Setting an XML namespace in Scala

gistfile1.scala

/*
 * Let's assume we have an XML fragment in Scala like:
 *  <entry xmlns="http://www.w3.org/2005/Atom">
 *   <content xmlns="http://mysite.com/xml/someSchema">
 *    Hello <name>you</name>!
 *   </content>
 *  </entry>
 * Then (_ \ "content") will leave you with:
 *  <content xmlns="http://mysite.com/xml/someSchema" xmlns="http://www.w3.org/2005/Atom">
 *   Hello <name>you</name>!
 *  </content>
 * which might cause problems at some places since you have two non-prefixed
 * namespaces mixed together. You might want to keep only one of them; the
 * function below does exactly that.
 */

import scala.xml._
import scala.xml.transform._

def setMainNamespace(x: Node, uri: String) = {
  // create a non-prefix namespace
  val ns = new NamespaceBinding(null, uri, TopScope)

  // create a RewriteRule that sets this as the only namespace
  object setNs extends RewriteRule {
    override def transform(n: Node): Seq[Node] = n match {
      case Elem(prefix, label, attribs, scope, children @ _*)  =>
        Elem(prefix, label, attribs, ns, children.map(setNs) :_*)
      case other => other
    }
  }

  // call that rule
  setNs(x)
}

/*
 * Now, the following will work:
 *
 * scala> val x = <entry xmlns="http://www.w3.org/2005/Atom"><content xmlns="http://mysite.com/xml/someSchema">Hello <name>you</name>!</content></entry> \ "content"
 * x: scala.xml.NodeSeq = NodeSeq(<content xmlns="http://mysite.com/xml/someSchema" xmlns="http://www.w3.org/2005/Atom">Hello <name>you</name>!</content>)
 *
 * scala> setMainNamespace(x head, "http://mysite.com/xml/someOtherSchema")
 * res1: scala.xml.Node = <content xmlns="http://mysite.com/xml/someOtherSchema">Hello <name>you</name>!</content>
 *
 */