[xsl] Re2: how to get new position() of a node in a sorted result tree AND document function and counting

[David_Marston@xxxxxxxxx]

Rolling together Michael Kay's and Wendell Piez's answers on their
respective threads, plus using a technique favored here at Lotus, we
have a dual solution that uses no extension functions. This can be
accomplished if one of the XML documents can contain a list of all the
XML files to be read in:

<?xml version="1.0" ?>
<files>
  <a>idkey49a.xml</a>
  <a>idkey49b.xml</a>
  <a>idkey49c.xml</a>
  <a>idkey49d.xml</a>
</files>

As long as <a> nodes are not used for any other purpose, then
document(a)//b will be a single node-set, not an RTF, containing all
the <b> nodes from all four files. Then, you can set up iteration over
the whole set, and sort it:

<xsl:apply-templates select="document(a)//b">
   <xsl:sort select="."/>
</xsl:apply-templates>

And in the next template, the position() function will return what you
wanted, a contiguous numbering of all the <b> elements from all files:

<xsl:template match="b">
  <xsl:value-of select="position()"/><xsl:text>. </xsl:text>
  <xsl:value-of select="."/><xsl:text>&#10;</xsl:text>
</xsl:template>

Embellish the above as necessary.
.................David Marston


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