Question

我梦想“动态实例化案例类” - 并根据每个字段类型为字段提供一些虚拟数据（我稍后会为此创建一些规则）

到目前为止，我有一些代码适用于String的案例类; Long或Int ...如果有可能处理嵌入式案例类，我会有点坚持

所以我可以实例化case class RequiredAPIResponse (stringValue: String, longValue: Long, intVlaue: Int)

但不是外面的;外面的地方......

case class Inner (deep: String)
case class Outer (in : Inner)

代码是

 def fill[T <: Object]()(implicit mf: ClassTag[T]) : T = {

      val declaredConstructors = mf.runtimeClass.getDeclaredConstructors
      if (declaredConstructors.length != 1)
      Logger.error(/*T.toString + */" has " + declaredConstructors.length + " constructors --- only 1 currently supported.")
      val constructor = declaredConstructors.headOption.get

      val m = constructor.getParameterTypes.map(p => {
          Logger.info("getName " + p.getName +" --- getCanonicalName " + p.getCanonicalName)
          Logger.info(p.getCanonicalName)

        p.getCanonicalName match {
          case "java.lang.String" => /*"Name"->*/ val s : java.lang.String = "DEFAULT STRING"
            s
          case "long" => /*"Name"-> */ val l : java.lang.Long = new java.lang.Long(99)
            l
          case "int" => /*"Name"->*/ val i : java.lang.Integer = new java.lang.Integer(99)
            i
          case _ => /*"Name"->*/

            So around here I am stuck!
        //THIS IS MADE UP :) But I want to get the "Type" and recursively call fill     
            //fill[p # Type] <- not real scala code

            //I can get it to work in a hard coded manner
            //fill[Inner]

        }
      })

我觉得Scala: How to invoke method with type parameter and manifest without knowing the type at compile time?上的最后一个答案是答案的起点。所以不要使用T＆lt ;: Object;填写应该采用ClassTag还是TypeTag？

此代码从 - How can I transform a Map to a case class in Scala?开始 - 提到（正如Lift-Framework所做的那样）我确实有liftweb源代码;但到目前为止，在解开所有秘密方面都没有成功。

编辑---根据Imm的观点，我已经得到了以下代码（对他的答案进行了一些小的更新）

def fillInner(cls: Class[_]) : Object = {
    val declaredConstructors = cls.getDeclaredConstructors
    if (declaredConstructors.length != 1)
      Logger.error(/*T.toString + */ " has " + declaredConstructors.length + " constructors --- only 1 currently supported.")
    val constructor = declaredConstructors.headOption.get

    val m = constructor.getParameterTypes.map(p => {
      Logger.info("getName " + p.getName + " --- getCanonicalName " + p.getCanonicalName)
      Logger.info(p.getCanonicalName)

      p.getCanonicalName match {
        case "java.lang.String" => /*"Name"->*/ val s: java.lang.String = "DEFAULT STRING"
          s
        case "long" => /*"Name"-> */ val l: java.lang.Long = new java.lang.Long(99)
          l
        case "int" => /*"Name"->*/ val i: java.lang.Integer = new java.lang.Integer(99)
          i
        case _ => fillInner(p)
      }

    })

    constructor.newInstance(m: _*).asInstanceOf[Object]

  }

    def fill[T](implicit mf: ClassTag[T]) : T = fillInner(mf.runtimeClass).asInstanceOf[T]

谢谢，布伦特

Answer 1

你实际上并没有使用ClassTag，只有Class[_]，而且这些都不是类型安全的（它只是Java反射），所以只需传递{{1递归地：

Class[_]

但是你可以用类型安全的方式完成你想做的事情，也许可以使用Shapeless：

def fillInner(cls: Class[_]) : Any = {
  val declaredConstructors = cls.getDeclaredConstructors
  if (declaredConstructors.length != 1)
  Logger.error(/*T.toString + */" has " + declaredConstructors.length + " constructors --- only 1 currently supported.")
  val constructor = declaredConstructors.headOption.get

  val m = constructor.getParameterTypes.map(p => {
      Logger.info("getName " + p.getName +" --- getCanonicalName " + p.getCanonicalName)
      Logger.info(p.getCanonicalName)

    p.getCanonicalName match {
      case "java.lang.String" => /*"Name"->*/ val s : java.lang.String = "DEFAULT STRING"
        s
      case "long" => /*"Name"-> */ val l : java.lang.Long = new java.lang.Long(99)
        l
      case "int" => /*"Name"->*/ val i : java.lang.Integer = new java.lang.Integer(99)
        i
      case _ => fillInner(p)
    }
  })

def fill[T: ClassTag]: T = fillInner(classOf[T].runtimeClass).asInstanceOf[T]

然后通过隐式解决方案的魔力，您可以获得trait Supplier[T] { def supply: T } object Supplier[T] { implicit val intSupplier = new Supplier[Int] { def supply = 99 } implicit val stringSupplier = ... implicit val emptyHListSupplier = new Supplier[HNil] { def supply = HNil } implicit def consHListSupplier[H, T <: HList]( implicit headSupplier: Supplier[H], tailSupplier: Supplier[T]) = new Supplier[H :: T] { def supply = headSupplier.supply :: tailSupplier.supply } }个Supplier[(String :: HNil) :: Int :: HNil]左右的HList HList个Supplier最终只包含您已获得的值{{1}} S;你只需要一点点无形（在版本1或2中有所不同，而且我已经做了一段时间了，所以我不记得具体内容）来回转换那些和案例类。

Answer 2

如果仅在测试中使用它，最好的方法是使用Scala / Java反射。

使用宏的优点是编译速度更快。与使用scalacheck相关的库相比，优点是API更好。

进行一些设置。这是完整的工作代码，您可以将其复制到代码库中：

import scala.reflect.api
import scala.reflect.api.{TypeCreator, Universe}
import scala.reflect.runtime.universe._

object Maker {
  val mirror = runtimeMirror(getClass.getClassLoader)

  var makerRunNumber = 1

  def apply[T: TypeTag]: T = {
    val method = typeOf[T].companion.decl(TermName("apply")).asMethod
    val params = method.paramLists.head
    val args = params.map { param =>
      makerRunNumber += 1
      param.info match {
        case t if t <:< typeOf[Enumeration#Value] => chooseEnumValue(convert(t).asInstanceOf[TypeTag[_ <: Enumeration]])
        case t if t =:= typeOf[Int] => makerRunNumber
        case t if t =:= typeOf[Long] => makerRunNumber
        case t if t =:= typeOf[Date] => new Date(Time.now.inMillis)
        case t if t <:< typeOf[Option[_]] => None
        case t if t =:= typeOf[String] && param.name.decodedName.toString.toLowerCase.contains("email") => s"random-$arbitrary@give.asia"
        case t if t =:= typeOf[String] => s"arbitrary-$makerRunNumber"
        case t if t =:= typeOf[Boolean] => false
        case t if t <:< typeOf[Seq[_]] => List.empty
        case t if t <:< typeOf[Map[_, _]] => Map.empty
        // Add more special cases here.
        case t if isCaseClass(t) => apply(convert(t))
        case t => throw new Exception(s"Maker doesn't support generating $t")
      }
    }

    val obj = mirror.reflectModule(typeOf[T].typeSymbol.companion.asModule).instance
    mirror.reflect(obj).reflectMethod(method)(args:_*).asInstanceOf[T]
  }

  def chooseEnumValue[E <: Enumeration: TypeTag]: E#Value = {
    val parentType = typeOf[E].asInstanceOf[TypeRef].pre
    val valuesMethod = parentType.baseType(typeOf[Enumeration].typeSymbol).decl(TermName("values")).asMethod
    val obj = mirror.reflectModule(parentType.termSymbol.asModule).instance

    mirror.reflect(obj).reflectMethod(valuesMethod)().asInstanceOf[E#ValueSet].head
  }

  def convert(tpe: Type): TypeTag[_] = {
    TypeTag.apply(
      runtimeMirror(getClass.getClassLoader),
      new TypeCreator {
        override def apply[U <: Universe with Singleton](m: api.Mirror[U]) = {
          tpe.asInstanceOf[U # Type]
        }
      }
    )
  }

  def isCaseClass(t: Type) = {
    t.companion.decls.exists(_.name.decodedName.toString == "apply") &&
      t.decls.exists(_.name.decodedName.toString == "copy")
  }
}

而且，当您要使用它时，可以致电：

val user = Maker[User]
val user2 = Maker[User].copy(email = "someemail@email.com")

上面的代码生成任意和唯一的值。它们不是完全随机的。该API非常好。考虑到它使用反射，最好在测试中使用。

在此处阅读我们的完整博客文章：https://give.engineering/2018/08/24/instantiate-case-class-with-arbitrary-value.html

Scala 2.10：动态实例化案例类

2 个答案: