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作为 Web 服务接口的一部分，我们创建了不依赖于任何特定平台/供应商 API 的更轻量级的面向客户端的 API。

现在，这些服务特征的 impl 必须将这些携带传入数据的平台无关类型转换为供应商特定类型，以便 impl 方法继续按原样工作。请注意，我们无法更改供应商 API 上的任何代码。我想我们必须创建某种适配器/翻译 API，可以双向进行这种类型转换。一些供应商类型非常神秘。例如，更轻量级的服务接口可能有一个扁平类型来承载所有数据，而供应商 API 有层次类型，可以接受输入数据等。返回类型也是如此。在 Scala 中实现这一目标的最佳方法是什么？

编辑：示例

case class A(var1,var2,var3,var4,var5,var6)

public Class B{

  private C c;
  private D d;

}

案例类是一个扁平结构，包含所有需要的数据，而另一方面，来自类 A 的数据将填充到 B (var2,var3)、C(var4,var5)、D(var6) 等。我也忘了提到供应商 API 都是基于 Java 的。

谢谢，

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1 回答 1

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我想说您可能可以利用“拉皮条”模式在两者之间来回切换。这是我在应用程序中使用的一大块东西。我只转换一种方式，但您可以看到如何使用另一种隐式方式，您可以实现逻辑以返回原始/您的格式。

// implicits for pimping 
implicit def pimpUser(user: User) = PimpedUser(user)
implicit def pimpLimitedViewUser(user: LimitedViewUser) = PimpedLimitedViewUser(user)

@entity
case class User(@serializer(classOf[UUIDOptionSerializer]) @id @column(name = "id") override val id: Option[UUID] = None,
                @serializer(classOf[DateOptionSerializer]) @column(name = "created_on") override val createdOn: Option[Date] = None,
                @serializer(classOf[DateOptionSerializer]) @column(name = "modified_on") override val modifiedOn: Option[Date] = None,
                @serializer(classOf[StringOptionSerializer]) @column(name = "first_name") firstName: Option[String] = None,
                @serializer(classOf[StringOptionSerializer]) @column(name = "last_name") lastName: Option[String] = None,
                @serializer(classOf[StringOptionSerializer]) @column(name = "email_address") emailAddress: Option[String] = None,
                @serializer(classOf[StringOptionSerializer]) @column(name = "password_salt") passwordSalt: Option[String] = None,
                @serializer(classOf[StringOptionSerializer]) @column(name = "password_hash") passwordHash: Option[String] = None,
                @serializer(classOf[IntOptionSerializer]) @column(name = "extension") extension: Option[Int] = None,
                devices: Set[Device] = Set.empty,
                @serializer(classOf[RingModeSerializer]) @column(name = "ring_mode") ringMode: RingMode = Waterfall) extends IdBaseEntity[UUID] {
  def this() = this(None, None, None, None, None, None, None, None, None, Set.empty[Device], Waterfall)

  override def equals(other: Any): Boolean =
    other match {
      case that: User => (that canEqual this) && (this.id match {
        case Some(uuid) => that.id match {
          case Some(thatUuid) => uuid == thatUuid
          case None => false
        }
        case None => false
      })
      case _ => false
    }

  override def hashCode: Int = this.id match {
    case Some(uuid) => 41 * (41 + uuid.hashCode())
    case None => super.hashCode
  }

  override def canEqual(other: Any): Boolean = other.isInstanceOf[User]

  def asAvailable: User = {
    User.devices.set(this)(devices.map(_.asAvailable))
  }

  def asUnavailable: User = {
    User.devices.set(this)(devices.map(_.asUnavailable))
  }
}

object User {
  implicit val userIso = Iso.hlist(User.apply _, User.unapply _)

  val id = Lens[User] >> _0
  val createdOn = Lens[User] >> _1
  val modifiedOn = Lens[User] >> _2
  val firstName = Lens[User] >> _3
  val lastName = Lens[User] >> _4
  val emailAddress = Lens[User] >> _5
  val passwordSalt = Lens[User] >> _6
  val passwordHash = Lens[User] >> _7
  val extension = Lens[User] >> _8
  val devices = Lens[User] >> _9
  val ringMode = Lens[User] >> _10
}

case class LimitedViewUser(id: Option[UUID], createdOn: Option[Date], modifiedOn: Option[Date], firstName: Option[String], lastName: Option[String]) {
  def this(user: User) = this(user.id, user.createdOn, user.modifiedOn, user.firstName, user.lastName)
}

case class PimpedUser(underlying: User) {
  def limited = LimitedViewUser(underlying)
}

case class PimpedLimitedViewUser(underlying: LimitedViewUser) {
  def normal = /* do something here to transform back to yours/the normal one */
}

// As long as those implicits are brought into scope, you will be able to do this:

val user: User = /* ... */
val limitedViewUser: LimitedViewUser = user.limited
val normalUser: User = limitedViewUser.normal
于 2013-08-08T22:10:22.497 回答