Re: [xsl] Sorting Two Dimensional Table

[David Carlisle <davidc@xxxxxxxxx>]

> Is it possible to do a sort in a XPath so that it would feed the
> sequence in a sorted manner ?

XPath 1 does not have a sequence data type so you can not store a sorted
sequence, it has node sets, which are intrinsically unorderd.
If you need to store an ordered set of things in xpath 1 you need to
make copies of the nodes under under a parent node so that sibling order
records the ordering. You then to use the node set extension to access
the nodes.

XPath 2 has a (ordered) sequence type as its primitive datatype,
replacing node sets.



So where in xslt/xpath 2 you can go

 <xsl:variable name="k1" as="xs:string+">
    <xsl:perform-sort select="x/titles/key">
      <xsl:sort select="."/>
    </xsl:perform-sort>
  </xsl:variable>

       <xsl:for-each select="$k1,titles/value">


and so iterate over a sorted sequence of nodes, in XSLt 1 you'd go


  <xsl:variable name="k1x">
    <xsl:for-each select="x/titles/key">
      <xsl:sort select="."/>
    </xsl:for-each>
    <xsl:copy-of select="x/titles/value"/>
  </xsl:variable>

  <xsl:variable name="k1" select="msxsl:node-set($k1x)/*"/>


       <xsl:for-each select="$k1">


Which would work, but copying nodes is relatively expensive and if you
repeatedly copied at every iteration of the sorting it would get
exponentially so.

David

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