Question

我在应用程序中使用外部API，而反序列化是通过Kotlinx序列化程序包完成的，但是当api结果是多个值的Int数组和单个值的原始int时，我遇到了问题。我如何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>)

插槽具有单一值API时返回{slots: 1} 当广告位具有多个值API时，返回{ slots: [1,2,3,4]}

Answer 1

可以使用自定义序列化器完成

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 年的答案，因为自 2019 年以来类和方法名称发生了一些变化：

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，避免其他数据类型崩溃

2 个答案: