scala - xml transformation using variable and RewriteRule

Question

I'm playing with scala xml transformation and my below program doesn't give me expected output.

import scala.xml.{Elem, Node, Text}
import scala.xml.transform.{RewriteRule, RuleTransformer}

object XmlTransform extends App {
  val name = "contents"
  val value = "2"

  val InputXml : Node =
    <root>
      <subnode>1</subnode>
      <contents>1</contents>
    </root>

  val transformer = new RuleTransformer(new RewriteRule {
    override def transform(n: Node): Seq[Node] = n match {
      case elem @ Elem(prefix, label, attribs, scope, _) if elem.label == name =>
        Elem(prefix, label, attribs, scope, false, Text(value))

      case other => other
    }
  })
  println(transformer(InputXml))
}

It prints xml without any transformation.

<root>
  <subnode>1</subnode>
  <contents>1</contents>
</root>

If I replace(though I didn't want that) name variable in "case if" statement like

case elem @ Elem(prefix, label, attribs, scope, _) if elem.label == "contents" =>
        Elem(prefix, label, attribs, scope, false, Text(value))

it prints out expected transformed xml

<root>
  <subnode>1</subnode>
  <contents>2</contents>
</root>

What am I doing wrong here?

Answer 1

The problem is that the match is defined inside the RewriteRule that happens to have a name field (in my test it had the value "<function1>"). This field shadows your name variable in the outer scope. Renaming your variable solves the problem.