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Re: xslt 1, node sets in variables
Subject: Re: xslt 1, node sets in variables|
From: Joelle Tegwen <tegwe002@xxxxxxx>
Date: Mon, 03 Aug 2009 08:24:40 -0500
Wow, thank you for explaining this to me in such detail. That was very
helpful. I did get this to work. I found I didn't need the xsl:copy-of
I never thought about doing a select statement like that; it hadn't
occurred to me to try it like that. I can adjust the xpath statement to
ensure I don't have logic errors like you described. I do like that
better than the choose statement (which I really don't care to use). So
I think in the end, I'll use that. But I can see that there might come a
time when I need to use the rtf version.
Thank you much for your time, patience and encouragement.
Wendell Piez wrote:
At 05:13 PM 7/31/2009, you wrote:
When I do what you suggested, it doesn't generate any errors, but I
still don't get the result as a node-set. The copy-of statement dumps
the xml of the tree to the output. Also,
count(exsl:node-set($projects)) returns 1 (It should return 26 when
there is no filter).
Sorry, I forgot about this crucial detail.
The data object being converted to the node set is a result tree
fragment. As such, it has a root. So when you count it, or rather the
node set resulting from converting it, (the count() function counts a
set of nodes), you get 1. It's a little tree.
You are probably looking for count(node-set($projects)/*).
You could also clarify things by using an extra variable, so:
<xsl:when test="$active_filter = 'active'">
<xsl:when test="$active_filter = 'archived'">
<xsl:when test="$active_filter = 'new'">
<xsl:variable name="projects" select="exsl:node-set($projects-rtf)/*"/>
In this case, the variable $projects-rtf contains a result tree
fragment, and $projects contains copies (as nodes) of the actual
Now, all this having been said, I'm going to take another step back to
see whether you even need an RTF to begin with. Why not just bind the
project elements themselves to your variable, instead of making copies
Assuming things are not more complicated than you've shown, you might
be able to do this as:
select="project[$active_filter = 'active'][@active!=0] |
project[$active_filter = 'archived'][@active=0] |
project[$active_filter = 'new'][@new=1] |
project[not($active_filter = 'active' or
$active_filter = 'archived' or
$active_filter = 'new')]"/>
Then you don't even need to use the node-set() function, since the
variable is already a node set (this time the actual nodes from the
source tree, not copies).
This works by using XPath predicates to emulate your conditional
statements. And while it may be more cumbersome to read (if you don't
prefer your logic in XPath anyway: some do), it simplifies things
considerably both in your code and in the processing it specifies.
But note I say "might" work since these conditionals are not actually
mutually exclusive. For example, if you have a
project[@new=1][@active=0], you will get it both for your 'new' filter
and your 'archived' filter. You can deal with this either by
controlling your input data to preclude such anomalous cases, or by
extending your filtering logic further.
Keep asking about mysterious points, as this represents a big advance.
Wendell Piez mailto:wapiez@xxxxxxxxxxxxxxxx
Mulberry Technologies, Inc. http://www.mulberrytech.com
17 West Jefferson Street Direct Phone: 301/315-9635
Suite 207 Phone: 301/315-9631
Rockville, MD 20850 Fax: 301/315-8285
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