RE: [xsl] grouping headers

["Corey Spitzer" <cspitzer at cisco dot com>]

first, you'll want to alphabetize the list based on city name so run this
stylesheet on the xml:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>

<xsl:template match="people">
  <people>
  <xsl:apply-templates select="person"><xsl:sort
select="city"/></xsl:apply-templates>
  </people>
</xsl:template>

<xsl:template match="person">
	<person>
	<city><xsl:value-of select="city"/></city>
	<name><xsl:value-of select="name"/></name>
</person>
</xsl:template>

</xsl:stylesheet>

Then, do this:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>

<xsl:template match="people">

  <xsl:apply-templates select="person"/>

</xsl:template>

<xsl:template match="person">

<xsl:if test="not(preceding-sibling::person[1]/city=city)">

  <xsl:text disable-output-escaping="yes">
</xsl:text> <!-- hard return to make it look pretty -->

  <xsl:value-of select="city"/>

  <xsl:text disable-output-escaping="yes">
</xsl:text> <!-- hard return to make it look pretty -->

</xsl:if>

<xsl:value-of select="name"/>

<xsl:text disable-output-escaping="yes">
</xsl:text> <!-- hard return to make it look pretty -->

</xsl:template>

</xsl:stylesheet>









-----Original Message-----
From: owner-xsl-list@lists.mulberrytech.com
[mailto:owner-xsl-list@lists.mulberrytech.com]On Behalf Of Jeroen
Janssen
Sent: Thursday, August 23, 2001 4:24 PM
To: xsl-list@lists.mulberrytech.com
Subject: [xsl] grouping headers


Sorry for the subject line, I couldn't think of anything better :-) I'm
trying to do the following:

I have something like this xml:

<people>
    <person>
        <city>Amsterdam</city>
        <name>John Doe</name>
    </person>

    <person>
        <city>Amsterdam</city>
        <name>Jane Doe</name>
    </person>

    <person>
        <city>London</city>
        <name>Jim Doe</name>
    </person>
</people>

And would like this result in html:

Amsterdam
John Doe
Jane Doe

London
Jim Doe

I hope I'm making myself clear, I searched the archives for an answer but I
don't really know what I'm looking for...




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