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?

I'm shocked at how lousy the library seems to be in this regard. — Daniel C. Sobral
– Daniel C. Sobral, Commented Apr 3, 2010 at 0:32
Many good answers here. Also see stackoverflow.com/a/23092226/35274 — Philippe
– Philippe, Commented Aug 4, 2016 at 21:06

Daniel C. Sobral · Accepted Answer · 2010-04-07 23:02:26Z

15

+150

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

edited Apr 7, 2010 at 23:02

answered Apr 5, 2010 at 23:08

Daniel C. Sobral

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3 Comments

Dave Over a year ago

The design of the library does seem to have made some strange choices. You've come up with something more versatile than I did so... points for that.

David Leppik Over a year ago

I tried this out in Scala 2.9.1. A few minor things: the .toSeq wrapping the RewriteRule seems to be redundant, as a Node is a Seq[Node]. Also the attributes end up reversed.

David Leppik Over a year ago

Here's my fix: def unchainMetaData(m: MetaData): Iterable[GenAttr] = m.flatMap(decomposeMetaData).toList.reverse

Magomed Abdurakhmanov · Accepted Answer · 2013-10-03 16:57:57Z

12

This is how you can do it using Scala 2.10:

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

val xml1 = <a><b attr1="100" attr2="50"></b></a>

val rule1 = new RewriteRule {
  override def transform(n: Node) = n match {
    case e @ <b>{_*}</b> => e.asInstanceOf[Elem] % 
      Attribute(null, "attr1", "200", 
      Attribute(null, "attr2", "100", Null))
    case _ => n 
  }
}

val xml2 = new RuleTransformer(rule1).transform(xml1)

answered Oct 3, 2013 at 16:57

Magomed Abdurakhmanov

1,9741 gold badge16 silver badges14 bronze badges

Comments

Steve · Accepted Answer · 2010-12-02 11:08:30Z

So if I were in your position, I think what I'd really want to be writing is something like:

case elem: Elem => elem.copy(attributes=
  for (attr <- elem.attributes) yield attr match {
    case attr@Attribute("attr1", _, _) =>
      attr.copy(value=attr.value.text.toInt * 2)
    case attr@Attribute("attr2", _, _) =>
      attr.copy(value=attr.value.text.toInt * -1)
    case other => other
  }
)

There are two reasons this won't work out of the box:

Attribute doesn't have a useful copy method, and
Mapping over a MetaData yields an Iterable[MetaData] instead of a MetaData so even something as simple as elem.copy(attributes=elem.attributes.map(x => x)) will fail.

To fix the first problem, we'll use an implicit to add a better copy method to Attribute:

implicit def addGoodCopyToAttribute(attr: Attribute) = new {
  def goodcopy(key: String = attr.key, value: Any = attr.value): Attribute =
    Attribute(attr.pre, key, Text(value.toString), attr.next)
}

It can't be named copy since a method with that name already exists, so we'll just call it goodcopy. (Also, if you're ever creating values that are Seq[Node] instead of things that should be converted to strings, you could be a little more careful with value, but for our current purposes it's not necessary.)

To fix the second problem, we'll use an implicit to explain how to create a MetaData from an Iterable[MetaData]:

implicit def iterableToMetaData(items: Iterable[MetaData]): MetaData = {
  items match {
    case Nil => Null
    case head :: tail => head.copy(next=iterableToMetaData(tail))
  }
}

Then you can write code pretty much like what I proposed at the beginning:

scala> val elem = <b attr1 = "100" attr2 = "50"/>
elem: scala.xml.Elem = <b attr1="100" attr2="50"></b>

scala> elem.copy(attributes=
     |   for (attr <- elem.attributes) yield attr match {
     |     case attr@Attribute("attr1", _, _) =>
     |       attr.goodcopy(value=attr.value.text.toInt * 2)
     |     case attr@Attribute("attr2", _, _) =>
     |       attr.goodcopy(value=attr.value.text.toInt * -1)
     |     case other => other
     |   }
     | )
res1: scala.xml.Elem = <b attr1="200" attr2="-50"></b>

Erik Kaplun · Accepted Answer · 2014-10-24 13:08:16Z

1

With the help of Scalate's Scuery and its CSS3 selectors and transforms:

def modAttr(name: String, fn: Option[String] => Option[String])(node: Node) = node match {
  case e: Elem =>
    fn(e.attribute(name).map(_.toString))
      .map { newVal => e % Attribute(name, Text(newVal), e.attributes.remove(name)) }
      .getOrElse(e)
}

$("#foo > div[bar]")(modAttr("bar", _ => Some("hello")))

— this transforms e.g. this

<div id="foo"><div bar="..."/></div>

into

<div id="foo"><div bar="hello"/></div>`

edited Oct 24, 2014 at 13:08

answered Jul 1, 2014 at 15:59

Erik Kaplun

38.5k15 gold badges102 silver badges113 bronze badges

3 Comments

millhouse Over a year ago

Scuery now lives on github

Erik Kaplun Over a year ago

Thanks for that, good news also (means it's alive); updated the answer.

Erik Kaplun Over a year ago

...too bad the issue tracker stayed on Assembla — it could have been migrated to Github as well!

Peter L · Accepted Answer · 2017-05-25 06:49:18Z

I found it easier to create a separate XML snippet and merge. This code fragment also demonstrates removing elements, adding extra elements and using variables in an XML literal:

val alt = orig.copy(
  child = orig.child.flatMap {
    case b: Elem if b.label == "b" =>
      val attr2Value = "100"
      val x = <x attr1="200" attr2={attr2Value}/>  //////////////////// Snippet
      Some(b.copy(attributes = b.attributes.append(x.attributes)))

    // Will remove any <remove-me some-attrib="specific value"/> elems
    case removeMe: Elem if isElem(removeMe, "remove-me", "some-attrib" -> "specific value") => 
      None

    case keep => Some(keep)
  }
    ++
      <added-elem name="..."/>

// Tests whether the given element has the given label
private def isElem(elem: Elem, desiredLabel: String, attribValue: (String, String)): Boolean = {
  elem.label == desiredLabel && elem.attribute(attribValue._1).exists(_.text == attribValue._2)
}

For other new-comers to Scala XML, you'll also need to add a separate Scala module to use XML in scala code.

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