject);
// 或者指定格式
string formattedJson = JsonConvert.SerializeObject(jObject, Formatting.Indented);

string originalJson = "{\"name\":\"李四\",\"age\":30}";
JObject jObject = JObject.Parse(originalJson);

// 修改内容
jObject["age"] = 31;

// 转回字符串
string updatedJson = jObject.ToString();

JObject jObject = new JObject
{
    ["users"] = new JArray
    {
        new JObject { ["name"] = "用户1", ["id"] = 1 },
        new JObject { ["name"] = "用户2", ["id"] = 2 }
    }
};

string jsonString = jObject.ToString(Formatting.Indented);

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

class Program
{
    static void Main()
    {
        // 创建 JObject
        JObject person = new JObject
        {
            ["firstName"] = "王",
            ["lastName"] = "小明",
            ["age"] = 28,
            ["hobbies"] = new JArray("阅读", "游泳", "编程"),
            ["address"] = new JObject
            {
                ["street"] = "长安街",
                ["city"] = "北京"
            }
        };
        
        // 转换为紧凑格式字符串
        string compactJson = person.ToString(Formatting.None);
        Console.WriteLine("紧凑格式:");
        Console.WriteLine(compactJson);
        
        Console.WriteLine("\n美化格式:");
        // 转换为美化格式字符串
        string prettyJson = person.ToString(Formatting.Indented);
        Console.WriteLine(prettyJson);
    }
}