Zipping zippers in Anti-XML

问题

In this question, the asker wants to transform documents like this:

<text>
  The capitals of Bolivia are <blank/> and <blank/>.
</text>

Into this:

<text>
  The capitals of Bolivia are <input name="blank.1"> and <input name="blank.2">.
</text>

As I noted in my answer there, Anti-XML's zippers provide a clean solution to this problem. The following, for example, would work for renaming the blank elements:

import com.codecommit.antixml._

val q = <text>The capitals of Bolivia are <blank/> and <blank/>.</text>.convert

(q \\ "blank").map(_.copy(name = "input")).unselect

Unfortunately the following doesn't work:

(q \\ "blank").zipWithIndex.map { case (el, i) => el.copy(
  name = "input",
  attrs = Attributes("name" -> "blank.%d".format(i + 1))
)}.unselect

Because of course once we've zipWithIndex-ed the zipper we no longer have a zipper, just an IndexedSeq—we can't have a Zipper[(Node, Int)] because the definition is trait Zipper[+A <: Node] ....

Is there a clean way to use zip or zipWithIndex on an Anti-XML zipper, do some other operations with map, etc., and end up with something that's still a zipper?

回答1:

I can't think of a direct way to achieve what you need, but if you're willing to go for a lower level function, you can use a fold, e.g.:

val blanks = q \\ "blank"

(0 until blanks.size).foldLeft(blanks) {case (z, i) => z.updated(i, z(i).copy(
  name = "input",
  attrs = Attributes("name" -> "blank.%d".format(i + 1)))
)}.unselect

Note that the zipper is a random access container, so efficiency shouldn't be an issue in this case.

来源：https://stackoverflow.com/questions/11143558/zipping-zippers-in-anti-xml

标签

xml

scala

zipper

anti-xml