使用 Gson 将嵌套可变映射反序列化为 Kotlin

Question

由于 MutableMap 对象，我无法使用 Gson 将 JSON 转换为 Kotlin 数据 class。数据class

data class MyAction(
    @Key("action") var action: String = "default",
    @Key("data") var data: MutableMap<String, Any> = mutableMapOf()
)

data 映射中的值是几个 types.I 尝试使用 TypeToken 和泛型的值，如在中一样，但没有用。收到的例子 json:

{"action":"playVideo","data":{"media":{"id":15060328,"url" :"http://url_to_get_item","name":"item name","shortDescription":"short desc"}
{"action":"setSpeed","data":{"value":1}}
{"action":"getProperty","data":{"value":"position"}}

Answer 1

以这种方式设计数据不是一个好的做法，但如果您无法控制后端，这里有一个如何反序列化的示例

class MyDeserializer : JsonDeserializer<MyAction>{
override fun deserialize(json: JsonElement, typeOfT: Type, context: JsonDeserializationContext): MyAction {
    val myAction = MyAction()
    val action = json.asJsonObject.get("action")
    val data = json.asJsonObject.get("data")

    myAction.action = context.deserialize<String>(action, String::class.java)

    val myMap = mutableMapOf<String, Any>()
    data.asJsonObject.keySet().forEach {
        when (it) {
            is String -> { myMap[it] = context.deserialize(data.asJsonObject.get(it), String::class.java) }
            is MyCustomObject1 -> { myMap[it] = context.deserialize(data.asJsonObject.get(it), MyCustomObject1::class.java) }
            is MyCustomObject2 -> { myMap[it] = context.deserialize(data.asJsonObject.get(it), MyCustomObject2::class.java) }
            else -> myMap[it] = context.deserialize(data.asJsonObject.get(it), Any::class.java)
        }
    }

    myAction.data = myMap
    return myAction
}

}

别忘了注册你的解串器

fun getSmartGson() = GsonBuilder().registerTypeAdapter(MyAction::class.java, MyDeserializer())

使用 Gson 将嵌套可变映射反序列化为 Kotlin

Deserialize nested mutable map from with Gson to Kotlin

android

json

gson

kotlin

data-class