scala条件累加

我正在尝试实现一个函数，该函数从字符串中提取“占位符”列表。开始占位符和结束占位符的分隔符字符为$。我有一些想法如何实现它使用一个var来启用/禁用累加，我试图做的是在没有任何var的情况下实现它

对于像这样的字符串

val stringToParse = "ignore/me/$aaa$/once-again/ignore/me/$bbb$/still-to-be/ignored

结果应该是

Seq("aaa", "bbb")

有什么提示吗

下面是一个使用var的解决方案

import fiddle.Fiddle, Fiddle.println
import scalajs.js
import scala.collection.mutable.ListBuffer

@js.annotation.JSExportTopLevel("ScalaFiddle")
object ScalaFiddle {
  // $FiddleStart


  val stringToParse = "ignore/me/$aaa$/once-again/ignore/me/$bbb$/still-to-be/ignored"

  class StringAccumulator {

    val accumulator: ListBuffer[String] = new ListBuffer[String]
    val sb: StringBuilder = new StringBuilder("")
    var open:Boolean = false

    def next():Unit = {
      if (open) {
        accumulator.append(sb.toString)
        sb.clear
        open = false
      } else {
        open = true
      }
    }

    def accumulateIfOpen(charToAccumulate: Char):Unit = {
      if (open) sb.append(charToAccumulate)
    }

    def get(): Seq[String] = accumulator.toList
  }

  def getPlaceHolders(str: String): Seq[String] = {
    val sac = new StringAccumulator

    str.foreach(chr => {
      if (chr == '$') {
        sac.next()
      } else {
        sac.accumulateIfOpen(chr)
      }
    })

    sac.get
  }

  println(getPlaceHolders(stringToParse))
  // $FiddleEnd
}

---

这个解决方案够了吗

scala> val stringToParse = "ignore/me/$aaa$/once-again/ignore/me/$bbb$/still-to-be/ignored"
stringToParse: String = ignore/me/$aaa$/once-again/ignore/me/$bbb$/still-to-be/ignored

scala> val P = """\$([^\$]+)\$""".r
P: scala.util.matching.Regex = \$([^\$]+)\$

scala> P.findAllIn(stringToParse).map{case P(s) => s}.toSeq
res1: Seq[String] = List(aaa, bbb)

---

我将向您介绍两种解决方案。第一个是你所做的最直接的翻译。在Scala中，如果您听到单词

acculate

，它通常会转换为

fold

或

reduce

的变体

def

然而，这似乎相当复杂，因为听起来应该是一项简单的任务。我们可以使用正则表达式，但它们很可怕，所以让我们避免使用它们。事实上，经过一点思考，这个问题实际上变得相当简单

 def extractValuesSimple(s: String)=
  {
    s.split('$').//split the string on the $ character
      dropRight(1).//Drops the rightmost item, to handle the case with an odd number of $
      zipWithIndex.filter{case (str, index) => index % 2 == 1}.//Filter out all of the even indexed items, which will always be outside of the matching $
      map{case (str, index) => str}.toList//Remove the indexes from the output
  }

---

需要明确的是，在本例中，您希望所有字符串都位于匹配的$？你能发布你目前拥有的代码吗，这样我们就知道你开始的方向了？我已经用var更新了这个问题

 def extractValuesSimple(s: String)=
  {
    s.split('$').//split the string on the $ character
      dropRight(1).//Drops the rightmost item, to handle the case with an odd number of $
      zipWithIndex.filter{case (str, index) => index % 2 == 1}.//Filter out all of the even indexed items, which will always be outside of the matching $
      map{case (str, index) => str}.toList//Remove the indexes from the output
  }