从反射的Java类实例中获取ClassTag

时间:2014-05-05 18:34:47

标签: java scala reflection

是否可以从通过反射获得的Java ClassTag实例中获取Class信息?

这是情况。我有一个看起来像这样的Scala case class

case class Relation[M : ClassTag](id: UUID, 
                                  model: Option[M] = None)

它被这样使用(虽然有更多的类相互关联):

case class Organization(name: String)

case class Person(firstName: String, 
                  lastName: String,
                  organization: Relation[Organization])

我要做的是以编程方式使用看起来像这样的东西构建这些关系的树:

private def generateFieldMap(clazz: Class[_]): Map[String, Class[_]] = {
    clazz.getDeclaredFields.foldLeft(Map.empty[String, Class[_]])((map, field) => {
        map + (field.getName -> field.getType)
    })
}

private def getRelationModelClass[M : ClassTag](relationClass: Class[_ <: Relation[M]]): Class[_] = {
    classTag[M].runtimeClass
}

def treeOf[M: ClassTag](relations: List[String]): Map[String, Any] = {
    val normalizedRelations = ModelHelper.normalize(relations)
    val initialFieldMap = Map("" -> generateFieldMap(classTag[M].runtimeClass))
    val relationFieldMap = relations.foldLeft(initialFieldMap)((map, relation) => {
        val parts = relation.split('.')
        val parentRelation = parts.dropRight(1).mkString(".")
        val relationClass = map(parentRelation)(parts.last)
        val relationModelClass = relationClass match {
            case clazz: Class[_ <: Relation[_]] => getRelationModelClass(clazz)
            case _ => throw ProcessStreetException("cannot follow non-relation: " + relation)
        }
        val fieldMap = generateFieldMap(relationModelClass)
        map + (relation -> fieldMap)
    })
    relationFieldMap
}

val relations = List("organization")
val tree = treeOf[Person](relations)

这不会编译。我收到这个错误:

[error] Foo.scala:148: not found: type _$12
[error]                 case clazz: Class[_ <: Relation[_]] => getRelationModelClass(clazz)
[error]                                   ^
[error] one error found
[error] (compile:compile) Compilation failed

基本上,当我拥有的是ClassTag时,我想要做的就是能够访问Class信息。这可能吗?

2 个答案:

答案 0 :(得分:5)

是的,绝对可能也很容易:

val clazz = classOf[String]
val ct = ClassTag(clazz)  // just use ClassTag.apply() method

在您的示例中,您想要像这样调用getRelationModelClass方法:

getRelationModelClass(clazz)(ClassTag(clazz))

这是可能的,因为[T: ClassTag]语法隐式创建了第二个参数列表,如(implicit ct: ClassTag[T])。通常它由编译器填充,但没有任何东西阻止你明确地使用它。

您也不需要同时将此clazz的类AND类标记传递给该方法。您甚至没有在其正文中使用显式类对象。只需传递类标记,就足够了。

答案 1 :(得分:1)

我最终使用TypeTags和Scala反射API完成了我的目标。以下是必要的更改。

首先,更改Relation类以使用TypeTag。

case class Relation[M : TypeTag](id: UUID, 
                                 model: Option[M] = None)

然后更改其余代码以使用Scala反射API:

private def generateFieldMap(tpe: Type): Map[String, Type] =
    tpe.members.filter(_.asTerm.isVal).foldLeft(Map.empty[String, Type])((map, field) => {
        map + (member.name.toString.trim -> member.typeSignature)
    })

private def getRelationModelType(tpe: Type): Type = 
    tpe match { case TypeRef(_, _, args) => args.head }

def treeOf[M: TypeTag](relations: List[String]): Map[String, Any] = {
    val normalizedRelations = ModelHelper.normalize(relations)
    val initialFieldMap = Map("" -> generateFieldMap(typeTag[T].tpe))
    val relationFieldMap = relations.foldLeft(initialFieldMap)((map, relation) => {
        val parts = relation.split('.')
        val parentRelation = parts.dropRight(1).mkString(".")
        val relationType = map(parentRelation)(parts.last)
        val relationModelType = getRelationModelType(relationType)
        val fieldMap = generateFieldMap(relationModelType)
        map + (relation -> fieldMap)
    })
    relationFieldMap
}