为什么不将其解码为ADT类型?

时间:2019-05-05 16:17:10

标签: scala circe

我正在尝试将以下字符串派生为适当的ADT类型:

res6: String = {"raw":"Hello","status":{"MsgSuccess":{}}} 

并使用circe库。

ADT类型如下:

sealed trait MsgDoc {
}

final case class MsgPreFailure(raw: String, reasons: Chain[String]) extends MsgDoc

final case class MsgProceed(raw: String, status: MsgStatus) extends MsgDoc

MsgStatus类型:

sealed trait MsgStatus {

}

case object MsgSuccess extends MsgStatus

final case class MsgFailure(reasons: Chain[String]) extends MsgStatus

final case class MsgUnknown(reason: String) extends MsgStatus

还有,我试着开车:

object MsgDocDerivation {

  import shapeless.{Coproduct, Generic}

  implicit def encodeAdtNoDiscr[A, Repr <: Coproduct](implicit
                                                      gen: Generic.Aux[A, Repr],
                                                      encodeRepr: Encoder[Repr]
                                                     ): Encoder[A] = encodeRepr.contramap(gen.to)

  implicit def decodeAdtNoDiscr[A, Repr <: Coproduct](implicit
                                                      gen: Generic.Aux[A, Repr],
                                                      decodeRepr: Decoder[Repr]
                                                     ): Decoder[A] = decodeRepr.map(gen.from)
}

和执行:

object Main extends App {


  val json = MsgProceed("Hello", MsgSuccess).asJson
  println(json)
  val adt = decode[MsgDoc](json.noSpaces)
  println(adt)

} 

结果是:

{
  "raw" : "Hello",
  "status" : {
    "MsgSuccess" : {

    }
  }
}
Left(DecodingFailure(CNil, List())) 

如您所见,它不能正确decode

可以找到源代码https://gitlab.com/playscala/adtjson

2 个答案:

答案 0 :(得分:7)

我不确定MsgDocDerivation这东西的用途是什么-似乎没有必要并且分散注意力-但我认为关键的问题是circ的编码(和解码)是由静态类型驱动的,而不是编码(或解码)的值的运行时类。这意味着以下两个JSON值将有所不同:

val value = MsgProceed("Hello", MsgSuccess)

val json1 = value.asJson
val json2 = (value: MsgDoc).asJson

对于您而言,以下对我来说很好:

import cats.data.Chain

sealed trait MsgStatus
case object MsgSuccess extends MsgStatus
final case class MsgFailure(reasons: Chain[String]) extends MsgStatus
final case class MsgUnknown(reason: String) extends MsgStatus

sealed trait MsgDoc
final case class MsgPreFailure(raw: String, reasons: Chain[String]) extends MsgDoc
final case class MsgProceed(raw: String, status: MsgStatus) extends MsgDoc

import io.circe.generic.auto._, io.circe.jawn.decode, io.circe.syntax._

val value: MsgDoc = MsgProceed("Hello", MsgSuccess)
val json = value.asJson

val backToValue = decode[MsgDoc](json.noSpaces)

请注意,json与您所看到的不同:

scala> json
res0: io.circe.Json =
{
  "MsgProceed" : {
    "raw" : "Hello",
    "status" : {
      "MsgSuccess" : {

      }
    }
  }
}

scala> backToValue
res1: Either[io.circe.Error,MsgDoc] = Right(MsgProceed(Hello,MsgSuccess))

这是因为我执行了从MsgProceedMsgDoc的(类型安全)转换。无论如何,这通常就是您使用ADT的方式-您不会传递静态类型为case类子类型的值,而是传递为sealed trait基类型的值。

答案 1 :(得分:3)

问题是MsgProceed类型和MsgProceed类型的MsgDoc被编码为不同的json。因此,您尝试从错误的json解码MsgDoc

  val json = MsgProceed("Hello", MsgSuccess).asJson
  println(json)
  //{
  //  "raw" : "Hello",
  //  "status" : {
  //    "MsgSuccess" : {
  //
  //    }
  //  }
  //}
  val json1 = (MsgProceed("Hello", MsgSuccess): MsgDoc).asJson
  println(json1)
  //{
  //  "MsgProceed" : {
  //    "raw" : "Hello",
  //    "status" : {
  //      "MsgSuccess" : {
  //
  //      }
  //    }
  //  }
  //}
  val adt0 = decode[MsgProceed](json.noSpaces)
  println(adt0)
  //Right(MsgProceed(Hello,MsgSuccess))
  val adt1 = decode[MsgDoc](json1.noSpaces)
  println(adt1)
  //Right(MsgProceed(Hello,MsgSuccess))
  val adt = decode[MsgDoc](json.noSpaces)
  println(adt)
  //Left(DecodingFailure(CNil, List()))