如何在一个对象中序列化相同类型的对象？

Question

我想将以下 JSON 片段结构化/转换为 Kotlin 数据类，以便我可以在进一步表示中使用它。 有点期望玩家列表和每个对象都是玩家，现在我发现很难序列化相同类型的这种情况。

"Players": {
        "63084": {
          "Position": "1",
          "Name_Full": "Imam-ul-Haq",
          "Batting": {
            "Style": "LHB",
            "Average": "60.66",
            "Strikerate": "81.54",
            "Runs": "910"
          },
          "Bowling": {
            "Style": "LB",
            "Average": "0",
            "Economyrate": "0",
            "Wickets": "0"
          }
        },
        "57492": {
          "Position": "2",
          "Name_Full": "Fakhar Zaman",
          "Batting": {
            "Style": "LHB",
            "Average": "55.25",
            "Strikerate": "96.64",
            "Runs": "1326"
          },
          "Bowling": {
            "Style": "SLO",
            "Average": "88",
            "Economyrate": "4.75",
            "Wickets": "1"
          }
        },
        "59429": {
          "Position": "3",
          "Name_Full": "Babar Azam",
          "Batting": {
            "Style": "RHB",
            "Average": "50.6",
            "Strikerate": "84.4",
            "Runs": "2328"
          },
          "Bowling": {
            "Style": "OB",
            "Average": "0",
            "Economyrate": "0",
            "Wickets": "0"
          }
        }
}

可以做这样的事情：

 @SerializedName("63084")
  val player: Player

甚至对每个 id 重复都感觉效率不高，甚至每个玩家的 id 都是动态的和不同的。 这里肯定漏掉了一些重要的东西。

API： https : //cricket.yahoo.net/sifeeds/cricket/live/json/sapk01222019186652.json

Answer 1

您使用的 API 有点不标准，但您可以将 Players 字段映射到 Map<String, Player> ，例如：

data class Player(
    @SerializedName("Position")
    val position: String,
    @SerializedName("Name_Full")
    val name: String 
...)

进而：

@SerializedName("Players")
val player: Map<String, Player>