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如何play.api.libs.Reads为我的People案例类创建一个?

scala> type Id = Long
defined type alias Id

scala> case class People(names: Set[Id])
defined class People

scala>   implicit val PeopleReads: Reads[People] = (
     |     (__ \ "names").read[Set[Id]])(People)
<console>:21: error: overloaded method value read with alternatives:
  (t: Set[Id])play.api.libs.json.Reads[Set[Id]] <and>
  (implicit r: play.api.libs.json.Reads[Set[Id]])play.api.libs.json.Reads[Set[Id]]
 cannot be applied to (People.type)
           (__ \ "names").read[Set[Id]])(People)
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1 回答 1

3

(...)(People)语法是为您构建参数列表(从技术上讲,它是一个Builder,而不是列表)而设计的,and并且想要将People构造函数提升到应用函子中,Reads以便您可以将其应用于这些参数。

例如,如果您的People类型如下所示:

case class People(names: Set[Id], home: String)

你可以写:

implicit val PeopleReads: Reads[People] = (
  (__ \ "names").read[Set[Id]] and
  (__ \ "home").read[String]
)(People)

但是,在您的情况下, for 的构造函数People只有一个参数,并且您没有使用and,所以您没有 a Builder[Reads[Set[Id] ~ String],您只有一个普通的 old Reads[Set[Id]]

这很好,因为这意味着你不需要奇怪的应用函子语法——你只需要map

implicit val PeopleReads = (__ \ "names").read[Set[Id]].map(People)

你完成了。

于 2014-01-30T16:24:22.453 回答