ktor 动态密钥序列化和 json 预处理

Question

在反序列化过程之前，Ktor 是否有改变 json 的方法？在下面的示例中，有一个带有动态键的 json ，我需要删除其中的“token”键，因为它是字符串类型而不是其他键的对象类型。当我运行代码时出现以下错误：

Exception in thread "main" kotlinx.serialization.json.internal.JsonDecodingException: Expected class kotlinx.serialization.json.JsonObject as the serialized body of shell.remoting.Market, but had class kotlinx.serialization.json.JsonLiteral

我不确定是否还有其他更好的方法。任何建议将不胜感激，谢谢

object MarketMapSerializer :
    JsonTransformingSerializer<Map<String, Market>>(MapSerializer(String.serializer(), Market.serializer())) {
    override fun transformSerialize(element: JsonElement): JsonElement =
        JsonObject(element.jsonObject.filterNot { (k, _) ->
            k == "token"
        })
}

@Serializable
data class Market(
    val stations: List<Station>,
)

@Serializable
data class Station(
    @JsonNames("iata_code")
    val iataCode: String,
    val destinations: List<String>,
)


fun main() {
    val jsonString = """
        {
            "Republica Dominicana": {
                "stations": [
                    {
                        "iata_code": "PUJ",
                        "destinations": [
                            "ADZ",
                            "BAQ",
                            "VVC"
                        ]
                    }
                ]
            },
            "Brasil": {
                "stations": [
                    {
                        "iata_code": "AJO",
                        "destinations": [
                            "ADZ",
                            "BAQ",
                            "VVC"
                        ]
                    }
                ]
            },
            "token": "eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJzdWIiOiJkb3RSRVogQVBJIiwianRpIjoiYTVhZDM4NmYtOGViNy0yOWQ5LWZkMGYtM2Q3MzQwZmRhOGI2IiwiaXNzIjoiZG90UkVaIEFQSSJ9.V2YfXCt9r7Tzae9SYSoj-qIyxjRc9YoE2XPoIQQNI9U"
        }
    """.trimIndent()
    println(
        Json.decodeFromString(
            MarketMapSerializer,
            jsonString
        )
    )
}

Answer 1

只需将 transformSerialize 替换为 transformDeserialize，因为您正在进行反序列化。

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