Subject: 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) From: Ed Blachman <EdB@xxxxxxxxxxx> Date: Fri, 4 Aug 2000 14:22:33 -0400 |
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@xxxxxxxxxxxxxxxxxxxx] > Sent: Friday, August 04, 2000 12:31 PM > To: xsl-list@xxxxxxxxxxxxxxxx > > 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. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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