用于案例对象的Play Framework JSON格式

Question

我有一组从一个特征继承的case对象,如下所示:

  sealed trait UserRole
  case object SuperAdmin extends UserRole
  case object Admin extends UserRole
  case object User extends UserRole

我想将其序列化为JSON,我只使用了Format机制:

implicit val userRoleFormat: Format[UserRole] = Json.format[UserRole]

但不幸的是,编译器不满意,它说:

No unapply or unapplySeq function found

我的案例对象出了什么问题？

Answer 1

好吧，我想出了要做什么！

这里是：

  implicit object UserRoleWrites extends Writes[UserRole] {
    def writes(role: UserRole) = role match {
      case Admin => Json.toJson("Admin")
      case SuperAdmin => Json.toJson("SuperAdmin")
      case User => Json.toJson("User")
    }
  }

Answer 2

另一种选择是override def toString这样的：

文件：Status.scala

    package models

    trait Status
    case object Active extends Status {
      override def toString: String = this.productPrefix
    }
    case object InActive extends Status {
      override def toString: String = this.productPrefix
    }

this.productPrefix 会给你案例对象名称

文件：Answer.scala

package models

import play.api.libs.json._

case class Answer(
    id: Int,
    label: String,
    status: Status
) {
  implicit val answerWrites = new Writes[Answer] {
    def writes(answer: Answer): JsObject = Json.obj(
      "id" -> answer.id,
      "label" -> answer.label,
      "status" -> answer.status.toString
    )
  }
  def toJson = {
    Json.toJson(this)
  }
}

文件：Controller.scala

import models._
val jsonAnswer = Answer(1, "Blue", Active).toJson
println(jsonAnswer)

你得到：

{"id":1,"label":"Blue","status":"Active"}

希望这可以帮助！