I had thought that node-set() was a trivial operation before, but the
Mozilla devs keep barking about how it's non-trivial and difficult to
hack into Transformiix. It got me thinking.
In XSLT 1.0, when a RTF is converted to a node-set, how are IDs and keys
computed? Will the following be true?
<xsl:variable name="rtf">
<node id="1"/>
<node id="2"/>
</xsl:variable>
<xsl:if test="generate-id(exsl:node-set($rtf)/node[(_at_)id=1]) =
generate-id(exsl:node-set($rtf)/node[(_at_)id=2])">true</xsl:if>
Keys... well, if your context node's document is the source document,
there's no question, but will the key access the variable's nodes if I
use something like this?
<xsl:for-each select="exsl:node-set($rtf)">
content
</xsl:for-each>
If so, are the keys recomputed every time it's called, or only once,
when the node-set is constructed for the first time?
I guess the common part of what I'm trying to ask is, once you have a
RTF captured in a variable, does node-set() return a newly-constructed
node-set, or a reference to the node-set corresponding to that variable?
It'd be nice to know. (I have an experimental alternate FXSL
implementation that leans _heavily_ on node-set(), but I've been really
hesitant to start using it because I have no idea how its performance
will be compared to "pure" XSLT 1.0.)
~ Scott
-----Original Message-----
From: David Carlisle [mailto:davidc(_at_)nag(_dot_)co(_dot_)uk]
Sent: Thursday, October 11, 2007 9:44 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] More addition questions: Using node-set()
and I do this 2-3 times for different problem sets in say the same
template,
node-set in most implementations is cheap (in some it's essentially a
no-op) In XSLT2 the distingction between result tree fragment and node
set goes away (and with it the need for a function)
David
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