我有Json:
{
"216632": {
"prefiks": "",
"anyid": "0",
"tel": "123456789",
"firma": "Brand",
"fio": "Name",
"comment": "Text",
"address": "Address",
"kv": "99",
"adres_id": "11130",
"gps": "99.666666 66.777777",
"oplata": 0,
"tovarName": "Simple.",
"tovar": "23",
"fish": "0",
"finish": "1",
"utro": "0",
"unik": "0",
"chek": "0",
"bezcont": "0",
"courier_price": "0",
"sort": "13"
},
"246578":{
......
......
......
键216322可以有任何名字(215423,654345 ...)如何在数据类中序列化它?我有:
@kotlinx.serialization.Serializable
data class ServerData(
val **UUID** : FromServer
)
@kotlinx.serialization.Serializable
data class FromServer(
val adres: String,
val adres_id: String,
val anyid: String,
val bezcont: String,
val chek: String,
val comment: String,
val courier_price: String,
val finish: String,
val fio: String,
val firma: String,
val fish: String,
val kv: String,
val oplata: Int,
val prefiks: String,
val sort: String,
val tel: String,
val tovar: String,
val unik: String,
val utro: String
)
UUID自然是错误的选项,我该怎么做才对?应该用什么代替UUID
我使用KotlinX序列化
我到处都找不到信息
1条答案
按热度按时间nzrxty8p1#
您可能需要分两步解析它:首先,将整个Json-String解析为JsonObject。然后可以查询它的所有属性,然后为
FromServer
解析每个属性。另一种方法是,如果您使用Gson或Moshi,则可以为这些json编写自定义适配器。