Re: Finding the highest node value

I'd sort the list in reverse order and then select the
first position.

<xsl:template match="/">
  <xsl:for-each select="population/country">
    <xsl:sort select="population" order="descending"/>
      <xsl:if test="position()=1">
        <xsl:value-of select="name"/>:
        <xsl:value-of select="population"/>
      </xsl:if>
  </xsl:for-each>
</xsl:template>



On Thu, Oct 25, 2001 at 05:22:32PM +0200, Antoine
Quint wrote:

> Say I have an XML fragment going like this:
>
> =====================
>
> <population unit="million">
>   <country>
>     <name>China</name>
>     <population>1273</population>
>     <continent key="2"/>
>   </country>
>   <!-- some more <country> -->
>   <country>
>     <name>India</name>
>     <population>1033</population>
>     <continent key="2"/>
>   </country>
> </population>
>
> =====================
>
> What I would like to do is store in a <xsl:variable>
the highest
> <population> value?
>
> How would one go around this?

Here's one way to do it:

<xsl:template name="find-max">
 <xsl:param name="nodeset"/>
 <xsl:param name="max"/>
 <xsl:choose>
  <xsl:when test="count($nodeset)=0">
   <xsl:value-of select="$max"/>
  </xsl:when>
  <xsl:otherwise>
   <xsl:call-template name="find-max">
    <xsl:with-param name="nodeset">
     <xsl:value-of select="$nodeset"/>
    </xsl:with-param>
    <xsl:with-param name="max">
     <xsl:value-of select="$max"/>
    </xsl:with-param>
   </xsl:call-template>
  </xsl:otherwise>
 </xsl:choose>
</xsl:template>

<xsl:template match="population">
 <xsl:variable name="maxpopulation">
  <xsl:call-template name="find-max">
   <xsl:with-param name="nodeset"
                  
select="country/population[position()&gt;1]"/>
   <xsl:with-param name="max"
select="number(country/population[1])"/>
  </xsl:call-template>
 </xsl:variable>
</xsl:template>

Tim.
*/


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