Question

大家好，我在SCALA上创建了以下案例类：

sealed abstract class Value;
  case class U(name: String) extends Value
  case class L(name: String) extends Value
  case class B(name: String) extends Value

  sealed abstract class Term
  case class Var(name: String) extends Term //variable name
  case class Val(value: Value) extends Term //value

sealed abstract class Pattern //patterns
  case class BGP(subject: Term, predicate: Term, obj: Term) extends Pattern
  case class And( pat1: Pattern, pat2: Pattern) extends Pattern
  case class Filter(pred: Predicate, pattern: Pattern ) extends Pattern


def function(p: Pattern): Unit = p match { 
    case BGP(Var(x), Val(y), Val(z)) => {
      val con:conv = new conv()
      val valor:Value = Val(y).value
}

然后，正如您所看到的，BGP包含Term并扩展到pattern，Val包含Values并扩展到Term，而U，L，B包含Strings并扩展为Value，在我的函数中我想访问包含U或L或B案例类的字符串，变量valor = Val（y）.value包含一个U类，例如，当我写valor.XXXX时不会出现我名称选项。最大的问题是如何从U访问String名称？

Answer 1

您只需在Value上定义，即btw可能是trait。

sealed trait Value {
  def name: String
}
case class U(name: String) extends Value
case class L(name: String) extends Value
case class B(name: String) extends Value

Scala多个嵌套案例条款

1 个答案: