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Re: [xsl] Using xsl:include when converting xsl to xsl
Subject: Re: [xsl] Using xsl:include when converting xsl to xsl From: Jeni Tennison <mail@xxxxxxxxxxxxxxxx> Date: Tue, 27 Feb 2001 12:12:30 +0000 |
Hi Tim, > I expected the xsl:template name="body", which is the HTML header > information contained in functions.xsl, to be added to the resulting > xsl file but this didn't happen. The code outputted what I expected it to output, but that doesn't mean that it's what you expected it to do :) The only thing 'wrong' is that the x:namespace-alias result-prefix should be 'x' rather than 'xsl': <x:namespace-alias stylesheet-prefix="xsl" result-prefix="x"/> so that the XSLT elements that you're producing are placed in the correct namespace when they're output. I think what you're after is a stylesheet that takes an existing stylesheet with xsl:include elements in it as its source, and expands the xsl:include elements in the same way as a normal processor would do to create a result that's also an XSLT stylesheet. Here is such a stylesheet: ---- <?xml version="1.0"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" indent="yes"/> <xsl:strip-space elements="*" /> <xsl:preserve-space elements="xsl:text" /> <!-- copy elements by default --> <xsl:template match="*"> <xsl:copy> <xsl:copy-of select="@*" /> <xsl:apply-templates /> </xsl:copy> </xsl:template> <!-- where there are include elements, apply templates to (i.e. copy) the content of the xsl:stylesheet element in the document that they reference --> <xsl:template match="xsl:include"> <xsl:apply-templates select="document(@href)/*/*" /> </xsl:template> </xsl:stylesheet> ---- Note that the above *doesn't* deal properly with xsl:import elements, but I guess that if you're having to do this with xsl:include then you're not using xsl:import anyway. I hope that helps, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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