How to transform to a stylesheet?

["Richman, Jeremy" <jrichman@xxxxxxxxxxxx>]

I am trying to create an XSL stylesheet that is used to transform from an
old-XSL-draft-conforming XSL stylesheet
to an October-draft-conforming XSL stylesheet.

The scope is lots smaller than that sounds, because the stylesheets I'm
transforming FROM in fact use very
few XSL features.  I don't have to worry about changes to XPath, etc.

But there seems to be an impasse when it comes to having an xsl:template
whose content
instantiates an xsl:stylesheet element with appropriate attributes (e.g.
xmlns:xsl="...").

The pertinent part of my stylesheet looks like this:

<xsl:template match="xsl:stylesheet>
    <xsl:copy>
       <xsl:attribute
name="xmlns:xsl">http://www.w3.org/XSL/Transform/1.0</xsl:attribute>
       <xsl:apply-templates select="node()"/>
    </xsl:copy>
 </xsl:template>

But neither LotusXSL nor XT like the <xsl:attribute> element because its
name attribute's prefix isn't defined (of course).
The XSL draft has some commentary about just this thing (section 7.1.3) but
it concludes:
   ...[this xsl:attribute]...will not result in a namespace declaration
being output.

Whatever that means.

Anyone have a solution?

Jeremy





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