我正在使用play框架,并且有一个抽象类:

 abstract class Base{...}


在随播对象中有自己的隐式JSON编写器

object Base {
   implicit val baseWrites: Writes[Base] = (...)(unlift(Base.unapply))
}


我将这个抽象类子类化:

case class SubClass{...}


在其伴随对象中也有自己的隐式JSON编写器

object SubClass {
   implicit val subClassWrites: Writes[SubClass] = (...)(unlift(SubClass.unapply))
}


当我尝试使用Json.toJson(SubClass)序列化子类对象时,出现错误:

[error]  both value subClassWrites in object SubClass of type => play.api.libs.json.
Writes[models.SubClass]
[error]  and value baseWrites in object Base of type =>
play.api.libs.json.Writes[models.Base]
[error]  match expected type play.api.libs.json.Writes[models.SubClass]
[error]  Ok(Json.toJson(SubClass.find(id)))


有什么办法可以消除歧义?

最佳答案

因为Writes具有变量类型参数A,您将发生碰撞:

trait Writes[-A] extends AnyRef


这意味着Writes[Base]Writes[SubClass]的子类-您可以在需要Writes[Base]的地方使用Writes[SubClass]

问题在这里:

val base: Base = new SubClass(...)
val jsBase = Json.toJson(base)


因此,Writes[Base]应该能够序列化SubClass的实例。在这种情况下,您可以使用ADT

sealed trait Base
object Base {
  implicit val baseWrites: Writes[Base] =
    new Writes[Base]{
      def writes(o: Base): JsValue = o match {
        case s: SubClass => SubClass.writes.writes(s)
        case s: SubClass2 => SubClass2.writes.writes(s)
      }
    }
}

case class SubClass(...) extends Base
object SubClass {
  val writes: Writes[SubClass] = (...)(unlift(SubClass.unapply))
}

case class SubClass2(...) extends Base
object SubClass2 {
  val writes: Writes[SubClass2] = (...)(unlift(SubClass2.unapply))
}


使用sealed关键字,如果match不够详尽,则会收到警告。

07-26 06:12