Re: xsl:sort

Dave,

>Does a sort use the key *only* for sorting,
>i.e. can I keep the function and its children together
>durin the sort?
>
>  <xsl:template match="faqindex">
>     <H2>Index for XSLT FAQ Website</H2>
>     <!-- Sort the functions -->
>       <xsl:variable name="fns">
>	<xsl:for-each select="functions[.!='']">
>	  <xsl:sort data-type="text" select="pair/word"/>
>	</xsl:for-each>
>       </xsl:variable>
>
><!-- now convert that to a node-set and step through it -->
>
>     <xsl:for-each select="xt:node-set($fns)">*
>       <xsl:value-of select="child::*/text()"/> <br />
>     </xsl:for-each>
>   </xsl:template>

You've probably slapped your head and sent a mail retracting this question
that's going to cross this reply in the post, but anyway:

You've said:

  <xsl:variable name="fns">
    <xsl:for-each select="functions[.!='']">
      <xsl:sort data-type="text" select="pair/word"/>
    </xsl:for-each>
  </xsl:variable>

The xsl:for-each loops over each of the functions, sorted in terms of
pair/word and then... does nothing with them!  Since there's nothing there
that actually produces any output, then the variable $fns is set to an
empty rtf.  When it's passed to the next for-each, there's nothing to
iterate over.

What you wanted was:

  <xsl:variable name="fns">
    <xsl:for-each select="functions[.!='']">
      <xsl:sort data-type="text" select="pair/word"/>
      <xsl:copy-of select="." />
    </xsl:for-each>
  </xsl:variable>

i.e. to produce a copy of each of the sorted functions to process later on.

[I guess there's a good reason for storing the sorted list and using an
extension function to access it rather than just doing:

  <xsl:for-each select="functions[.!='']">
    <xsl:sort data-type="text" select="pair/word" />*
    <xsl:value-of select="child::*/text()"/> <br />
  </xsl:for-each>]

I hope that this helps,

Jeni

Dr Jeni Tennison
Epistemics Ltd * Strelley Hall * Nottingham * NG8 6PE
tel: 0115 906 1301 * fax: 0115 906 1304 * email: jeni.tennison@xxxxxxxxxxxxxxxx


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