请关注我,在OP有意义之前有一些背景.我正在使用Slick 3.1.x和光滑的代码生成器.btw整个源代码可以在play-authenticate-usage-scala github项目中找到.对于这个项目,我希望有一个光滑的通用Dao,以避免为每个模型重复相同的样板代码.
我有一个postgres sql脚本,使用evolutions在这里创建数据库: 1.sql
然后,我调用生成以下数据模型的生成器: Tables.scala
为了能够为模型类提供通用的dao slick实现,我需要它们遵守一些基本的抽象,例如
id例如dao的需要findByIddef copyWithNewId(id : PK) : Entity[PK].这是dao实现的必要条件,createAndFetch它持久保存新实体并id在一个步骤中检索自动生成的PK.这copyWithNewId是OP的要点.请注意,它被调用,copyWithNewId而不是copy为了避免无限递归.为了能够实现允许插入并立即获取自动生成的GenericDaoAutoIncImpl,id实体行需要一个copy(id = id)来自<Model>Rowcase类的方法,在定义GenericDaoAutoIncImpl它时尚未知道它.相关实施如下:
override def createAndFetch(entity: E): Future[Option[E]] = {
val insertQuery = tableQuery returning tableQuery.map(_.id)
into ((row, id) => row.copyWithNewId(id))
db.run((insertQuery += entity).flatMap(row => findById(row.id)))
}
这需要我copyWithNewId在每个AutoInc id生成的模型中实现该方法,这是不好的,例如
// generated code and modified later to adapt it for the generic dao
case class UserRow(id: Long, ...) extends AutoIncEntity[Long] with Subject {
override def copyWithNewId(id : Long) : Entity[Long] = this.copy(id = id)
}
但是,如果我可以 - 使用一些Scala技巧 - 定义我的<Model>Rowcase类的子类,它是可复制的并且复制自身除了传递的idie IdCopyable,copy(id = id)然后我不需要copyWithNewId为每个<Model>Row 生成的case类一遍又一遍地实现.
有没有办法抽象或"拉出" copy(id = id)包含id属性的任何案例类的重构？还有其他推荐的解决方案吗？
更新1以下几乎总结了我遇到的问题:
scala> abstract class BaseA[A <: BaseA[_]] { def copy(id : Int) : A }
defined class BaseA
scala> case class A(id: Int) extends BaseA[A]
<console>:12: error: class A needs to be abstract, since method copy in class BaseA of type (id: Int)A is not defined
case class A(id: Int) extends BaseA[A]
^
scala> case class A(id: Int); val a = A(5); a.copy(6)
defined class A
a: A = A(5)
res0: A = A(6)
更新2使用下面提出的解决方案,我得到以下编译错误:
[error] /home/bravegag/code/play-authenticate-usage-scala/app/dao/GenericDaoAutoIncImpl.scala:26: could not find implicit value for parameter gen: shapeless.Generic.Aux[E,Repr]
[error] val insertQuery = tableQuery returning tableQuery.map(_.id) into ((row, id) => row.copyWithNewId(id))
[error] ^
[error] /home/bravegag/code/play-authenticate-usage-scala/app/dao/GenericDaoAutoIncImpl.scala:27: value id is not a member of insertQuery.SingleInsertResult
[error] db.run((insertQuery += entity).flatMap(row => findById(row.id)))
[error] ^
[error] two errors found
更新3使用和调整下面提出的镜头解决方案我得到以下编译器错误:
import shapeless._, tag.@@
import shapeless._
import tag.$at$at
/**
* Identifyable base for all Strong Entity Model types
* @tparam PK Primary key type
* @tparam E Actual case class EntityRow type
*/
trait AutoIncEntity[PK, E <: AutoIncEntity[PK, E]] extends Entity[PK] { self: E =>
//------------------------------------------------------------------------
// public
//------------------------------------------------------------------------
/**
* Returns the entity with updated id as generated by the database
* @param id The entity id
* @return the entity with updated id as generated by the database
*/
def copyWithNewId(id : PK)(implicit mkLens: MkFieldLens.Aux[E, Symbol @@ Witness.`"id"`.T, PK]) : E = {
(lens[E] >> 'id).set(self)(id)
}
}
然后我得到以下编译器错误:
[error] /home/bravegag/code/play-authenticate-usage-scala/app/dao/GenericDaoAutoIncImpl.scala:26: could not find implicit value for parameter mkLens: shapeless.MkFieldLens.Aux[E,shapeless.tag.@@[Symbol,String("id")],PK]
[error] val insertQuery = tableQuery returning tableQuery.map(_.id) into ((row, id) => row.copyWithNewId(id))
[error] ^
[error] /home/bravegag/code/play-authenticate-usage-scala/app/dao/GenericDaoAutoIncImpl.scala:27: value id is not a member of insertQuery.SingleInsertResult
[error] db.run((insertQuery += entity).flatMap(row => findById(row.id)))
[error] ^
通过无形,您可以抽象出案例类.
如果假设每个id都是a Long并且是case类的第一个参数,它可能如下所示:
scala> import shapeless._, ops.hlist.{IsHCons, Prepend}
import shapeless._
import ops.hlist.{IsHCons, Prepend}
scala> trait Copy[A <: Copy[A]] { self: A =>
| def copyWithId[Repr <: HList, Tail <: HList](l: Long)(
| implicit
| gen: Generic.Aux[A,Repr],
| cons: IsHCons.Aux[Repr,Long,Tail],
| prep: Prepend.Aux[Long :: HNil,Tail,Repr]
| ) = gen.from(prep(l :: HNil, cons.tail(gen.to(self))))
| }
defined trait Copy
scala> case class Foo(id: Long, s: String) extends Copy[Foo]
defined class Foo
scala> Foo(4L, "foo").copyWithId(5L)
res1: Foo = Foo(5,foo)
它也可能以更清洁的方式; 我对无形编程还不是很精通.而且我很确定它也可以用于id参数列表中任何位置的任何类型的案例类.见下文第2段.
您可能希望将此逻辑封装在可重用的类型类中:
scala> :paste
// Entering paste mode (ctrl-D to finish)
import shapeless._, ops.hlist.{IsHCons, Prepend}
sealed trait IdCopy[A] {
def copyWithId(self: A, id: Long): A
}
object IdCopy {
def apply[A: IdCopy] = implicitly[IdCopy[A]]
implicit def mkIdCopy[A, Repr <: HList, Tail <: HList](
implicit
gen: Generic.Aux[A,Repr],
cons: IsHCons.Aux[Repr,Long,Tail],
prep: Prepend.Aux[Long :: HNil,Tail,Repr]
): IdCopy[A] =
new IdCopy[A] {
def copyWithId(self: A, id: Long): A =
gen.from(prep(id :: HNil, cons.tail(gen.to(self))))
}
}
// Exiting paste mode, now interpreting.
import shapeless._
import ops.hlist.{IsHCons, Prepend}
defined trait IdCopy
defined object IdCopy
scala> def copy[A: IdCopy](a: A, id: Long) = IdCopy[A].copyWithId(a, id)
copy: [A](a: A, id: Long)(implicit evidence$1: IdCopy[A])A
scala> case class Foo(id: Long, str: String)
defined class Foo
scala> copy(Foo(4L, "foo"), 5L)
res0: Foo = Foo(5,foo)
您仍然可以将copyWithId方法放在您的案例类可以扩展的特征中,如果这对您很重要:
scala> trait Copy[A <: Copy[A]] { self: A =>
| def copyWithId(id: Long)(implicit copy: IdCopy[A]) = copy.copyWithId(self, id)
| }
defined trait Copy
scala> case class Foo(id: Long, str: String) extends Copy[Foo]
defined class Foo
scala> Foo(4L, "foo").copyWithId(5L)
res1: Foo = Foo(5,foo)
重要的是,通过使用上下文边界或隐式参数,将类型类实例从使用站点传播到需要它的位置.
override def createAndFetch(entity: E)(implicit copy: IdCopy[E]): Future[Option[E]] = {
val insertQuery = tableQuery returning tableQuery.map(_.id)
into ((row, id) => row.copyWithId(id))
db.run((insertQuery += entity).flatMap(row => findById(row.id)))
}
Shapeless还提供可用于此目的的镜头.这样,您可以更新id具有某个id字段的任何案例类的字段.
scala> :paste
// Entering paste mode (ctrl-D to finish)
sealed trait IdCopy[A,ID] {
def copyWithId(self: A, id: ID): A
}
object IdCopy {
import shapeless._, tag.@@
implicit def mkIdCopy[A, ID](
implicit
mkLens: MkFieldLens.Aux[A, Symbol @@ Witness.`"id"`.T, ID]
): IdCopy[A,ID] =
new IdCopy[A,ID] {
def copyWithId(self: A, id: ID): A =
(lens[A] >> 'id).set(self)(id)
}
}
def copyWithId[ID, A](a: A, elem: ID)(implicit copy: IdCopy[A,ID]) = copy.copyWithId(a, elem)
// Exiting paste mode, now interpreting.
defined trait IdCopy
defined object IdCopy
copyWithId: [ID, A](a: A, elem: ID)(implicit copy: IdCopy[A,ID])A
scala> trait Entity[ID] { def id: ID }
defined trait Entity
scala> case class Foo(id: String) extends Entity[String]
defined class Foo
scala> def assignNewIds[ID, A <: Entity[ID]](entities: List[A], ids: List[ID])(implicit copy: IdCopy[A,ID]): List[A] =
| entities.zip(ids).map{ case (entity, id) => copyWithId(entity, id) }
assignNewIds: [ID, A <: Entity[ID]](entities: List[A], ids: List[ID])(implicit copy: IdCopy[A,ID])List[A]
scala> assignNewIds( List(Foo("foo"),Foo("bar")), List("new1", "new2"))
res0: List[Foo] = List(Foo(new1), Foo(new2))
请注意,在使用的方法中assignNewIds,如何请求copyWithId类型类的实例IdCopy[A,ID]作为隐式参数.这是因为在使用它时copyWithId需要隐式实例IdCopy[A,ID]在范围内.您需要从使用站点传播隐式实例,在这里使用具体类型,例如Foo,从调用链一直到copyWithId调用的位置.
您可以将隐式参数视为方法的依赖项.如果方法具有类型的隐式参数,则IdCopy[A,ID]在调用它时需要满足该依赖关系.通常,这也会对调用它的方法产生相同的依赖性.
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