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目前，当我结合@JsonIdentityInfo和@JsonTypeInfo. 下面的 Kotlin 代码在最后一行抛出异常。它dog1AndDog1Json按预期将实例序列化为 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 main 函数，得到以下输出：

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

其次是这个例外：

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>

似乎杰克逊希望animal2属性中的对象具有 @class 属性来找到要反序列化的正确类类型。但它已被@JsonIdentityInfo注释替换为 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？

我不确定杰克逊是否不支持这个用例，或者我做错了什么（我所希望的）。或者它可能是一个错误？

4

1 回答 1

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我设法通过以下方式使其工作：

  1. 从子类和子类中删除@JsonIdentityInfo注释DogCar
  2. 将 a 添加@JsonIdentityInfo到Animal基类
于 2019-12-10T08:20:59.737 回答