[xslt] fixed/default attribute value from xml schema

[Steffen Malmgaard Mortensen <smm futarque com>]

Hi all
I'm new to xml/xsl so please bear with me if I'm being stupid. For my
application I'm using xml schema to describe the xml files and xslt to
transform them into different output variants - pretty standard.
In my schema I set default/fixed values for some attributes that I then
use during the xslt. My problem is the xsltproc/libxslt is not able to
find these default/fixed values in the schema. 

Well, I have googled a little and found the following:

1) In the w3c xpath spec (http://www.w3.org/TR/xpath#attribute-nodes) it
says that "The XML Recommendation does not require an XML processor to
read an external DTD or an external parameter unless it is validating".

2) According to a mail in the archive this works for xml files described
by a DTD (test case 53)

My question is: Is this supposed to work for transformation of xml files
described by an xml schema?? I also tried to write a python script that
both validates the xml file and applies the style sheet - same result.

Any ideas/help??

Best regards,
Steffen

Attachment: test_document.xml
Description: application/xml

Attachment: test_schema.xsd
Description: application/xml

Attachment: test_stylesheet.xsl
Description: application/xslt

#! /usr/bin/python
import libxml2
import libxslt

styledoc = libxml2.parseFile("test_stylesheet.xsl")
style = libxslt.parseStylesheetDoc(styledoc)

ctxt_parser = libxml2.schemaNewParserCtxt("test_schema.xsd")
ctxt_schema = ctxt_parser.schemaParse()
ctxt_valid  = ctxt_schema.schemaNewValidCtxt()

doc = libxml2.parseFile("test_document.xml")
ret = doc.schemaValidateDoc(ctxt_valid)

if ret != 0:
    print "error doing schema validation"
else:
    result = style.applyStylesheet(doc, None)
    output = style.saveResultToString(result)
    print output
style.freeStylesheet()
doc.freeDoc()
result.freeDoc()