Re: [xsl] Generating a list of items NOT present in source XML

[Jeni Tennison <jeni at jenitennison dot com>]

Hi Greg,

> My query originates from a collection of Star Wars toys that I have
> at home. Each toy is manufactured with a series index value, and I
> am missing some of the individual products. Since I'm a geek and
> have keyed my collection into XML, I would like for XSL to generate
> a listing of all the missing indices.

That has to be the geekiest thing I have *ever* heard ;) Kudos.

> It would be supremely cool if the type of indexed data could be
> changed dynamically to some pre-determinable "growing" type (e.g.
> a,b,c,d,e,...; 1,2,3,4,5,... ; i,ii,iii,iv,v,vi,...).

You could manage that if you used xsl:number to generate the numbers
against which you tested. So if you amended the previous template I
shows you so that it was:

<xsl:template name="fill-in-gaps">
  <xsl:param name="count" select="1" />
  <xsl:param name="format" select="'1'" />
  <xsl:param name="nodes" select="/.." />
  <xsl:if test="$nodes">
    <xsl:variable name="number">
      <xsl:number value="$count" format="{$format}" />
    </xsl:variable>
    <xsl:choose>
      <xsl:when test="$number = $nodes[1]">
        ...
      </xsl:when>
      <xsl:otherwise>
        ...
      </xsl:otherwise>
    </xsl:choose>
  </xsl:if>
</xsl:template>

Cheers,

Jeni

---
Jeni Tennison
http://www.jenitennison.com/


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