Subject: RE: [xsl] constructing the Node Sets From: "Roger Glover" <glover_roger@xxxxxxxxx> Date: Wed, 29 Jan 2003 17:10:46 -0600 |
David Carlisle wrote: > > 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. My bad. I misread the original poster's variable setting. > > 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. I will address this in a separate message. > > 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) Right. Again, my mistake in reading the original poster's variable setting. -- Roger Glover glover_roger@xxxxxxxxx XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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