Question

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

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

data映射中的

值有几种类型。我尝试使用TypeToken和Generics，如here所示，但没有用。接收到的json的示例：

{“ action”：“ playVideo”，“ data”：{“ media”：{“ id”：15060328，“ url”：“ http://url_to_get_item”，“ name”：“商品名称“，” shortDescription“：”简短说明“}
{“ 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习惯，但是如果您对后端没有控制权，那么这里是一个如何反序列化此示例的示例

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
}

}

不要忘记注册反序列化器

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

从Gson反序列化嵌套的可变地图到Kotlin

1 个答案: