Question

我有以下示例代码：

import java.util.UUID

import shapeless.LabelledGeneric
import shapeless.record._
import shapeless.syntax.singleton._

object LabelTest extends App {

  case class IncomingThing(name: String, age: Int)
  case class DatabaseIncomingThing(name: String, age: Int, id: UUID)

  val genIncoming = LabelledGeneric[IncomingThing]
  val genDatabase = LabelledGeneric[DatabaseIncomingThing]

  val thing = IncomingThing("John", 42)

  val structuralVersionOfIncomingThing = genIncoming.to(thing)

  val updated = genDatabase.from(structuralVersionOfIncomingThing + ('id ->> UUID.randomUUID()))

  println(updated) // DatabaseIncomingThing(John,42,a45081f2-4ed5-4d2b-8fd9-4d8986875ed7)

}

这很不错，因为我不必编写用于复制从IncomingThing到DatabaseIncomingThing的所有字段的样板。但是，我希望不必同时维护这两种类型，因为两者之间的关系非常明确（一个拥有id，另一个没有）。

是否可以通过添加或删除一个字段从给定的案例类创建类型？我想像

type IncomingThing = withoutField[DatabaseIncomingThing]('id)

或类似的东西。

Answer 1

代替DatabaseIncomingThing

val updated: DatabaseIncomingThing = 
  genDatabase.from(structuralVersionOfIncomingThing + ('id ->> UUID.randomUUID()))

您可以使用原始的HList

val updated1: FieldType[Witness.`'name`.T, String] :: FieldType[Witness.`'age`.T, Int] :: FieldType[Witness.`'id`.T, UUID] :: HNil = 
  structuralVersionOfIncomingThing + ('id ->> UUID.randomUUID())
val updated2: FieldType[Witness.`'name`.T, String] :: FieldType[Witness.`'age`.T, Int] :: HNil = 
  updated1 - 'id

在文字级别

implicitly[Remover.Aux[FieldType[Witness.`'name`.T, String] :: FieldType[Witness.`'age`.T, Int] :: FieldType[Witness.`'id`.T, UUID] :: HNil,
  Witness.`'id`.T,
  (UUID, FieldType[Witness.`'name`.T, String] :: FieldType[Witness.`'age`.T, Int] :: HNil)]]

implicitly[Updater.Aux[
  FieldType[Witness.`'name`.T, String] :: FieldType[Witness.`'age`.T, Int] :: HNil,
  FieldType[Witness.`'id`.T, UUID],
  FieldType[Witness.`'name`.T, String] :: FieldType[Witness.`'age`.T, Int] :: FieldType[Witness.`'id`.T, UUID] :: HNil]]

您可以创建类型类

trait WithoutField[A, K] {
  type Out <: HList
}

object WithoutField {
  type Aux[A, K, Out0 <: HList] = WithoutField[A, K] { type Out = Out0 }
  def instance[A, K, Out0 <: HList]: Aux[A, K, Out0] = new WithoutField[A, K] { type Out = Out0 }

  implicit def mkWithoutField[A, L <: HList, K, T, L1 <: HList](implicit
    labelledGeneric: LabelledGeneric.Aux[A, L],
    remover: Remover.Aux[L, K, (T, L1)]): Aux[A, K, L1] = instance
}

并使用它

def foo[Out <: HList](implicit withoutField: WithoutField.Aux[DatabaseIncomingThing, Witness.`'id`.T, Out]) = {
  // now you can use type Out inside
  type IncomingThing = Out
  ???
}

在Scala中使用Shapeless从案例类中创建“丰富”类型

1 个答案: