Transforming XML using RewriteRule and RuleTransformer

transform.scala

// Import the magic libraries

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

// Source xml. In Scala, xml is literal.

val xml = 
  <user>
    <email>joe@example.com</email>
    <password>secret</password>

    <features>
      <feature id="one"/>
      <feature id="two"/>
      <feature id="three"/>
    </features>

    <foo>
      <bar id="qux" name="caddr"/>
      <bar id="baz" name="cdddr"/>
    </foo>
  </user>

// The problem: replace the "features" list in the above XML
// with a new features list: ["a", "b", "c", "d"]

// Define a function to take a dom and replace the existing
// list of features with a new list of features.

def replaceFeatures(dom: Node, features: List[String]): Node = {

  // You can define objects (or classes) inside functions.

  object replaceFeatures extends RewriteRule {
    // The rule replaces any "features" element with a new
    // features element we build for the occasion.

    override def transform(n: Node): Seq[Node] = n match {

      // match on the "features" element

      case Elem(prefix, "features", attribs, scope, _*) =>

        // XML literals again...

        // Can embed scala inside an XML literal: In this case,
        // apply an anonymous function over the list of features
        // in the original parameter list of the replaceFeatures
        // function. The func turns a feature name into a feature
        // node.

        <features>
          { features map { d => <feature id={d}/> } }
        </features>

      // That which we cannot speak of, we must pass over
      // in silence....

      case other =>
        other
    }

  }

  // Subclass a RuleTransformer (because it's abstract), handing
  // it a vararg list of rules to use (in this case, just one).

  object transform extends RuleTransformer(replaceFeatures)

  // Do the transform. (A scala function's return value is the
  // value of the last expression (and everything's an expression).

  transform(xml)
}

// Transform the xml, replacing the existing feature list
// with a new list constructed from our feature list.

val newXml = replaceFeatures(xml, List("a", "b", "c", "d"))

// Just for fun...
//
// Create a partial function pp so we don't have to have a nasty object
// floating around. pp(xml) is much better than printer.format(xml), 
// eh? The underscore stands for the expected value.

val pp = new PrettyPrinter(123, 2).format(_:Node)

// Print before and after

println("before = \n" + pp(xml))
println("after  = \n" + pp(newXml))

I had a problem similar to the above at work. Basically, I need to take an existing XML doc and alter some values so that I could then save it back to a data store. If you've ever had to work with DOM manipulation, you know it's pretty painful. I really like the Scala solution. Also, I love the "partial function" thing at the end. I really need to do more of that in my own code.

Output

before =

joe@example.com
secret

after =

<user>
  <email>joe@example.com</email>
  <password>secret</password>
  <features>
    <feature id="a"></feature>
    <feature id="b"></feature>
    <feature id="c"></feature>
    <feature id="d"></feature>
  </features>
  <foo>
    <bar name="caddr" id="qux"></bar>
    <bar name="cdddr" id="baz"></bar>
  </foo>
</user>

I love that when I search for Scala examples, your gists keep popping up! Thanks again!

Hey I am interested in your code for doing xml transformations, but I am completely new to scala can you help explain the process of running this file?

Does this work if there iare XML entities present in the file?