[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: constructing the Node Sets
David Carlisle wrote: > > I expected that the intersection of a and b will produce a count of 2, > > ah. In that case you asked the wrong question. > > Your node sets are disjoint, no node has both @name='Asm' and @name='Cmp' > so the intersection of the sets $a and $b is empty. > > You don't want the intesection of the sets, you want all elements of $a > that have a string value equal to some member of $b, that's easier than > the question you asked, > > $a[.=$b] Err... I think that you mean: $a[@obid = $b/@obid] Right? The standard for "equality" is that the "obid" attributes have the same value. Your expression will result in "$a" all over again, since all the elements in "$a" and "$b" have empty string values. > except that if you have the possibility of duplicate obid values you may > need to remove duplicates as well. > > Note that > $a[ count( . | $b ) = count( $b ) ] > is the same as > $b[ count( . | $a ) = count( $a ) ] > it really is the nodes in both $a and $b, Well, except possibly for order, as I noted earlier. > but > > $a[.=$b] $a[@obid = $b/@obid] > is not the same > as > > > $b[.=$a] $b[@obid = $a/@obid] > one is the set of nodes in $a the other is the set in $b. Right, in the first expression all the "Class" elements will have a "name" attribute with the value "Asm". In the second expression all the "Class" elements will have a "name" attribute with the value "Cmp". -- Roger Glover glover_roger@xxxxxxxxx XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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