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}]}
我怎么能不包含“__type”:“Child:#TestJson”?我们在一些查询中返回数百个这些元素,并且会添加额外的文本。
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;
}
}
答案 0 :(得分:13)
正如我在上一个问题中告诉你的那样,我不知道,但是一些研究让我相信以下内容可能达到你想要的效果:
var settings = new DataContractJsonSerializerSettings();
settings.EmitTypeInformation = EmitTypeInformation.Never;
var serializer = new DataContractJsonSerializer(yourType, settings);