Re: [xsl] Automatically generate xpath

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

----- Original Message ----- From: "David Carlisle" <davidc@xxxxxxxxx>
To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
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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