This is preliminary, but helps.
__________________
<?xml version="1.0" ?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="xml"/>
<xsl:key name="bookMarkStart" match="p" use="bookmark-start/@id" />
<xsl:key name="bookMarkEnd" match="p" use="bookmark-end/@id" />
<xsl:template match="root">
<xsl:apply-templates />
</xsl:template>
<xsl:template match="p[count(. | key('bookMarkStart',
bookmark-start/@id)[1]) = 1]">
<xsl:if test="generate-id()=generate-id(key('bookMarkStart',
bookmark-start/@id))">
<xsl:variable name="bib_number"
select="substring(bookmark-start/@id, 4,
string-length(bookmark-start/@id))"/>
<xsl:variable name="begin1"
select="bookmark-start[(_at_)id=concat('bib',$bib_number)]/ancestor-or-self::p"/>
<xsl:variable name="end1"
select="../p/bookmark-end[(_at_)id=concat('bib',$bib_number)]/ancestor-or-self::p
" />
<bib-reference>
<xsl:attribute name="id"><xsl:value-of
select="concat('bib',$bib_number)"/></xsl:attribute>
<xsl:copy-of
select="($begin1/following-sibling::node()|$begin1)[(.|following-sibling::*)
[generate-id()=generate-id($end1)]]"/>
</bib-reference>
</xsl:if>
</xsl:template>
</xsl:stylesheet>
__________________
-----Original Message-----
From: aaron apigo [mailto:aaronjose_apigo2001(_at_)yahoo(_dot_)com]
Sent: Tuesday, September 27, 2005 6:40 AM
To: xsltforum
Subject: RE: [xsl] Another <xsl:key> problem
Hi,
sorry for incomplete data.
it is supposed to be like this.
<xsl:template match="bookmark-start">
<xsl:variable name="bib_number"
select="substring(@id, 4, string-length(@id))"/>
Other question is: why two keys; "contribution"
and "h_issue" are
identical?
oopps, sorry, my fault, no other reason for that, it
supposed to be "contribution" only, then i should used
it again for getting the nodes for variable "sample".
my another problem about that is, i don't know how can
i process the nodes for each key, e.g.,
<xsl:for-each select="$biblio[$contrib_sample]">
....................
</xsl:for-each>
is this correct?
regards.
aaron
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