java 使 @JsonTypeInfo 属性可选

Make @JsonTypeInfo property optional

提问by Mark Peters

I'm using @JsonTypeInfoto instruct Hymanson to look in the @classproperty for concrete type information. However, sometimes I don't want to have to specify @class, particularly when the subtype can be inferred given the context. What's the best way to do that?

我@JsonTypeInfo用来指示 Hymanson 在@class属性中查找具体类型信息。但是，有时我不想指定@class，特别是当可以根据上下文推断子类型时。 最好的方法是什么？

Here's an example of the JSON:

这是 JSON 的示例：

{ 
    "owner": {"name":"Dave"},
    "residents":[
        {"@class":"Hymansonquestion.Dog","breed":"Greyhound"},
        {"@class":"Hymansonquestion.Human","name":"Cheryl"},
        {"@class":"Hymansonquestion.Human","name":"Timothy"}
    ]
}

and I'm trying to deserialize them into these classes (all in Hymansonquestion.*):

我正在尝试将它们反序列化为这些类（全部在 中Hymansonquestion.*）：

public class Household {
    private Human owner;
    private List<Animal> residents;

    public Human getOwner() { return owner; }
    public void setOwner(Human owner) { this.owner = owner; }
    public List<Animal> getResidents() { return residents; }
    public void setResidents(List<Animal> residents) { this.residents = residents; }
}

public class Animal {}

public class Dog extends Animal {
    private String breed;
    public String getBreed() { return breed; }
    public void setBreed(String breed) { this.breed = breed; }
}

public class Human extends Animal {
    private String name;
    public String getName() { return name; }
    public void setName(String name) { this.name = name; }
}

using this config:

使用这个配置：

@JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "@class")
private static class AnimalMixin {
}

//...

ObjectMapper objectMapper = new ObjectMapper();
objectMapper.getDeserializationConfig().addMixInAnnotations(Animal.class, AnimalMixin.class);
Household household = objectMapper.readValue(json, Household.class);
System.out.println(household);

As you can see, the owner is declared as a Human, not an Animal, so I want to be able to omit @classand have Hymanson infer the type as it normally would.

如您所见，所有者被声明为人类，而不是动物，因此我希望能够@class像往常一样省略并让Hyman逊推断类型。

When I run this though, I get

当我运行这个时，我得到

org.codehaus.Hymanson.map.JsonMappingException: Unexpected token (END_OBJECT), 
   expected FIELD_NAME: missing property '@class' that is to contain type id  (for class Hymansonquestion.Human)

Since "owner" doesn't specify @class.

由于“所有者”没有指定@class.

Any ideas? One initial thought I had was to use @JsonTypeInfoon the property rather than the type. However, this cannot be leveraged to annotate the element type of a list.(Incorrect, see answer)

有任何想法吗？我最初的一个想法是使用@JsonTypeInfo属性而不是类型。 但是，这不能用于注释列表的元素类型。（不正确，见答案）

采纳答案by StaxMan

You probably should not actually do this -- it seems like a micro-optimization for special cases, complicating life -- but if you really think you do want to, you can try adding @JsonTypeInfooverride on Human. Annotations are inheritable in Hymanson, and you can override definitions. In this case which one gets used then depends on declared type: so anything declared as Humanwould see annotation on Human; and anything declared as Animalonly one in `Animal.

您可能实际上不应该这样做——这似乎是对特殊情况的微优化，使生活复杂化——但如果你真的认为你愿意，你可以尝试@JsonTypeInfo在 Human 上添加覆盖。注释在 Hymanson 中是可继承的，您可以覆盖定义。在这种情况下，使用哪个取决于声明的类型：因此声明为的任何内容Human都会在Human;上看到注释。以及任何Animal在“Animal”中声明为唯一的东西。

One tricky case is the root value (value you directly serialize): since there is no declared type, it will use the runtime type. And this will probably not work the way you want.

一个棘手的情况是根值（您直接序列化的值）：由于没有声明类型，它将使用运行时类型。这可能不会以您想要的方式工作。

Another possibility is sub-classing AnnotationIntrospector: you can change the handling of @JsonTypeInfothere as well; just see what HymansonAnnotationIntrospectordoes, override behavior as applicable.

另一种可能性是子类化AnnotationIntrospector：您也可以更改@JsonTypeInfo那里的处理方式；看看有什么HymansonAnnotationIntrospector作用，根据情况覆盖行为。

回答by Mark Peters

So it turns out that I had misunderstood the Javadoc for @JsonTypeInfo. When I said in my question

所以事实证明我误解了Javadoc 的@JsonTypeInfo. 当我在我的问题中说

One initial thought I had was to use @JsonTypeInfo on the property rather than the type. However, this cannot be leveraged to annotate the element type of a list.

我最初的一个想法是在属性上使用 @JsonTypeInfo 而不是类型。但是，这不能用于注释列表的元素类型。

I was basing that on this quote from the Javadoc:

我是基于 Javadoc 中的这句话：

When used for properties (fields, methods), this annotation applies to values: so when applied to structure types (like Collection, Map, arrays), will apply to contained values, not the container; for non-structured types there is no difference. (...) There is no per-property way to force type information to be included for type of container (structured type); for container types one has to use annotation for type declaration.

当用于属性（字段、方法）时，此注释适用于值：因此，当应用于结构类型（如 Collection、Map、数组）时，将应用于包含的值，而不是容器；对于非结构化类型，没有区别。(...) 没有针对每个属性的方法来强制包含容器类型（结构化类型）的类型信息；对于容器类型，必须使用注释进行类型声明。

I somehow misread that to mean the opposite; that the type info would apply to the container and not the elements. Obviously that was wrong. So I was able to fix this by using the following:

我不知何故误读了相反的意思；类型信息将应用于容器而不是元素。显然这是错误的。所以我能够通过使用以下方法来解决这个问题：

public class Household {
   //...

   @JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "@class")
   public void setResidents(List<Animal> residents) { 
      this.residents = residents; 
   }
}

Now @classis only required for Animals that are specified in the residentsproperty.

Now@class仅对residents属性中指定的 Animals 是必需的。