The result of a path expression (an expression involving "/") is always in
document order.
I'm not sure exactly what you want, but you might be able to get it by
replacing
A/B
with
for $a in A return $a/B
which doesn't sort into document order.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Kenneth Stephen
[mailto:marvin(_dot_)the(_dot_)cynical(_dot_)robot(_at_)gmail(_dot_)com]
Sent: 27 November 2005 01:43
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Node ordering question with preceding axis
Hi,
I have the following testcase:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="2.0">
<xsl:output method="xml" omit-xml-declaration="yes" />
<xsl:variable name="forest">
<tree>One</tree>
<tree>Two</tree>
<tree><tree>Three</tree></tree>
<tree><tree>Four</tree><tree>Five</tree></tree>
<tree><tree>Six</tree><tree>Seven</tree><tree>Eight</tree></tree>
</xsl:variable>
<xsl:template match="/">
<xsl:value-of select="$forest//tree[text() =
'Eight']/preceding::*" />
</xsl:template>
</xsl:stylesheet>
When run, this code produces:
One Two Three Three FourFive Four Five Six Seven
I'm not entirely clear about the expected behaviour of preceding
axis with temporary trees, but I was expecting that the for the last
"forest" of nodes, the node ordering would be "Seven" followed by
"Six" since preceding is a reverse axis. What am I missing?
Thanks,
Kenneth
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