XML parsing with Scala: Equivalent to "getElementByTagName(name)" in JS

XML parsing in Scala doesn't seem to be as easy and straightforward as it should be.

What I needed was something that behaved similar to document.getElementsByTagName(name) in JavaScript, but for my purposes all I needed was the first element of a particular tag-name. Here is what I ended up with:

import scala.xml.{Document, Elem, Node}
import scala.xml.parsing.ConstructingParser
def _getFirstMatchingElementByName(search: String, n: Node): Option[Node] = {
    if (n.label == search) {
        Some(n)
    } else {
        var i = 0
        var result: Option[Node] = None
        try {
            while (result == None) {
                result = _getFirstMatchingElementByName(search, n.child(i))
                i += 1
            }
        } catch {
            case e: IndexOutOfBoundsException => None
        }
        result
    }
}

It basically recurses through until a match is found or all possibilities are exhausted.

Now that the feature which required that I have this ability has been released I have reviewed this a little more and it really bugs me. I'm sure there are many Java libraries available to help开发者_如何转开发 parse XML, but given the native support that Scala has for generating XML (i.e. it can pretty much just be inlined anywhere), I am curious if I am missing something.

Is there a better way to do this in Scala?

---

You doing it wrong!
all I needed was the first element of a particular tag-name
given this xml:

val page = 
  <root>
    <need>text1</need>
    <doesnotneed>text2</doesnotneed>
    <doesnotneed>text3</doesnotneed>
    <need>text4</need>
  </root>

Now calling this code will give you list of all nodes with given tag name:

scala> page \\ "need"
res3: scala.xml.NodeSeq = NodeSeq(<need>text1</need>, <need>text4</need>)

To get only first one:

scala> page \\ "need" head
res4: scala.xml.Node = <need>text1</need>

P.S. deep-first element would be treated as head.