Scala 案例 Class

Question

下面是代码片段：

package caseClassExp

sealed trait Resources  {
  def fullpath :String
}

case class Folder (name :String ,
    path :Option[String] = None) extends Resources {
    def fullpath :String = path match {
      case Some(p) => List(p, name).mkString("/")
      case None => s"./$name"
    }
}

case class File (name :String ,
      folder :Option[Folder] = None ) extends Resources {
  def fullpath :String = folder match {
    case Some(f) => List(f.fullpath, name).mkString("/")
    case None => s"./$name"    
  }
}

object caseClass {
  def main(agrs:Array[String]):Unit = {
 
val resources = Seq[Resources] (
File("ex1.Scala",Some(Folder("example",Some("~/Dev"))))
Folder("temp")
Folder("bin",Some("/usr"))
File(".Clouder")
)  

resources foreach {
  case f :File => println(s"File: ${f.fullpath}")
  case f:Folder  => println(s"FOlder : ${f.fullpath}")
 }
}
}

当我在 main 中调用文件和文件夹方法（定义为 case class）时出现错误，我不确定我做错了什么。

Answer 1

所以你的顺序不正确。

您应该在序列中的每个元素后加上一个逗号。

val resources = Seq[Resources] (
  File("ex1.Scala",Some(Folder("example",Some("~/Dev")))),
  Folder("temp"),
  Folder("bin",Some("/usr")),
  File(".Clouder")
)

结果：

File: ~/Dev/example/ex1.Scala
FOlder : ./temp
FOlder : /usr/bin
File: ./.Clouder

所以一切都很好 :) 除了那个逗号

在您的代码中，我的建议是向资源添加类型并仅使用 Seq 来创建序列。代码将更具可读性。

val resources: Seq[Resources] = Seq(
  File("ex1.Scala",Some(Folder("example",Some("~/Dev")))),
  Folder("temp"),
  Folder("bin",Some("/usr")),
  File(".Clouder")
)

Scala 案例 Class

Scala case Class

scala

case-class