Question

我有一个变量 user ，其类型为VertexProperty。当我在spark-shell中输入 user 时，我明白了，

res67: VertexProperty = User(273,0,A806683,Gustavo,Wilcox,127,NULL,Employee,1,0,0,1)

我可以通过user.id访问用户的id，但是我不能以这种方式访问offset，userCode和其他属性。为什么会这样？我如何访问它们？我无法理解OOP的这些属性。提前谢谢。

VertexProperty定义如下：

class VertexProperty(val id:Long) extends Serializable
case class User(val userId:Long, var offset:Int, val userCode:String, val Name:String, val Surname:String, val organizational_unit:String, val UME:String, val person_type:String, val SOD_HIGH:String, val SOD_MEDIUM:String, val SOD_LOW:String, val Under_mitigated:String) extends VertexProperty(userId)
case class Account(val accountId:Long, var offset:Int, val userCode:String, val userId:String, val account_creation_date:String, var disabled:String, var forcechangepwd:String, var pwdlife:String, var numberloginerror:String, var lastchangepwd:String, var lastlogin:String, var lastwronglogin:String, var state:String, var expire:String, var last_cert_time:String, var creation_date:String, var creation_user:String,var challenge_counter:String, var challenge_failed_attempt:String) extends VertexProperty(accountId) //Check if userCode is actually the code in this example.
case class Application(var applicationId:Long, var offset:Int, var Name:String, var Description:String, var Target:String, var Owner:String, var Ownercode:String, var Creation_date:String, var Creation_user:String) extends VertexProperty(applicationId)
case class Entitlement(val entitlementId:Long, var offset:Int, val Name:String, var Code:String, var Description:String, var Type:String, var Application:String, var Administrative:String, var Parent_ID:String, var Owner_code:String, var Scope_type:String, var Business_name:String, var Business_policy:String, var SOD_high:String, var SOD_medium:String, var SOD_low:String) extends VertexProperty(entitlementId)

Answer 1

user的静态类型为VertexProperty。这意味着编译器只知道它在编译时有一个id字段，除非你向下转换到其他类型，否则不会让你访问任何其他字段。

这有效，但被认为是不好的做法：

user.asInstanceOf[User].offset

如果实际上不是ClassCastException，它也会在运行时抛出User。

这稍微好一点，不会抛出异常：

user match {
  case user: User => user.offset
  case _ => // handle other cases
}

但实现目标的最佳方法是重构代码，以使user的静态类型真正User。

如何访问从父类

1 个答案: