1

使用 kotlinx.serialization 多态性,我想得到

{"type":"veh_t","owner":"Ivan","bodyType":"cistern","carryingCapacityInTons":5,"detachable":false}

但我明白了

{"type":"kotlin.collections.LinkedHashMap","owner":"Ivan","bodyType":"cistern","carryingCapacityInTons":5,"detachable":false}

我使用以下型号

interface Vehicle {
    val owner: String
}

@Serializable
@SerialName("veh_p")
data class PassengerCar(
    override val owner: String,
    val numberOfSeats: Int
) : Vehicle

@Serializable
@SerialName("veh_t")
data class Truck(
    override val owner: String,
    val body: Body
) : Vehicle {
    @Serializable
    data class Body(
        val bodyType: String,
        val carryingCapacityInTons: Int,
        val detachable: Boolean
        //a lot of other fields
    )    
}

我应用以下Json

inline val VehicleJson: Json get() = Json(context = SerializersModule {
        polymorphic(Vehicle::class) {
            PassengerCar::class with PassengerCar.serializer()
            Truck::class with TruckKSerializer
        }
    })

我使用序列化器 TruckKSerializer 因为服务器采用扁平结构。同时,在应用程序中我想使用一个对象 Truck.Body。对于扁平化,我根据这些类中的文档使用 JsonOutput 和 JsonInput覆盖fun serialize(encoder: Encoder, obj : T)和在 Serializator 中。fun deserialize(decoder: Decoder): T

object TruckKSerializer : KSerializer<Truck> {
    override val descriptor: SerialDescriptor = SerialClassDescImpl("Truck")

    override fun serialize(encoder: Encoder, obj: Truck) {
        val output = encoder as? JsonOutput ?: throw SerializationException("This class can be saved only by Json")
        output.encodeJson(json {
            obj::owner.name to obj.owner
            encoder.json.toJson(Truck.Body.serializer(), obj.body)
                .jsonObject.content
                .forEach { (name, value) ->
                    name to value
                }
        })
    }

    @ImplicitReflectionSerializer
    override fun deserialize(decoder: Decoder): Truck {
        val input = decoder as? JsonInput
            ?: throw SerializationException("This class can be loaded only by Json")
        val tree = input.decodeJson() as? JsonObject
            ?: throw SerializationException("Expected JsonObject")
        return Truck(
            tree.getPrimitive("owner").content,
            VehicleJson.fromJson<Truck.Body>(tree)
        )
    }
}

最后,我使用stringify(serializer: SerializationStrategy<T>, obj: T)

VehicleJson.stringify(
    PolymorphicSerializer(Vehicle::class),
    Truck(
        owner = "Ivan",
        body = Truck.Body(
            bodyType = "cistern",
            carryingCapacityInTons = 5,
            detachable = false
        )
    )
)

我最终得到{"type":"kotlin.collections.LinkedHashMap", ...},但我需要{"type":"veh_t", ...} 如何获得正确的类型?我想使用多态性Vehicle并使用 Truck.Body.serializer() 对 Body 对象进行编码以展平。

通过这种序列化,PassengerCar 类运行良好。

VehicleJson.stringify(
    PolymorphicSerializer(Vehicle::class),
    PassengerCar(
        owner = "Oleg",
        numberOfSeats = 4
    )
)

结果正确:

{"type":"veh_p","owner":"Oleg","numberOfSeats":4}

我认为问题在于自定义序列化程序TruckKSerializer。我注意到如果我在覆盖的fun serialize(encoder: Encoder, obj : T)下一个代码中使用

encoder
            .beginStructure(descriptor)
            .apply { 
                //...
            }
            .endStructure(descriptor)

我得到了正确的类型,但无法使用它的序列化程序展平对象 Truck.Body。

4

1 回答 1

0

打开和关闭组合的正确方法{} 是这段代码

val composite = encoder.beginStructure(descriptor)
// use composite instead of encoder here
composite.endStructure(descriptor)

你应该能够使用序列Body.encodeSerializable(Body.serializer(), body)

并始终传递描述符,否则它将退回到 json 字典的 LinkedhashMap 之类的东西

于 2019-10-21T17:59:52.617 回答