自定义JsonConverter WriteJson不会改变子属性的序列化

Mic*_*uso 12 c# serialization json converter json.net

我总觉得JSON序列化程序实际遍历整个对象的树,并在它遇到的每个接口类型对象上执行自定义JsonConverter的WriteJson函数 - 不是这样.

我有以下类和接口:

public interface IAnimal
{
    string Name { get; set; }
    string Speak();
    List<IAnimal> Children { get; set; }
}

public class Cat : IAnimal
{
    public string Name { get; set; }
    public List<IAnimal> Children { get; set; }        

    public Cat()
    {
        Children = new List<IAnimal>();
    }

    public Cat(string name="") : this()
    {
        Name = name;
    }

    public string Speak()
    {
        return "Meow";
    }       
}

 public class Dog : IAnimal
 {
    public string Name { get; set; }
    public List<IAnimal> Children { get; set; }

    public Dog()
    {
        Children = new List<IAnimal>();   
    }

    public Dog(string name="") : this()
    {
        Name = name;
    }

    public string Speak()
    {
        return "Arf";
    }

}
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为了避免JSON中的$ type属性,我编写了一个自定义JsonConverter类,其WriteJson是

public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
    JToken t = JToken.FromObject(value);

    if (t.Type != JTokenType.Object)
    {
        t.WriteTo(writer);                
    }
    else
    {
        IAnimal animal = value as IAnimal;
        JObject o = (JObject)t;

        if (animal != null)
        {
            if (animal is Dog)
            {
                o.AddFirst(new JProperty("type", "Dog"));
                //o.Find
            }
            else if (animal is Cat)
            {
                o.AddFirst(new JProperty("type", "Cat"));
            }

            foreach(IAnimal childAnimal in animal.Children)
            {
                // ???
            }

            o.WriteTo(writer);
        }
    }
}
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在这个例子中,是的,一只狗可以为孩子养猫,反之亦然.在转换器中,我想插入"type"属性,以便将其保存到序列化中.我有以下设置.(动物园只有一个名字和一个IAnimals列表.为了简洁和懒惰,我没有把它包括在内;))

Zoo hardcodedZoo = new Zoo()
            {   Name = "My Zoo",               
                Animals = new List<IAnimal> { new Dog("Ruff"), new Cat("Cleo"),
                    new Dog("Rover"){
                        Children = new List<IAnimal>{ new Dog("Fido"), new Dog("Fluffy")}
                    } }
            };

            JsonSerializerSettings settings = new JsonSerializerSettings(){
                ContractResolver = new CamelCasePropertyNamesContractResolver() ,                    
                Formatting = Formatting.Indented
            };
            settings.Converters.Add(new AnimalsConverter());            

            string serializedHardCodedZoo = JsonConvert.SerializeObject(hardcodedZoo, settings);
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serializedHardCodedZoo 序列化后具有以下输出:

{
  "name": "My Zoo",
  "animals": [
    {
      "type": "Dog",
      "Name": "Ruff",
      "Children": []
    },
    {
      "type": "Cat",
      "Name": "Cleo",
      "Children": []
    },
    {
      "type": "Dog",
      "Name": "Rover",
      "Children": [
        {
          "Name": "Fido",
          "Children": []
        },
        {
          "Name": "Fluffy",
          "Children": []
        }
      ]
    }
  ]
}
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类型属性显示在Ruff,Cleo和Rover上,但不适用于Fido和Fluffy.我猜WriteJson不是递归调用的.我如何在那里获得那种类型的财产?

顺便说一句,为什么它不是像我期望的那样的驼峰式IAnimals?

Bri*_*ers 19

您的转换器未应用于子对象的原因是因为JToken.FromObject()在内部使用了一个新的串行器实例,而不知道您的转换器.有一个允许你传入序列化程序的重载,但是如果你这样做会有另一个问题:因为你在转换器中并且你正在JToken.FromObject()尝试序列化父对象,你将进入无限递归环.(JToken.FromObject()调用串行器,调用转换器,调用JToken.FromObject()等)

若要解决此问题,您必须手动处理父对象.您可以使用一些反射来枚举父属性,而不会有太多麻烦:

public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
    JObject jo = new JObject();
    Type type = value.GetType();
    jo.Add("type", type.Name);

    foreach (PropertyInfo prop in type.GetProperties())
    {
        if (prop.CanRead)
        {
            object propVal = prop.GetValue(value, null);
            if (propVal != null)
            {
                jo.Add(prop.Name, JToken.FromObject(propVal, serializer));
            }
        }
    }
    jo.WriteTo(writer);
}
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小提琴:https://dotnetfiddle.net/sVWsE4

  • 但值得注意的是,这可能会导致忽略其他基于序列化的属性.例如,如果您将[JsonIgnore]放在属性上,它仍将添加上面的代码. (4认同)