Question

我有一个Kotlin密封类-Pet和两个子类-Dog和Cat。我的应用程序需要传输以JSON序列化的宠物集合。为了区分子类，我使用Jackson @JsonTypeInfo和@JsonSubTypes批注。下面的清单：

@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.PROPERTY, property = "type")
@JsonSubTypes(
    JsonSubTypes.Type(value = Dog::class, name = "dog"),
    JsonSubTypes.Type(value = Cat::class, name = "cat")
)
sealed class Pet { abstract val name: String }

data class Dog(override val name: String): Pet()
data class Cat(override val name: String): Pet()

单个实例已正确序列化和反序列化：

    @Test
    fun `serialize dog`() {
        val dog = Dog("Kevin")
        val dogJson = objectMapper.writeValueAsString(dog)

        JsonAssert.assertEquals(dogJson, """{"type":"dog","name":"Kevin"}""")
        val newDog = objectMapper.readValue<Dog>(dogJson)
    }

当一系列宠物被序列化和反序列化时出现问题：

    @Test
    fun `serialize dog and cat`() {
        val pets: Set<Pet> = setOf(Dog("Kevin"), Cat("Marta"))
        val petsJson = objectMapper.writeValueAsString(pets)

        JsonAssert.assertEquals(petsJson, """[{"name":"Kevin"},{"name":"Marta"}]""")
        val newPets = objectMapper.readValue<Set<Pet>>(petsJson)
    }

Jackson在序列化过程中吞下了type属性，由于objectMapper无法readValue：

com.fasterxml.jackson.databind.exc.InvalidTypeIdException: Missing type id when trying to resolve subtype of [simple type, class s3.moria.workflows.common.model.Pet]: missing type id property 'type'
 at [Source: (String)"[{"name":"Kevin"},{"name":"Marta"}]"; line: 1, column: 17] (through reference chain: java.util.HashSet[0])

任何想法如何解决这个问题？还是解决方法？

Jackson版本：2.9.0

Answer 1

这实际上不是错误，而是功能。对于具有泛型的集合，Jackson将忽略您的子类型注释。这里有一个讨论：

https://github.com/FasterXML/jackson-databind/issues/1816

以下2个“修复”对我有用，并且所需的设置比上面的答案要少（我认为我们也许正在使用不同的杰克逊版本，但是我无法让杰克逊为子类使用非默认构造函数，所以我用lateinit重写了子类定义）

一种解决此问题的方法是：

Create your own set writer

{
    "data": [
        {
            "key1": "value1",
            "key2": "value1",
            "key3": "value1",
            "key4": "value4"
        },
        {
            "key2": "value1",
            "key3": "value1",
            "key4": "value4"
        },
        {
            "key1": "value1",
            "key4": "value4"
        },
        {
            "key4": "value4"
        }
    ]
}

这些测试通过了（再次，JSONAssert对我来说是不同的方法签名）

@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.PROPERTY, property = "type")
@JsonSubTypes(
    JsonSubTypes.Type(value = Dog1::class, name = "dog"),
    JsonSubTypes.Type(value = Cat1::class, name = "cat")
)
sealed class Pet1 {
    abstract val name: String
}

class Dog1 : Pet1() {
    override lateinit var name: String
}

class Cat1 : Pet1() {
    override lateinit var name: String
}

您还可以使用这种方法： Workaround without custom serializers/deserializers

您的代码如下：

package com.example.demo

import com.fasterxml.jackson.databind.ObjectMapper
import com.fasterxml.jackson.module.kotlin.readValue
import org.junit.jupiter.api.Test
import org.skyscreamer.jsonassert.JSONAssert

internal class PetTest1 {

    private var objectMapper = ObjectMapper()

    @Test
    fun `serialize dog`() {
        val dog = Dog1()
        dog.name = "Kevin"
        val dogJson = objectMapper.writeValueAsString(dog)

        JSONAssert.assertEquals(dogJson, """{"type":"dog","name":"Kevin"}""", true)
        val newDog = objectMapper.readValue<Dog1>(dogJson)
    }

    @Test
    fun `serialize dog and cat with mapper`() {
        val dog = Dog1()
        dog.name = "Kevin"
        val cat = Cat1()
        cat.name = "Marta"
        val pets: Set<Pet1> = setOf(dog, cat)
        val petCollectionType = objectMapper.typeFactory
            .constructCollectionType(Set::class.java, Pet1::class.java)

        val petsJson = objectMapper.writer().forType(petCollectionType).writeValueAsString(pets)

        JSONAssert.assertEquals(
            petsJson, """[{"type":"dog","name":"Kevin"},{"type":"cat","name":"Marta"}]""", true
        )
        val newPets = objectMapper.readValue<Set<Pet1>>(petsJson)
    }
}

以下测试通过

@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.EXISTING_PROPERTY)
@JsonSubTypes(
    JsonSubTypes.Type(value = Dog::class, name = "dog"),
    JsonSubTypes.Type(value = Cat::class, name = "cat")
)
sealed class Pet {
    abstract val jacksonMarker: String
        @JsonProperty("@type")
        get
    abstract val name: String
}

class Dog : Pet() {
    override val jacksonMarker: String
        get() = "dog"
    override lateinit var name: String
}

class Cat : Pet() {
    override val jacksonMarker: String
        get() =  "cat"
    override lateinit var name: String
}

Answer 2

您应该进行的更改
首先，在JsonTypeInfo，您需要将 visibility设置为true 然后在反序列化器中可以使用 type 属性。

然后您需要实施PetDeserializer

下面是一个示例： Pet.kt

    import com.fasterxml.jackson.annotation.JsonSubTypes
    import com.fasterxml.jackson.annotation.JsonTypeInfo
    import com.fasterxml.jackson.core.JsonParser
    import com.fasterxml.jackson.core.JsonProcessingException
    import com.fasterxml.jackson.databind.DeserializationContext
    import com.fasterxml.jackson.databind.JsonNode
    import com.fasterxml.jackson.databind.annotation.JsonDeserialize
    import com.fasterxml.jackson.databind.deser.std.StdDeserializer
    import java.io.IOException
    import kotlin.jvm.Throws
    
    @JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.PROPERTY, property = "type",visible=true)
    @JsonSubTypes(
        JsonSubTypes.Type(value = Dog::class, name = "dog"),
        JsonSubTypes.Type(value = Cat::class, name = "cat")
    )
    
    
    @JsonDeserialize(using = PetDeserializer::class)
    sealed class Pet {
        abstract val name: String
    }
    
    data class Dog(override val name: String) : Pet()
    data class Cat(override val name: String) : Pet()
    
    class PetDeserializer @JvmOverloads constructor(vc: Class<*>? = Pet::class.java) :
        StdDeserializer<Pet?>(vc) {
        @Throws(IOException::class, JsonProcessingException::class)
        override fun deserialize(jp: JsonParser, ctxt: DeserializationContext): Pet {
            val node = jp.codec
                .readTree<JsonNode>(jp)
            val itemName = node["name"]
                .asText()
            val type = node["type"]
                .asText()
            return when (type) {
                "dog" -> Dog(itemName)
                "cat" -> Cat(itemName)
                else -> throw Error("unknown type")
            }
        }
    
        companion object {
            private const val serialVersionUID = 1883547683050039861L
        }
    }

以及 PetTest.kt

import com.fasterxml.jackson.databind.ObjectMapper
import org.junit.Test
import org.skyscreamer.jsonassert.JSONAssert


class HelloTest {

    val objectMapper = ObjectMapper()

    @Test
    fun `serialize dog`() {
        val dog = Dog("Kevin")
        val dogJson = objectMapper.writeValueAsString(dog)
        JSONAssert.assertEquals("""{"type":"dog","name":"Kevin"}""", dogJson, false)
        val dogType = objectMapper.typeFactory.constructType(Dog::class.java)
        if (objectMapper.canDeserialize(dogType)) {
            ObjectMapper().readValue<Dog>(dogJson, Dog::class.java)
        } else {
              throw Error("deserializer not loaded")

        }
    }

}

Maven依赖项：

        <dependency>
            <groupId>com.fasterxml.jackson.datatype</groupId>
            <artifactId>jackson-datatype-jsr310</artifactId>
            <version>2.9.4</version>
        </dependency>
        <dependency>
            <groupId>com.fasterxml.jackson.datatype</groupId>
            <artifactId>jackson-datatype-joda</artifactId>
            <version>2.9.4</version>
        </dependency>
        <dependency>
            <groupId>com.fasterxml.jackson.core</groupId>
            <artifactId>jackson-core</artifactId>
            <version>2.9.4</version>
        </dependency>

        <dependency>
            <groupId>org.skyscreamer</groupId>
            <artifactId>jsonassert</artifactId>
            <version>1.5.0</version>
        </dependency>

Kotlin和Jackson-尝试解析简单类型的子类型时缺少类型ID

2 个答案: