Re: [xsl] my node position within a certain context, efficiently?

On Sun, Oct 26, 2008 at 12:37 AM, Syd Bauman <Syd_Bauman(_at_)brown(_dot_)edu> 
wrote:
> This results in all of the <thing> elements being counted
> irrespective of their position within a <box>, though, whereas I had
> asked to count their positions with respect to their ancestor <box>.
Here's another approach ...

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
                       version="2.0">

 <xsl:output method="xml" indent="yes" />

 <xsl:template match="/">
   <result>
     <xsl:apply-templates select="//box//thing"/>
   </result>
 </xsl:template>

 <xsl:template match="box//thing">
   <thing myCount="{count(ancestor::box[1]//thing[. &lt;&lt; current()]) + 1}">
     <xsl:value-of select="."/>
   </thing>
 </xsl:template>

</xsl:stylesheet>

Here I have used the XPath 2.0 node comparison operator <<.


-- 
Regards,
Mukul Gandhi

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