Scala XML parsing with Xpath

Question

I'm trying to parse XML in Scala using scala.xml.XML and I would like to avoid having to pass strings via XPath to an XML Elem in order to get a NodeSeq.

Having to specify strings like elem \\ "tag1" \\ "tag2" \\ "@tagN" in multiple places seems fragile and repetitive.

Most of the Scala XML parsing tutorials have you parsing with XPath hardcoded string jank - elem \\ "tag1" \\ "tag2" \\ "@tagN" - everywhere.

Is it possible to write a method that takes a searches an Elem through a Seq[String] via XPath and returns a NodeSeq?

The method would look something like:

def getNodeFromXml(elem: Elem, tags: Seq[String]): NodeSeq = {
  Returns a NodeSeq elem \\ tags(0) \\ tags(1) \\ ... \\ tags(tags.length)
}

Thanks

som-snytt · Accepted Answer · 2015-03-08 19:43:30Z

2

Something like:

scala> import xml._
import xml._

scala> val x = <top><middle><bottom>text</bottom></middle></top>
x: scala.xml.Elem = <top><middle><bottom>text</bottom></middle></top>

scala> val names = Seq("middle","bottom")
names: Seq[String] = List(middle, bottom)

scala> val ns: NodeSeq = x
ns: scala.xml.NodeSeq = <top><middle><bottom>text</bottom></middle></top>

scala> names.foldLeft(ns)((e,s) => e \\ s)
res2: scala.xml.NodeSeq = NodeSeq(<bottom>text</bottom>)

answered Mar 8, 2015 at 19:43

som-snytt

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1 Comment

Anthony Over a year ago

This is much cleaner than the approach I was trying. I was trying to use recursion passing the NodeSeq + tail of the Seq down and when seq.length was 1, returning the final NodeSeq. I was trying to use collection built-ins for this but never thought to use foldLeft. Thank you for sharing!

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