How to change attribute on Scala XML Element

Question

I have an XML file that I would like to map some attributes of in with a script. For example:

<a>
  <b attr1 = "100" attr2 = "50"/>
</a>

might have attributes scaled by a factor of two:

<a>
  <b attr1 = "200" attr2 = "100"/>
</a>

This page has a suggestion for adding attributes but doesn't detail a way to map a current attribute with a function (this way would make that very hard): http://www.scalaclass.com/book/export/html/1

What I've come up with is to manually create the XML (non-scala) linked-list... something like:

// a typical match case for running thru XML elements:
case  Elem(prefix, e, attributes, scope, children @ _*) => {
 var newAttribs = attributes
 for(attr <- newAttribs)  attr.key match {
  case "attr1" => newAttribs = attribs.append(new UnprefixedAttribute("attr1", (attr.value.head.text.toFloat * 2.0f).toString, attr.next))
  case "attr2" => newAttribs = attribs.append(new UnprefixedAttribute("attr2", (attr.value.head.text.toFloat * 2.0f).toString, attr.next))
  case _ =>
 }
 Elem(prefix, e, newAttribs, scope, updateSubNode(children) : _*)  // set new attribs and process the child elements
}

Its hideous, wordy, and needlessly re-orders the attributes in the output, which is bad for my current project due to some bad client code. Is there a scala-esque way to do this?

Answer 1

Ok, best effort, Scala 2.8. We need to reconstruct attributes, which means we have to decompose them correctly. Let's create a function for that:

import scala.xml._

case class GenAttr(pre: Option[String], 
                   key: String, 
                   value: Seq[Node], 
                   next: MetaData) {
  def toMetaData = Attribute(pre, key, value, next)
}

def decomposeMetaData(m: MetaData): Option[GenAttr] = m match {
  case Null => None
  case PrefixedAttribute(pre, key, value, next) => 
    Some(GenAttr(Some(pre), key, value, next))
  case UnprefixedAttribute(key, value, next) => 
    Some(GenAttr(None, key, value, next))
}

Next, let's decompose the chained attributes into a sequence:

def unchainMetaData(m: MetaData): Iterable[GenAttr] = 
  m flatMap (decomposeMetaData)

At this point, we can easily manipulate this list:

def doubleValues(l: Iterable[GenAttr]) = l map {
  case g @ GenAttr(_, _, Text(v), _) if v matches "\\d+" => 
    g.copy(value = Text(v.toInt * 2 toString))
  case other => other
}

Now, chain it back again:

def chainMetaData(l: Iterable[GenAttr]): MetaData = l match {
  case Nil => Null
  case head :: tail => head.copy(next = chainMetaData(tail)).toMetaData
}

Now, we only have to create a function to take care of these things:

def mapMetaData(m: MetaData)(f: GenAttr => GenAttr): MetaData = 
  chainMetaData(unchainMetaData(m).map(f))

So we can use it like this:

import scala.xml.transform._

val attribs = Set("attr1", "attr2")
val rr = new RewriteRule {
  override def transform(n: Node): Seq[Node] = (n match {
    case e: Elem =>
      e.copy(attributes = mapMetaData(e.attributes) {
        case g @ GenAttr(_, key, Text(v), _) if attribs contains key =>
          g.copy(value = Text(v.toInt * 2 toString))
        case other => other
      })
    case other => other
  }).toSeq
}
val rt = new RuleTransformer(rr)

Which finally let you do the translation you wanted:

rt.transform(<a><b attr1="100" attr2="50"></b></a>)

All of this could be simplified if:

• Attribute actually defined prefix, key and value, with an optional prefix
• Attribute was a sequence, not a chain
• Attribute had a map, mapKeys, mapValues
• Elem had a mapAttribute