我有这样的类:
class MyDate
{
int year, month, day;
}
class Lad
{
string firstName;
string lastName;
MyDate dateOfBirth;
}
我想把一个Lad对象变成一个JSON字符串，就像这样:
{
"firstName":"Markoff",
"lastName":"Chaney",
"dateOfBirth":
{
"year":"1901",
"month":"4",
"day":"30"
}
}
(没有格式)。我找到了这个链接，但是它使用的命名空间不在. net 4中。我还听说过JSON。NET，但是他们的网站似乎暂时宕机了，而且我不喜欢使用外部DLL文件。
除了手动创建JSON字符串写入器，还有其他选项吗?
当前回答
Wooou !使用JSON框架真的更好:)
下面是我使用Json的例子。网(http://james.newtonking.com/json):
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
}
}
}
测试:
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。
其他回答
另一个解决方案是使用内置的System.Text.Json(。NET Core 3.0+)，其中对象是自给自足的，不公开所有可能的字段:
通过测试:
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);
}
}
}
使用Json。Net库，你可以从Nuget包管理器下载。
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);
反序列化到Object:
var obj = Newtonsoft.Json.JsonConvert.DeserializeObject<Lad>(jsonString );
如果它们不是很大，那就导出为JSON。
这也使得它在所有平台上都是可移植的。
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);
}
在System.Text.Json命名空间中有一个新的JSON序列化器。它包含在。net Core 3.0共享框架中，并且是针对。net Standard或。net framework或。net Core 2.x的项目的NuGet包。
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);
}
}
在这个例子中，要序列化的类有属性而不是字段;System.Text.Json序列化器目前不序列化字段。
文档:
System.Text.Json概述 如何使用System.Text.Json
你可以通过使用Newtonsoft.json来实现这一点。安装Newtonsoft。json从NuGet。然后:
using Newtonsoft.Json;
var jsonString = JsonConvert.SerializeObject(obj);
