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Re: [xsl] sort by predefined order

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2008/5/20 Michael Kay <mike@xxxxxxxxxxxx>:
> <xsl:sort select="index-of($sortorder, @ID)"/>
>
> where
>
> <xsl:variable name="sortorder"
>  select="for $t in tokenize(/source/sortorder, ',\s*') return
> xs:integer($t)"/>
>
> This isn't going to be very efficient if sortorder is a long list, in that
> case you probably want to use keys to speed it up.

I would've tended towards that route anyway - not using sort at all.
So modifying the above a bit you get:

<xsl:key name="element-by-ID" match="element" use="@ID"/>

<xsl:for-each select="for $t in tokenize(/source/sortorder, ',\s*')
return xs:integer($t)">
  <xsl:apply-templates select="key('element-by-ID', .)"/>
</xsl:for-each>

or perhaps

<xsl:apply-templates select="for $t in tokenize(/source/sortorder,
',\s*') return key('element-by-ID', xs:integer($t))"/>


-- 
Andrew Welch
http://andrewjwelch.com
Kernow: http://kernowforsaxon.sf.net/

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