[英]Scala Anorm Why is scala complaining about result not having a trait?
I haven't seen anywhere this is done but I have the following: 我还没有看到完成此操作的任何地方,但有以下几点:
abstract class Player { val user_id: Long }
case class AlivePlayer(user_id: Long, target_id: Long) extends Player
case class DeadPlayer(user_id: Long) extends Player
def getPlayers(game: Game):Option[List[Player]] = DB.withConnection { implicit connect =>
SQL(
"""
SELECT * FROM game_participants WHERE game_id = {game_id})
"""
).on(
'game_id -> game.id
).as(Game.participantParser *)
}
In this case, I get an compiler error with the following message 在这种情况下,出现以下消息时出现编译器错误
error: type mismatch; found : anorm.ResultSetParser[List[Product with Serializable with models.Player]] required: anorm.ResultSetParser[Option[List[models.Player]]]
).as(Game.participantParser *)
Why is it not sufficient for me to specify the return type solely as Option[List[Player]]
? 为什么仅将返回类型指定为
Option[List[Player]]
还是不够的?
Given the error message, it sounds that Game.participantParser
is not returning an Option[List[Player]]
but List[Player]
. 给定错误消息,听起来
Game.participantParser
不是返回Option[List[Player]]
而是List[Player]
。 IMHO, getPlayers
should just return a List[Player]
that may be empty. 恕我直言,
getPlayers
应该只返回可能为空的List[Player]
。
If you want to return an Option[]
, your will have to add Some(...)
in front of your list or just return None
if the list is empty. 如果要返回
Option[]
,则必须在列表前面添加Some(...)
,如果列表为空,则只返回None
。
And as said in the comments, without the Game.participantParser
definition it is not possible to help you furthermore. 如评论中所述,如果没有
Game.participantParser
定义,就无法进一步帮助您。
Hope this helps. 希望这可以帮助。
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