Subject: Re: [xsl] Automatically generate xpath From: "Liron" <magilam@xxxxxxxxxxxxxxxx> Date: Tue, 31 Jan 2006 02:38:28 +0100 |
David, thank you VERY much! In any case, I thing I should head back to the books first
Thank you Liron
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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