Subject: How to transform to a stylesheet? From: "Richman, Jeremy" <jrichman@xxxxxxxxxxxx> Date: Thu, 28 Oct 1999 16:19:21 -0400 |
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 XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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