Select all child elements of a node in scala

Question

I want to select the first child Elem of a Node named "a". What I've got now is:

(xml \ "a")(0).child.collect {case e: Elem => e}

This is quite verbose. I was looking for something like:

xml \ "a" \ "*"

Is this possible in scala?

Doretha

Answer 1

You can't do anything with the existing \ or \\ methods on NodeSeq. But you can extend NodeSeq with a new \* method (note the lack or space character), as per the pimp-your-library pattern:

import xml.{NodeSeq, Elem}

class ChildSelectable(ns: NodeSeq) {
  def \* = ns flatMap { _ match {                                     
    case e:Elem => e.child                                   
    case _ => NodeSeq.Empty                                  
  } }
}

implicit def nodeSeqIsChildSelectable(xml: NodeSeq) = new ChildSelectable(xml)

In the REPL, this then gives me:

scala> val xml = <a><b><c>xxx</c></b></a>
xml: scala.xml.Elem = <a><b><c>xxx</c></b></a>

scala> xml \*                                                                            
res7: scala.xml.NodeSeq = NodeSeq(<b><c>xxx</c></b>)

scala> xml \ "b" \*
res8: scala.xml.NodeSeq = NodeSeq(<c>xxx</c>)

scala> xml \ "b" \ "c" \*
res9: scala.xml.NodeSeq = NodeSeq(xxx)

Vadose

Answer 2

This is pretty close to what you are looking for:

import scala.xml.Elem

val xml = <a><b><c>HI</c></b></a>

println( xml )
println( xml \ "_" )
println( xml \ "b" )
println( xml \ "b" \ "_" )
println( xml \ "b" \ "c" )
println( xml \ "b" \ "c" \ "_")


<a><b><c>HI</c></b></a>
<b><c>HI</c></b>
<b><c>HI</c></b>
<c>HI</c>
<c>HI</c>
// Empty

Preshrunk

This is better than the accepted answer. "_" does exactly what was asked for, without the need to write a new method. Less code to maintain, less non-standard syntax to confuse the next guy looking at your code. – Armallas

dont overestimate this awnser. scala-xml is broken. Use something like github.com/lihaoyi/scalatags or github.com/chris-twiner/scalesXml – Moldavia

@GuillaumeMassé Can you please explain what do you mean by "scala-xml is broken". I am beginning to learn "scala-xml". – Bobbitt

@Mukeshprajapati see web.archive.org/web/20150801055953/http://anti-xml.org – Moldavia