RE: Standard XPath expression for the intersection of two node se ts (Was: RE: How can I test if an node included in a nodeset)

[Ed Blachman <EdB at trellix dot com>]

I think the answer here is that in some sense you're both right, but Oliver
is closer to standard usage:

http://mathworld.wolfram.com/SetDifference.html defines "set difference" in
a manner that matches Oliver's expression; it references in addition
something the "symmetric difference" of two sets
(http://mathworld.wolfram.com/SymmetricDifference.html) which matches Ken's
expression.

-- ed

> -----Original Message-----
> From: G. Ken Holman [mailto:gkholman@CraneSoftwrights.com]
> Sent: Friday, August 04, 2000 12:31 PM
> To: xsl-list@mulberrytech.com
> 
> At 00/08/04 10:59 +0200, Oliver Becker wrote:
> >I don't know if anybody has mentioned this before:
> >
> >If we add a single exclamation mark
> >    $nodes1[count(.|$nodes2)!=count($nodes2)]
> >it will become the difference of the two node-sets $nodes1 
> and $nodes2.
> 
> I disagree, though I thought the same at first when saw 
> Mike's great post 
> ... what you have will be those in $nodes1 not in $nodes2, 
> but not those in 
> $nodes2 not in $nodes1.
> 
> I figured you would need:
> 
>     (   $ns1[count(.|$ns2)!=count($ns2)]
>       | $ns2[count(.|$ns1)!=count($ns1)] )
> 
> to get the true difference as I understand it, that being all 
> nodes in both sets not in the other set.  I have an example below.


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