Strictness of XSLT stylesheet validation when no schema is referenced ?

Discussion in 'XML' started by Andy Dingley, Jul 8, 2004.

  1. Andy Dingley

    Andy Dingley Guest

    I'm writing XSLT to transform fntg-schema (our own project's document
    schema) into PartnerML, a HTML-like XML dialect used for mobile
    phones. I'm using XMLSPy 4.4 to do this.

    The schema for PartnerML is supplied in several modules, brought
    together by <xs:include> and <xs:import> statements. My understanding
    (and experiment) suggests that XMLSpy can't work with such a
    modularized schema.

    As a result, my XSLT documents have a header like this:

    <xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:fntg="http://www.futurenet.com"
    exclude-result-prefixes="fntg" >

    I use the usual XSL namespace, our own namespace for the input
    elements, and just use the default namespace for output. Everything
    works fine for _using_ the stylesheet, but XMLSpy complains that the
    stylesheet is invalid (it can't find the output elements). And when
    XMLSpy complains about this, it gets annoying (3 pop-up dialog clicks
    every time you try to save the file)


    What's going on here (strictly speaking) ? My understanding is that a
    document like my stylesheet is to be seen as "valid", because it's
    well-formed and it's valid as far as the referenced schemas do
    describe it. In the absence of a schema for the default namespace,
    then tools should be generous in what they permit.

    Now I'm not an XML language-lawyer, so I can't point at the relevant
    part of the W3C XML TR offhand. If anyone could do so, or tell me I'm
    either definitely right or definitely wrong, then I'd find that really
    helpful. Have I got it wrong, or is it XMLSpy ?


    Anyone have useful work-arounds ?


    Thanks
     
    Andy Dingley, Jul 8, 2004
    #1
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