David, thank you VERY much!
In any case, I thing I should head back to the books first
Thank you
Liron
----- Original Message -----
From: "David Carlisle" <davidc(_at_)nag(_dot_)co(_dot_)uk>
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Tuesday, January 31, 2006 1:48 AM
Subject: Re: [xsl] Automatically generate xpath
You showed me how to use union for two xpaths but how would I remove a
node
from a xpath?
you dont want to remove nodes from a xpath (which is an expression) but
from a node set (which is a value). So you are asking for set
difference: given two node sets $a and $b generate a set of nodes in $a
but not $b.
As I think someone showed earlier, in XPath2 this is
$a except $b
in XPath1 set difference isn't a standard operator but it's available
using the standard idiom:
$a[count(.|$b)!=count($b)]
David
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