How to change attribute on Scala XML Element

问题:

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

might have attributes scaled by a factor of two:

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 = 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?

回答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() 

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

回答2:

This is how you can do it using Scala 2.10:

import scala.xml._ import scala.xml.transform._  val xml1 =   val rule1 = new RewriteRule {   override def transform(n: Node) = n match {     case e @ {_*} => e.asInstanceOf[Elem] %        Attribute(null, "attr1", "200",        Attribute(null, "attr2", "100", Null))     case _ => n    } }  val xml2 = new RuleTransformer(rule1).transform(xml1) 

回答3:

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        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:

1. Attribute doesn't have a useful copy method, and
2. 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 =  elem: scala.xml.Elem =   scala> elem.copy(attributes=      |   for (attr       |       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 =  

回答4:

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

into

` 

回答5:

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 =   //////////////////// Snippet       Some(b.copy(attributes = b.attributes.append(x.attributes)))      // Will remove any  elems     case removeMe: Elem if isElem(removeMe, "remove-me", "some-attrib" -> "specific value") =>        None      case keep => Some(keep)   }     ++         // 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.

文章来源: How to change attribute on Scala XML Element