scala - XML insert/update

Question

do you know any Scala API to insert and (or) update Nodes according to XPath? e.g for a given Node and XPath, this API would create a copy of XML with new node

thanks

Xpath is a query language. As far as I remember you can't update or write with it. Do you mean xslt? — Jan, Oct 15 '12 at 11:52
XPath will be used to "select" a node and you provide a node that will be added as a child to this "parent" nod. btw. I know, that Scala XML is immutable, but is there any possible way, how to accomplish this task? — user1746915, Oct 15 '12 at 12:02
What about [this](http://stackoverflow.com/questions/4018300/substituting-xml-values-programatically-with-scala) question? It shows how to update a node with scala xml... — Jan, Oct 15 '12 at 12:06
In Scala's XML library nodes don't point to their parents, which makes the kind of thing you describe tricky. The [Anti-XML](http://anti-xml.org/) library provides this functionality through [zippers](http://en.wikipedia.org/wiki/Zipper_%28data_structure%29), where you use an XPath-like operator to navigate into the document, make your edits, and then `unselect` to get the edited document—see e.g. [this answer](http://stackoverflow.com/a/10421660/334519) for an example. — Travis Brown, Oct 15 '12 at 15:22
@TravisBrown: can that be an answer? Then this q wouldn't be categorized as "unanswered." — LarsH, Oct 15 '12 at 17:36

score 9 · Answer 1 · answered Apr 15 '14 at 19:02

You can use the RewriteRule to do this, 2.10.3 documentation.

val cats = <Cats>
  <Cat Name="Floyd"/>
  <Cat Name="Onyx"/>
</Cats>

Then suppose the RewriteRule

class AddCat(name: String) extends RewriteRule {
   override def transform(n: Node): Seq[Node] = n match {
     case e: Elem if e.label == "Cats" =>
       val cats = (e \\ "Cat")
       val newCat = <Cat Name={name}/>
       new Elem(e.prefix, "Cats", e.attributes, e.scope, e.minimizeEmpty, (cats ++ newCat).toSeq:_*)
     case x => x
   }
 }

Then you could do,

val rule = new RuleTransformer(new AddCat("Stevie"))
rule.transform(cats)
res2: Seq[scala.xml.Node] = List(<Cats><Cat Name="Floyd"/><Cat Name="Onyx"/><Cat Name="Stevie"/></Cats>)

Similarly if you wanted to change an attribute

class AddLastName(name: String, lastName: String) extends RewriteRule {
  override def transform(n: Node): Seq[Node] = n match {
    case e: Elem if e.label == "Cat" && (e \\ "@Name" text).equals(name) =>
      val cat: String = e.attributes("Name").head.text
      e % Attribute(None, "Name", Text(s"$name $lastName"), Null)
    case x => x
  }
}

val rule = new RuleTransformer(new AddLastName("Stevie", "Nicks"))
rule.transform(cats)
res3: Seq[scala.xml.Node] = List(<Cats><Cat Name="Floyd"/><Cat Name="Onyx"/><Cat Name="Stevie Nicks"/></Cats>)

Both of these approaches would do what you're looking for. The hard part is figuring out how to get at the children, then build the parent node.

This is one of the best answers out here on SO. I can't believe it does not get more upvotes and did not get marked as correct. I have looked and looked at many others, but this was the most helpful. Thanks a lot @tysonjh. — Philippe, Aug 04 '16 at 21:05

score 0 · Answer 2 · answered Dec 12 '12 at 16:18

What I would do is parse the xml using \\

convert the xml to a list using .toList.last.text

append what I need to and then collect my options: jsonOption collect here you can change some options again using orElse Some and then finally use a for yield to yield the xml I want.

Take advantage of scale's flexibility in data conversion. It doesn't have to be identical but I'm sure you can see the light...

scala - XML insert/update

2 Answers2

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