Subject: Accessing current node attributes
From: George Phillips <phillips@xxxxxxxxxxxx>
Date: Mon, 14 Sep 1998 11:32:20 -0700
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I'd like to write an XSL style sheet that reduces well-formed HTML
down to a simple list of links. Something like this nearly
sufficies:
<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/TR/WD-xsl">
<xsl:template match="a[attribute(href)]">
<A HREF="{.[attribute(href)]}">
<xsl:value-of expr="." />
</A>
</xsl:template>
</xsl:stylesheet>
But "{.[attribute(href)]}" doesn't get me the HREF attribute of the
current A node being processed. Not that I expect it to as it is
not XSL-grammatically correct. In fact, I can't see any way to do
this in XSL at all. Am I missing something or is this really not
possible?
-- George
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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