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Re: Is Anyone Interested In XPath 2.0 Use Cases or Requiremen

• To: xml-dev@l...
• Subject: Re: Is Anyone Interested In XPath 2.0 Use Cases or Requirements?
• From: "Christopher R. Maden" <crism@m...>
• Date: Tue, 09 Jul 2002 04:15:00 -0700
• In-reply-to: <200207080908.KAA13254@p...>
• References: < <8BD7226E07DDFF49AF5EF4030ACE0B7E06621DB4@r...><8BD7226E07DDFF49AF5EF4030ACE0B7E06621DB4@r...>

At 02:08 8/7/02, David Carlisle wrote:
>Yes of course. If you want value based set equality that's
>  not($b[. != $a]) and not($a[. != $b])

Um... (thinks, scribbles in *scratch*, checks spec)

ITYM
  not($b[not(. = $a)]) and not($a[not(. = $b)])

No?

$b[. = $a] returns all X in $b such that there is some Y in $a for which X = Y

$b[. != $a] returns all X in $b such that there is some Y in $a for which X 
!= Y

$b[not(. = $a)] returns all X in $b such that there is no node Y in $a for 
which X = Y

In other words, if the three nodes in $b have values "1", "2", and "3", as 
do the three nodes in $a:

$b[. = $a] will return 3 nodes, because for each node X in $b, there is a 
node Y in $a with the same value.

But $b[. != $a] will *also* return 3 nodes, because for each node X in $b, 
there is a node Y in $a with a *different* value.

(I know David knows this, but I wanted to convince myself... see the note 
in §3.4 of XPath 1.0 if curious for details.  David's solution will work if 
$a and $b each have only one node.)

~Chris
-- 
Christopher R. Maden, Principal Consultant, crism consulting
DTDs/schemas - conversion - ebooks - publishing - Web - B2B - training
<URL: http://crism.maden.org/consulting/ >
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