Re: How do I use DTD's with Scala 2.3.0's XML obje: msg#00512 lang.scala

Hi Raphael,

On 1/29/07, Raphael James Cohn <raphaelcohn@xxxxxxxxx> wrote:

Hi,

I'm using scala's wickedly cool XML object literals (great idea - no more need
for Freemarker or Velocity for XML - could we have them for other things, too,
please... like a compile time templating engine!) to generate XHTML reports for
rehersal. To create valid XHTML, I need to have a DTD before the first element.
However, the compiler complains if I have a method like this:-

private def template(category: TestCategory) =
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">

This won't work because I consciously left out DTDs from the parser. The biggest trouble is that DTDs would mess up Scala's "data model" (the XML people's slang for XML object mapping). They may only appear at the top :/ So there is a class scala.xml.Document (http://www.scala-lang.org/docu/files/api/scala/xml/Document.html) which can contains XML nodes but is not regarded as a node itself.

There are routines for writing out XML with a prefixed DTD declaration though, see Utility.toXML

<head>
<meta http-equiv="content-type" content="application/xhtml+xml; charset=UTF-8"/>
<title>{category.name} Testing Results</title>
<style type="text/css" media="all">
(SNIPPED lots of content)

Removing the DTD causes it to compile.

Does anybody know why?

Additionally, does the compiler or API understand XML and HTML literals? Is ther
a utility method I could use to convert strings?

The compiler parses your XML and gives you the objects, not more, not less (XHTML is ok, plain HTML is not).
Of course, the API offers methods for parsing XML, the easiest being ( http://www.scala-lang.org/docu/files/api/scala/xml/parsing/ConstructingParser.html)

import scala.io.Source
import scala.xml.parsing.ConstructingParser
val src = "" // there's also fromFile, fromInputStream, fromURL...
val prs = ConstructingParser.fromSource(src , true/*preserve whitespace*/)
val doc = prs.document // parses the document
val root= doc.docElem // the content node

If you know that your source does not have prologue (DTD, processing instructions or comments), you can directly tell the parser to grap the content node like this

val root = prs.element(scala.xml.TopScope)(0) // parse an element, assuming no namespace bindings.

The (0) at the end is necessary to get the "first" element, because the routine is designed to return a node sequence in some cases.

Many thanks

Raph

For information, the compile error (I'm using scalac in ant, with Scala 2.3.0) is:-

I don't think we would add support for parsing the DTD, it is very easy to achieve, but from a design point of view it adds nothing but causes trouble. Certainly we don't want to have scala.xml.Document a node (something that can appear everywhere in an xml document).

The alternative, making the parser clever enough to understand that something with a docype declaration is not a node, but a scala.xml.Document is also annoying, because then in order to do anything with it (traverse,transform,match,xpath...), one has to go through the docElem method first, which one may find confusing and inconvenient.

cheers,
Burak

--
Burak Emir
Research Assistant / PhD Candidate
Programming Methods Group
EPFL, 1015 Lausanne, Switzerland
http://lamp.epfl.ch/~emir

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Re: How do I use DTD's with Scala 2.3.0's XML object literals?: msg#00512

lang.scala