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Re: [xsl] Sorting and grouping

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sorry I mistakenly used following-sibling in my code example, here an update:

Make sure you first order your items, the next step is breaking it up in groups of say 5.
Don't try to start with grouping before sorting, that won't work!

I now first sort according to the at attribute, then I test within item wich position it has within the sorted items list.
is position mod $group is 1 than it's the first of the group, if it's zero, it's the last.

otherwise it's something in between.

Hope this helps.

RH

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
   <xsl:output method="xml" encoding="Windows-1252" />

<xsl:param name="group">5</xsl:param> <!-- grouping size-->

<xsl:variable name="items" select="count(//item)"/> <!-- total number of items-->

   <xsl:template match="list">
	    <xsl:apply-templates select="item">
	    	<xsl:sort select="@at" order="ascending"/>
	    </xsl:apply-templates>
	</xsl:template>

<xsl:template match="item">

<xsl:choose>
<xsl:when test="position() mod $group =1">
<xsl:comment>Begin new group</xsl:comment>
<p><xsl:value-of select="position()"/> - <xsl:apply-templates/></p>
</xsl:when>
<xsl:when test="position() mod $group =0">
<p><xsl:value-of select="position()"/> - <xsl:apply-templates/></p>
<xsl:comment>End group</xsl:comment>
</xsl:when>
<xsl:when test="not(position() mod $group =0) and position() = $items"> <!-- last item in sorted list-->
<p><xsl:value-of select="position()"/> - <xsl:apply-templates/></p>
<xsl:comment>End group</xsl:comment>
</xsl:when>
<xsl:otherwise>
<p><xsl:value-of select="position()"/> - <xsl:apply-templates/></p>
</xsl:otherwise>

</xsl:choose>

</xsl:template>

</xsl:stylesheet>

RH


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