杰克逊结合@JsonIdentityInfo 和@JsonTypeInfo 抛出 InvalidTypeIdException

Question

目前，当我合并 @JsonIdentityInfo 和 @JsonTypeInfo 时，我遇到了 Jackson 的问题。下面的 Kotlin 代码在最后一行抛出异常。它按预期将 dog1AndDog1Json 实例序列化为 Json，但随后在将其反序列化回实例时抛出异常。

package some.test

import com.fasterxml.jackson.annotation.*
import com.fasterxml.jackson.databind.ObjectMapper

@JsonTypeInfo(use = JsonTypeInfo.Id.CLASS)
@JsonSubTypes(JsonSubTypes.Type(value = Dog::class), JsonSubTypes.Type(value = Cat::class))
interface Animal {
    val name: String
}

@JsonIdentityInfo(generator = ObjectIdGenerators.IntSequenceGenerator::class)
data class Dog(@JsonProperty("name") override val name: String) : Animal

@JsonIdentityInfo(generator = ObjectIdGenerators.IntSequenceGenerator::class)
data class Cat(@JsonProperty("name") override val name: String) : Animal

data class TwoAnimals(@JsonProperty("animal1") val animal1: Animal, @JsonProperty("animal2") val animal2: Animal)

fun main() {
    val om = ObjectMapper();

    val dog1 = Dog("Dog1")
    val dog2 = Dog("Dog2")
    val cat1 = Cat("Cat1")

    val dog1AndDog2 = TwoAnimals(dog1, dog2)
    val dog1AndDog2Json = om.writerWithDefaultPrettyPrinter().writeValueAsString(dog1AndDog2)
    assert(dog1AndDog2 === om.readValue(dog1AndDog2Json, TwoAnimals::class.java)) // OK

    val dog1AndCat1 = TwoAnimals(dog1, cat1)
    val dog1AndCat2Json = om.writerWithDefaultPrettyPrinter().writeValueAsString(dog1AndCat1)
    assert(dog1AndCat1 === om.readValue(dog1AndCat2Json, TwoAnimals::class.java)) // OK

    val dog1AndDog1 = TwoAnimals(dog1, dog1)
    val dog1AndDog1Json = om.writerWithDefaultPrettyPrinter().writeValueAsString(dog1AndDog1)
    println(dog1AndDog1Json)
    assert(dog1AndDog1 === om.readValue(dog1AndDog1Json, TwoAnimals::class.java)) // DESERIALIZE FAILS
}

然后我运行主函数我得到以下输出：

{
  "animal1" : {
    "@class" : "some.test.Dog",
    "@id" : 1,
    "name" : "Dog1"
  },
  "animal2" : 1
}

紧随其后的是这个异常：

Exception in thread "main" com.fasterxml.jackson.databind.exc.InvalidTypeIdException: Missing type id when trying to resolve subtype of [simple type, class some.test.Animal]: missing type id property '@class' (for POJO property 'animal2')
 at [Source: (String)"{
  "animal1" : {
    "@class" : "some.test.Dog",
    "@id" : 1,
    "name" : "Dog1"
  },
  "animal2" : 1
}"; line: 7, column: 15] (through reference chain: some.test.TwoAnimals["animal2"])
    at com.fasterxml.jackson.databind.exc.InvalidTypeIdException.from(InvalidTypeIdException.java:43)
<truncated rest of stacktrace>

Jackson 似乎希望 animal2 属性处有一个 @class 属性的对象找到正确的 class 类型反序列化。但它已被 @JsonIdentityInfo 注释替换为一个 id。为什么 Jackson 不通过该 id 查找对象，然后检查该实例的 @class 属性？

我不确定 Jackson 是否不支持这个用例，或者我做错了什么（我所希望的）。或者这可能是一个错误？

Answer 1

我设法通过以下方式让它工作：

正在从 Dog 和 Car 子 classes

@JsonIdentityInfo

将 @JsonIdentityInfo 添加到 Animal 基础 class

杰克逊结合@JsonIdentityInfo 和@JsonTypeInfo 抛出 InvalidTypeIdException

Jackson combining @JsonIdentityInfo and @JsonTypeInfo throws InvalidTypeIdException

serialization

json

jackson

object-identity