<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="html" indent="yes" />
<xsl:template match="o">
<xsl:variable name="item" select="'item'"/>
<xsl:for-each select="CITATION">
<xsl:variable name="citationid">
<xsl:value-of select="@id"/>
</xsl:variable>
<xsl:element name="{concat('item',@id)}">
<xsl:for-each select="../CITEREF[(_at_)rid=$citationid]">
<xsl:element name="{concat('item',@id)}"/>
</xsl:for-each>
</xsl:element>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
for the xml file
<?xml version="1.0"?>
<o>
<CITATION id="c9"/>
<CITATION id="c10"/>
<CITEREF id="cr9-1" rid="c9"/>
<CITEREF id="cr9-2" rid="c10"/>
</o>
Does this help you?
sundar
-----Original Message-----
From: Shellenberger, Mark [mailto:MShellenberger(_at_)apa(_dot_)org]
Sent: Monday, April 28, 2003 6:12 PM
To: 'XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com'
Subject: [xsl] Creating internal lists
At least this is what I think I am trying to do.....
I am trying to take a large document and convert it into something much
smaller. The document has a References list and then citations within the
document body. I want to create a list where each citation in the
References is matched up with all instances of matching citation references
in the body.
The citation elements are like this:
<CITATION ID="c9">
The reference elements are like this:
<CITEREF ID="cr9-1" RID="c9">
I want the end result to be like this:
item 9
item9-1
item10
item10-1
item10-2
item10-3
item11
item11-1
item11-2
I am not even sure what to call this....thus looking within the FAQ/Archives
and at Developerworks, and XML.com/.org has been difficult at best.
Any suggestions or assistance is greatly appreciated. Heck just tell me
where to read about it and I will be eternally grateful.
Thanks,
Mark
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list