Re: [xsl] Dynamic numbering of lists in xslt

[Abel Braaksma <abel.online@xxxxxxxxx>]

David Carlisle wrote:

> >    <xsl:number />  <!-- will only count the matches consecutively -->
>
> Nope, it will number according to the input tree (but the advice to use
> templates rather than a xsl:choose stricture is good)

You are right, of course. Using position(), you can change this behavior 
through the apply-templates. Using the input from the OP, and my 
approach, the following is a way to do it (using xslt 2 for ease of use 
and not needing an input doc, call it on itself)

<xsl:stylesheet
   version="2.0"
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
  
   <xsl:output indent="yes"/>
 
   <xsl:variable name="source">
       <chapter1>
           <section>first section</section>
       </chapter1>
       <chapter1>
           <section>second section</section>
       </chapter1>
       <chapter1>
           <section>third section</section>
       </chapter1>
       <chapter1>
           <section>fourth section</section>
       </chapter1>
   </xsl:variable>

   <xsl:template match="/">
       <xsl:apply-templates
           select="$source/chapter1[not(section = 'second section')]" />
   </xsl:template>
  
   <xsl:template match="chapter1">
       <chap>
           <xsl:value-of select="concat(position(), '. ', section)" />
       </chap>
   </xsl:template>
  
</xsl:stylesheet>

Output:
<chap>1. first section</chap>
<chap>2. third section</chap>
<chap>3. fourth section</chap>


However, if the select-statement becomes more complex, other approaches 
may be better (not meaning xsl:choose). In addition, if XSLT 2.0 were an 
option, the select-attribute could be used to achieve the same goal.

If Saxon extensions can be used, an easy (but unwanted) quick fix is to 
use saxon's assignable variables.

-- Abel