Subject: RE: [xsl] constructing the Node Sets From: David Carlisle <davidc@xxxxxxxxx> Date: Wed, 29 Jan 2003 21:05:34 GMT |
> > Err... I think that you mean: > $a[@obid = $b/@obid] No I meant what I wrote. $a and $b in the original posting were sets of attribute nodes so @obit in your expression would always be an empty node set. > Well, except possibly for order, as I noted earlier. Not at all. Node sets are sets and as such have no ordering property. If two node sets have the same members they are indistinguishable in Xpath. > 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". as both $a and $b just contain obid attributes the subsets will similarly contain just these attribute nodes, but they will be different attribute nodes (but with the same value) David XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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