Hei Michael,
its the default namespace, the document starts with
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<?xml-stylesheet type="text/xsl" href="extractDublicates.xslt"?>
<html xmlns="http://www.w3.org/1999/xhtml">?
Georg Lenz
-----Original Message-----
From: Michael Kay [mailto:mhk(_at_)mhk(_dot_)me(_dot_)uk]
Sent: Montag, 27. Oktober 2003 18:57
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] Do a copy of a document but avoid duplicates in
certain lists of tags
Probably the "ul" element is in the XHTML namespace. You need to declare
a prefix for this namespace in your stylesheet and write
match="xhtml:ul".
Michael Kay
-----Original Message-----
From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of
Lenz, Georg
Sent: 27 October 2003 16:15
To: 'XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com'
Subject: [xsl] Do a copy of a document but avoid duplicates
in certain lists of tags
I have the following problem:
I want a perfect copy of an xhtml document but want avoid coping
duplicates "li"s in all "ul" lists.
I tried:
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="ul" >
<xsl:comment>From Here</xsl:comment>
<xsl:copy >
<xsl:for-each
select="li[not(.=following-sibling::li)]">
<xsl:copy>
<xsl:apply-templates
select="node()|@*"/>
</xsl:copy>
</xsl:for-each>
</xsl:copy>
</xsl:template>
but it does not work.
It does not even touch the ul template?
If the document node would be "ul" it works???
Any help available.
Thanks in advance
Mit freundlichem Gruß / Best Regards
Georg Lenz
Java IDE Core
SAP AG
Neurottstrasse 16
69190 Walldorf
T +49-6227-7-64235
F +49-6227-7-74235
E georg(_dot_)lenz(_at_)sap(_dot_)com
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