Fabien Tillier wrote:
<data>
<row>
<level>40</level>
<keycode>1254.45.12</keycode>
</row>
<row>
<level>50</level>
<keycode>1254.45.12.7</keycode>
</row>
<row>
<level>50</level>
<keycode>1254.45.12.8</keycode>
</row>
<row>
<level>50</level>
<keycode>1254.45.12.9</keycode>
</row>
<row>
<level>40</level>
<keycode>99.25.6</keycode>
</row>
<row>
<level>50</level>
<keycode>99.25.6.45</keycode>
</row>
<row>
<level>50</level>
<keycode>99.25.6.46</keycode>
</row>
</data>
What could be the XPath expression to get the maximum number of nodes of
level=50 in data those keycode starts like the level=40 line ?
Here the answer would be 3 as the maximum number of level = 50 nodes for
a given level = 40 is 3
The following outputs "3" for your sample:
<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="2.0">
<xsl:param name="l1" select="'50'"/>
<xsl:param name="l2" select="'40'"/>
<xsl:key name="k1" match="row" use="level"/>
<xsl:template match="/">
<xsl:variable name="groups" as="element(group)*">
<xsl:for-each select="key('k1', $l2)">
<group>
<xsl:copy-of select="key('k1', $l1)[starts-with(keycode,
current()/keycode)]"/>
</group>
</xsl:for-each>
</xsl:variable>
<xsl:value-of select="max($groups/count(row))"/>
</xsl:template>
</xsl:stylesheet>
--
Martin Honnen
http://msmvps.com/blogs/martin_honnen/
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