本文介绍了懒惰地解析非常大的xml的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有一个巨大的 xml 文件 (40 gbs).我想从中提取一些字段而不将整个文件加载到内存中.有什么建议吗?

I have a huge xml file (40 gbs). I would like to extract some fields from it without loading the entire file into memory. Any suggestions?

推荐答案

基于 SAXParser 教程的 XMLEventReader 快速示例 此处(由 Rinat Tainov 发布).

A quick example with XMLEventReader based on a tutorial for SAXParser here (as posted by Rinat Tainov).

我相信它可以做得更好,但只是为了显示基本用法:

I'm sure it can be done better but just to show basic usage:

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", "staff", "company") => println("Last Name: " + text)
      case List("nickname", "staff", "company") => println("Nick Name: " + text)
      case List("salary", "staff", "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)
}

这篇关于懒惰地解析非常大的xml的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

08-11 19:05