Re: [xsl] the nearest ancestor with the attribute
Subject: Re: [xsl] the nearest ancestor with the attribute|
From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx>
Date: Thu, 18 Jan 2001 10:20:50 -0500
At 01/01/18 13:07 +0000, Jeni Tennison wrote:
Since I started this confusion in the first place, can I summarise and
check that I've got it right now?
A node set is always unordered until you do anything with it like
iterate over it, get a value from it or whatever.
I find it is more succinct to say: "Nodes from the attribute and namespace
axes are always in an arbitrary implementation-defined order, while nodes
from all other axes are in proximity order within location step expressions
and in document order in location path expressions."
Note that attribute and namespace axes are *not* unordered (you can index
into them), just that the order cannot be relied upon. There is an
Since a predicate can be applied to a location path expression, as in the
union of two other path expressions, that predicate is always evaluated in
xsl:value-of and xsl:for-each deal with location path expressions (not
location step expressions), therefore their evaluation is done in document
This has proven to be a simple rule-of-thumb for students.
G. Ken Holman mailto:gkholman@xxxxxxxxxxxxxxxxxxxx
Crane Softwrights Ltd. http://www.CraneSoftwrights.com/s/
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Oliver Becker - Thu, 18 Jan 2001 16:49:23 +0100 (MET)