Question

我根据相应的Java类编写了以下Scala类：结果不好。它仍然看起来像Java一样，充满了vars，很长，而且在我看来并不是惯用的Scala。

我希望缩小这段代码，消除变量和@BeanHeader的东西。

这是我的代码：

    import scala.collection.immutable.Map 

     class ReplyEmail {

     private val to: List[String] = List()   
     private val toname: List[String] = List()
     private var cc: ArrayList[String] = new ArrayList[String]()

    @BeanProperty
    var from: String = _

    private var fromname: String = _

    private var replyto: String = _

    @BeanProperty
    var subject: String = _

    @BeanProperty
    var text: String = _

    private var contents: Map[String, String] = new scala.collection.immutable.HashMap[String, String]()

    @BeanProperty
    var headers: Map[String, String] = new scala.collection.immutable.HashMap[String, String]()

    def addTo(to: String): ReplyEmail = {
      this.to.add(to)
      this
    }

    def addTo(tos: Array[String]): ReplyEmail = {
      this.to.addAll(Arrays.asList(tos:_*))
      this
    }

    def addTo(to: String, name: String): ReplyEmail = {
      this.addTo(to)
      this.addToName(name)
    }

    def setTo(tos: Array[String]): ReplyEmail = {
      this.to = new ArrayList[String](Arrays.asList(tos:_*))
      this
    }

    def getTos(): Array[String] = {
      this.to.toArray(Array.ofDim[String](this.to.size))
    }

    def getContentIds(): Map[_,_] = this.contents

    def addHeader(key: String, `val`: String): ReplyEmail = {
      this.headers + (key -> `val`)
      this
    }

     def getSMTPAPI(): MyExperimentalApi = new MyExperimentalApi
      }

   }

= ---------------------

我感谢您帮助实现这一目标。 更新代码

我对代码做了一些小改动，比如引入Option [String]而不是String

case class ReplyEmail(
  to: List[String] = Nil,
  toNames: List[String] = Nil,
  cc: List[String],
  from: String,
  fromName: String,
  replyTo: String,
  subject: String,
  text: String,
  contents: Map[String, String] = Map.empty,
  headers: Map[String, String] = Map.empty) {
  def withTo(to: String): ReplyEmail = copy(to = this.to :+ to)
  def withTo(tos: List[String]): ReplyEmail = copy(to = this.to ++ to)
  def withTo(to: Option[String], name: Option[String]) = copy(to = this.to :+ to, toNames = toNames :+ name)

  def setTo(tos: List[String]): ReplyEmail = copy()
  def withHeader(key: String, value: String) = copy(headers = headers + (key  -> value))
  def smtpAPI = new MyExperimentalApi

}

现在，我面临的唯一问题是这一行：错误是：类型不匹配：找到：列表[java.io.Serializable] required：List [String]

def withTo(to: Option[String], name: Option[String]) = copy(to = this.to :+ to, toNames = toNames :+ name)

Answer 1

只需将其作为案例类。

case class ReplyEmail(
  to: List[String] = Nil,
  toNames: List[String] = Nil,
  cc: List[String],
  from: String,
  fromName: String,
  replyTo: String,
  subject: String,
  text: String,
  contents: Map[String, String] = Map.empty,
  headers: Map[String, String] = Map.empty) {

  def withTo(to: String) = copy(to = this.to :+ to)

  def withTo(to: List[String] = copy(to = this.to ++ to)

  def withTo(to: String, name: String) = copy(to = this.to :+ to, toNames = toNames :+ name)

  def withHeader(key: String, value: String) = copy(headers = headers + (key -> value))

  def smtpAPI = new MyExperimentalApi

}

如何消除Scala类中的变量？

1 个答案: