Subject: RE: [xsl] selecting the nearest preceding sibling From: Jarno.Elovirta@xxxxxxxxx Date: Tue, 13 May 2003 17:16:49 +0300 |
Hi, > OK - the node set of all the preceding siblings which > fit the criterium is given by > > preceding-sibling::tag[a] > > The nearest preceding sibling fitting the criterium is > the last one in that nodeset (the nodes are always in > document order)... that is, > > preceding-sibling::tag[a and last()] Wrong. Simple "last()" will always evaluate to a positive number, thus the above will compile to preceding-sibling::tag[a] Furthermore, nodes are always *processed* in document order, but preceding-sibling is a preceding axis and preceding-sibling::tag[a and position() = last()] will select the first node in document order. You can use (preceding-sibling::tag[a])[position() = last()] and then you'd get the first preceding sibling. Evan Lenz wrote an article about this stuff <http://www.biglist.com/lists/xsl-list/archives/200201/msg00108.html>. Cheers, Jarno - neuroticfish: wakemeup! (club-edit) XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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