Scala:在 XML 中获取所有叶节点及其路径的最简单方法是什么?

问题描述

限时删除！！

我目前正在实现一个 xml 的 DFS 遍历，以便它转到每个叶节点并生成到叶节点的路径.

I am currently implementing DFS traversal of an xml such that it goes to each leaf node and generates the path to the leaf node.

给定的 XML:

<vehicles>
  <vehicle>
    gg
  </vehicle>
  <variable>
  </variable>
</vehicles>

输出(类似):

Map("gg" -> "vehicles/vehicle", "" -> "vehicles/variable")

如果有一个可用的库可以做到这一点就好了，这样我就不必维护代码了.

谢谢.任何帮助表示赞赏.

It would be great if there is a library available that does this so I dont have to maintain the code.

Thanks. Any help is appreciated.

推荐答案

这里是一个使用标准scala xml库的解决方案，打印出路径图->节点文本"

Here is a solution using standard scala xml library, prints out a map of paths -> "node text"

import scala.xml._
val x = <div class="content"><a></a><p><q>hello</q></p><r><p>world</p></r><s></s></div>
var map = Map[String,String]()
def dfs(n: Seq[Node], brc: String): Unit =
        n.foreach(x => {
                        if(x.child.isEmpty){
                           if(x.text == ""){
                            map = map + (brc + x.label -> "")
                            dfs(x.child,brc)
                          }
                          else{
                            map = map + (brc + x.label + " " -> x.text)
                            dfs(x.child,brc)
                          }
                        }
                        else{
                          val bc = brc + x.label + ">"
                          dfs(x.child,bc)
                        }
                     }
               )

dfs(x,"")
print(map)

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