Kotlinx 序列化，避免在其他数据类型上崩溃

Question

我在应用程序中使用外部 API，而反序列化是使用 Kotlinx 序列化包完成的，当 api 结果是多个值的 Int 数组和原始 int 时，我遇到了问题对于单个值。我怎样才能避免在此过程中崩溃。有没有更好的方法来避免崩溃或创建数据类

例如：

import kotlinx.serialization.Serializable

@Serializable

data class Bookings (val slots: List<Int>)

当插槽具有单一值时 API returns {slots: 1} 当插槽有多个值时 API return { slots: [1,2,3,4]}

Answer 1

可以使用自定义序列化程序来完成：

import kotlinx.serialization.*
import kotlinx.serialization.json.Json
import kotlinx.serialization.json.JsonInput
import kotlinx.serialization.json.JsonObject
import kotlinx.serialization.json.int

@Serializable(with = BookingsSerializer::class)
data class Bookings(val slots: List<Int>)

@Serializer(forClass = Bookings::class)
object BookingsSerializer : KSerializer<Bookings> {
    override fun deserialize(decoder: Decoder): Bookings {
        val json = (decoder as JsonInput).decodeJson().jsonObject
        return Bookings(parseSlots(json))
    }

    private fun parseSlots(json: JsonObject): List<Int> {
        val slotsJson = json["slots"] ?: return emptyList()
        return try {
            slotsJson.jsonArray.content.map { it.int }
        } catch (e: Exception) {
            listOf(slotsJson.int)
        }
    }
}

@ImplicitReflectionSerializer
fun main() {
    val json = """{"slots": 1}"""
    val result = Json.parse<Bookings>(json)
    println(result) // prints Bookings(slots=[1])
}

Answer 2

我更新了@Andrei 2021 年的回答，因为 class 和方法名称自 2019 年以来发生了一些变化：

import kotlinx.serialization.*
import kotlinx.serialization.json.*
import kotlinx.serialization.encoding.Decoder

@Serializable(with = BookingsSerializer::class)
data class Bookings(val slots: List<Int>)

@Serializer(forClass = Bookings::class)
object BookingsSerializer : KSerializer<Bookings> {
    override fun deserialize(decoder: Decoder): Bookings {
        val json = (decoder as JsonDecoder).decodeJsonElement().jsonObject
        return Bookings(parseSlots(json))
    }

    private fun parseSlots(json: JsonObject): List<Int> {
        val slotsJson = json["slots"] ?: return emptyList()
        return try {
            slotsJson.jsonArray.map { it.jsonPrimitive.int }
        } catch (e: Exception) {
            listOf(slotsJson.jsonPrimitive.int)
        }
    }
}

val json = """{"slots": 1}"""
val result = Json.decodeFromString<Bookings>(json)
println(result.toString()) // prints Bookings(slots=[1])

Kotlinx 序列化，避免在其他数据类型上崩溃

Kotlinx Serialization, avoid crashes on other datatype

kotlin

kotlinx.serialization