scala – 懒惰地解析非常大的xml

我确信它可以做得更好但只是为了显示基本用法：

import scala.io.Source
import scala.xml.pull._

object Main extends App {
  val xml = new XMLEventReader(Source.fromFile("test.xml"))

  def printText(text: String,currNode: List[String]) {
    currNode match {
      case List("firstname","staff","company") => println("First Name: " + text)
      case List("lastname","company") => println("Last Name: " + text)
      case List("nickname","company") => println("Nick Name: " + text)
      case List("salary","company") => println("Salary: " + text)
      case _ => ()
    }
  }

  def parse(xml: XMLEventReader) {
    def loop(currNode: List[String]) {
      if (xml.hasNext) {
        xml.next match {
          case EvElemStart(_,label,_,_) =>
            println("Start element: " + label)
            loop(label :: currNode)
          case EvElemEnd(_,label) =>
            println("End element: " + label)
            loop(currNode.tail)
          case EvText(text) =>
            printText(text,currNode)
            loop(currNode)
          case _ => loop(currNode)
        }
      }
    }
    loop(List.empty)
  }

  parse(xml)
}