5

我需要将 KnownType 添加到下面的代码中才能成功序列化。当我这样做时，生成的 JSON 如下：

JSON form of Adult with 1 child: {"age":42,"name":"John","children":[{"__type":"
Child:#TestJson","age":4,"name":"Jane","fingers":10}]}

我如何让它不包括“__type”：“Child：#TestJson”？我们在一些查询中返回数百个这样的元素，并且额外的文本会加起来。

完整代码：

using System;
using System.Collections.Generic;
using System.IO;
using System.Runtime.Serialization;
using System.Runtime.Serialization.Json;

namespace TestJson
{
    class Program
    {
        static void Main(string[] args)
        {
            Adult parent = new Adult {name = "John", age = 42};

            MemoryStream stream1 = new MemoryStream();
            DataContractJsonSerializer ser = new DataContractJsonSerializer(typeof(Adult));
            ser.WriteObject(stream1, parent);

            stream1.Position = 0;
            StreamReader sr = new StreamReader(stream1);
            Console.Write("JSON form of Adult with no children: ");
            Console.WriteLine(sr.ReadToEnd());


            Child child = new Child { name = "Jane", age = 4, fingers=10 };

            stream1 = new MemoryStream();
            ser = new DataContractJsonSerializer(typeof(Child));
            ser.WriteObject(stream1, child);

            stream1.Position = 0;
            sr = new StreamReader(stream1);
            Console.Write("JSON form of Child with no parent: ");
            Console.WriteLine(sr.ReadToEnd());


            // now connect the two
            parent.children.Add(child);

            stream1 = new MemoryStream();
            ser = new DataContractJsonSerializer(typeof(Adult));
            ser.WriteObject(stream1, parent);

            stream1.Position = 0;
            sr = new StreamReader(stream1);
            Console.Write("JSON form of Adult with 1 child: ");
            Console.WriteLine(sr.ReadToEnd());
        }
    }

    [DataContract]
    [KnownType(typeof(Adult))]
    [KnownType(typeof(Child))]
    class Person
    {
        [DataMember]
        internal string name;

        [DataMember]
        internal int age;
    }

    [DataContract]
    class Adult : Person
    {
        [DataMember] 
        internal List<Person> children = new List<Person>();
    }

    [DataContract]
    class Child : Person
    {
        [DataMember]
        internal int fingers;
    }
}
4

1 回答 1

14

正如我在上一个问题中告诉你的那样，我不知道，但一些研究让我相信以下可能会达到你想要的效果：

var settings = new DataContractJsonSerializerSettings();
settings.EmitTypeInformation = EmitTypeInformation.Never;

var serializer = new DataContractJsonSerializer(yourType, settings);
于 2013-07-23T16:29:33.493 回答