如何在.NET中将C#对象转换为JSON字符串？

Question

我有这样的课程:

class MyDate
{
    int year, month, day;
}

class Lad
{
    string firstName;
    string lastName;
    MyDate dateOfBirth;
}

我想将一个Lad对象变成一个像这样的JSON字符串:

{
    "firstName":"Markoff",
    "lastName":"Chaney",
    "dateOfBirth":
    {
        "year":"1901",
        "month":"4",
        "day":"30"
    }
}

(没有格式化).我找到了这个链接,但它使用的是一个不在.NET 4中的命名空间.我也听说过JSON.NET,但是他们的网站目前似乎已经关闭了,我并不热衷于使用外部DLL文件.除了手动创建JSON字符串编写器之外还有其他选项吗？

Answer 1

因为我们都喜欢一个衬里

...这个依赖于Newtonsoft NuGet包,它比默认的序列化器更受欢迎.

Newtonsoft.Json.JsonConvert.SerializeObject(new {foo = "bar"})

文档:序列化和反序列化JSON

Answer 2

您可以使用JavaScriptSerializer该类(添加引用System.Web.Extensions):

using System.Web.Script.Serialization;

var json = new JavaScriptSerializer().Serialize(obj);

一个完整的例子:

using System;
using System.Web.Script.Serialization;

public class MyDate
{
    public int year;
    public int month;
    public int day;
}

public class Lad
{
    public string firstName;
    public string lastName;
    public MyDate dateOfBirth;
}

class Program
{
    static void Main()
    {
        var obj = new Lad
        {
            firstName = "Markoff",
            lastName = "Chaney",
            dateOfBirth = new MyDate
            {
                year = 1901,
                month = 4,
                day = 30
            }
        };
        var json = new JavaScriptSerializer().Serialize(obj);
        Console.WriteLine(json);
    }
}

Answer 3

使用Json.Net库,您可以从Nuget Packet Manager下载它.

序列化为Json字符串:

 var obj = new Lad
        {
            firstName = "Markoff",
            lastName = "Chaney",
            dateOfBirth = new MyDate
            {
                year = 1901,
                month = 4,
                day = 30
            }
        };

var jsonString = Newtonsoft.Json.JsonConvert.SerializeObject(obj);

反序列化为对象:

var obj = Newtonsoft.Json.JsonConvert.DeserializeObject<Lad>(jsonString );

Answer 4

使用DataContractJsonSerializer类:MSDN1,MSDN2.

我的例子:这里.

它也可以安全地从JSON字符串反序列化对象,不像JavaScriptSerializer.但我个人仍然喜欢Json.NET.

Answer 5

Wooou!使用JSON框架真的更好:)

这是我使用Json.NET(http://james.newtonking.com/json)的例子:

using System;
using System.Collections.Generic;
using System.Text;
using Newtonsoft.Json;
using System.IO;

namespace com.blogspot.jeanjmichel.jsontest.model
{
    public class Contact
    {
        private Int64 id;
        private String name;
        List<Address> addresses;

        public Int64 Id
        {
            set { this.id = value; }
            get { return this.id; }
        }

        public String Name
        {
            set { this.name = value; }
            get { return this.name; }
        }

        public List<Address> Addresses
        {
            set { this.addresses = value; }
            get { return this.addresses; }
        }

        public String ToJSONRepresentation()
        {
            StringBuilder sb = new StringBuilder();
            JsonWriter jw = new JsonTextWriter(new StringWriter(sb));

            jw.Formatting = Formatting.Indented;
            jw.WriteStartObject();
            jw.WritePropertyName("id");
            jw.WriteValue(this.Id);
            jw.WritePropertyName("name");
            jw.WriteValue(this.Name);

            jw.WritePropertyName("addresses");
            jw.WriteStartArray();

            int i;
            i = 0;

            for (i = 0; i < addresses.Count; i++)
            {
                jw.WriteStartObject();
                jw.WritePropertyName("id");
                jw.WriteValue(addresses[i].Id);
                jw.WritePropertyName("streetAddress");
                jw.WriteValue(addresses[i].StreetAddress);
                jw.WritePropertyName("complement");
                jw.WriteValue(addresses[i].Complement);
                jw.WritePropertyName("city");
                jw.WriteValue(addresses[i].City);
                jw.WritePropertyName("province");
                jw.WriteValue(addresses[i].Province);
                jw.WritePropertyName("country");
                jw.WriteValue(addresses[i].Country);
                jw.WritePropertyName("postalCode");
                jw.WriteValue(addresses[i].PostalCode);
                jw.WriteEndObject();
            }

            jw.WriteEndArray();

            jw.WriteEndObject();

            return sb.ToString();
        }

        public Contact()
        {
        }

        public Contact(Int64 id, String personName, List<Address> addresses)
        {
            this.id = id;
            this.name = personName;
            this.addresses = addresses;
        }

        public Contact(String JSONRepresentation)
        {
            //To do
        }
    }
}

考试:

using System;
using System.Collections.Generic;
using com.blogspot.jeanjmichel.jsontest.model;

namespace com.blogspot.jeanjmichel.jsontest.main
{
    public class Program
    {
        static void Main(string[] args)
        {
            List<Address> addresses = new List<Address>();
            addresses.Add(new Address(1, "Rua Dr. Fernandes Coelho, 85", "15º andar", "São Paulo", "São Paulo", "Brazil", "05423040"));
            addresses.Add(new Address(2, "Avenida Senador Teotônio Vilela, 241", null, "São Paulo", "São Paulo", "Brazil", null));

            Contact contact = new Contact(1, "Ayrton Senna", addresses);

            Console.WriteLine(contact.ToJSONRepresentation());
            Console.ReadKey();
        }
    }
}

结果:

{
  "id": 1,
  "name": "Ayrton Senna",
  "addresses": [
    {
      "id": 1,
      "streetAddress": "Rua Dr. Fernandes Coelho, 85",
      "complement": "15º andar",
      "city": "São Paulo",
      "province": "São Paulo",
      "country": "Brazil",
      "postalCode": "05423040"
    },
    {
      "id": 2,
      "streetAddress": "Avenida Senador Teotônio Vilela, 241",
      "complement": null,
      "city": "São Paulo",
      "province": "São Paulo",
      "country": "Brazil",
      "postalCode": null
    }
  ]
}

现在我将实现构造函数方法,该方法将接收JSON字符串并填充类的字段.

Answer 6

您可以使用Newtonsoft.json来实现。从Nuget安装Newtonsoft.json。然后：

using Newtonsoft.Json;

var jsonString = JsonConvert.SerializeObject(obj);

Answer 7

System.Text.Json名称空间中提供了一个新的JSON序列化器。它包含在.NET Core 3.0共享框架中，并且位于NuGet软件包中，用于针对.NET Standard或.NET Framework或.NET Core 2.x的项目。

示例代码：

using System;
using System.Text.Json;

public class MyDate
{
    public int year { get; set; }
    public int month { get; set; }
    public int day { get; set; }
}

public class Lad
{
    public string FirstName { get; set; }
    public string LastName { get; set; }
    public MyDate DateOfBirth { get; set; }
}

class Program
{
    static void Main()
    {
        var lad = new Lad
        {
            FirstName = "Markoff",
            LastName = "Chaney",
            DateOfBirth = new MyDate
            {
                year = 1901,
                month = 4,
                day = 30
            }
        };
        var json = JsonSerializer.Serialize(lad);
        Console.WriteLine(json);
    }
}

在此示例中，要序列化的类具有属性而不是字段。在System.Text.Json串行目前不序列字段。

说明文件：

• System.Text.Json概述
• 如何使用System.Text.Json

Answer 8

在 Lad 模型类中，添加对ToString()方法的重写 ，该方法返回 Lad 对象的 JSON 字符串版本。
注意：您需要导入System.Text.Json；

using System.Text.Json;

class MyDate
{
    int year, month, day;
}

class Lad
{
    public string firstName { get; set; };
    public string lastName { get; set; };
    public MyDate dateOfBirth { get; set; };
    public override string ToString() => JsonSerializer.Serialize<Lad>(this);
}

Answer 9

2023 年 10 月更新：

另一种使用内置System.Text.Json（.NET Core 3.0+）的解决方案，其中对象是自给自足的并且不会公开所有可能的字段：

通过测试（使用 NUnit）：

using NUnit.Framework;

namespace Intech.UnitTests
{
    public class UserTests
    {
        [Test]
        public void ConvertsItselfToJson()
        {
            var userName = "John";
            var user = new User(userName);

            var actual = user.ToJson();

            Assert.AreEqual($"{{\"Name\":\"{userName}\"}}", actual);
        }
    }
}

一个实现：

using System.Text.Json;
using System.Collections.Generic;

namespace Intech
{
    public class User
    {
        private readonly string name;

        public User(string name)
        {
            this.name = name;
        }

        public string ToJson()
        {
            var params = new Dictionary<string, string>{{"Name", name}};
            return JsonSerializer.Serialize(params);
        }
    }
}

Answer 10

如果您在ASP.NET MVC Web控制器中，则非常简单：

string ladAsJson = Json(Lad);

不敢相信没有人提到这一点。

Answer 11

如果它们不是很大，那么您的情况可能是将其导出为Json。这也使所有平台之间的可移植性

 using Newtonsoft.Json;
     [TestMethod]
        public void ExportJson()
        {
        double[,] b = new double[,] {
            { 110, 120, 130, 140, 150 },
            { 1110, 1120, 1130, 1140, 1150 },
            { 1000, 1, 5 ,9, 1000},
            {1110, 2, 6 ,10,1110},
            {1220, 3, 7 ,11,1220},
            {1330, 4, 8 ,12,1330} };


        string jsonStr = JsonConvert.SerializeObject(b);

        Console.WriteLine(jsonStr);

        string path = "X:\\Programming\\workspaceEclipse\\PyTutorials\\src\\tensorflow_tutorials\\export.txt";

        File.WriteAllText(path, jsonStr);
    }

Answer 12

我会投票给 ServiceStack 的 JSON Serializer：

using ServiceStack;

string jsonString = new { FirstName = "James" }.ToJson();

它也是 .NET 可用的最快的 JSON 序列化程序：http : //www.servicestack.net/benchmarks/