Subject: Re: [xsl] sorting question From: "Margarit A. Nickolov" <man@xxxxxxxxxxxxxxx> Date: Wed, 28 May 2003 12:35:41 +0300 |
lars_huttar@xxxxxxx said: > select="/*/sd:Service[sd:ServiceDescription/@ServiceName='Y' > and ../sd:Service[sd:ServiceID < current()/sd:ServiceID]]" it works. > But even if this worked, I don't think this would really be a win because it > could be much less efficient (order N*N) than your original approach (order N > * log N for sorting). I do not pursue any performance, just learning XSLT. > Don't think I helped much but nobody else had replied. Thanks a lot, it helped me. best regards, margarit nickolov. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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