[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: Simple XPath question
Actually, Sebastien was pretty close. Given : <PERSON firstname="fn1" lastname="ln1"/> <PERSON firstname="fn2" lastname="ln2"/> <PERSON firstname="fn3" lastname="ln3"/> <PERSON firstname="fn2" lastname="ln2"/> <PERSON firstname="fn3" lastname="ln3"/> <PERSON firstname="fn4" lastname="ln4"/> <PERSON firstname="fn3" lastname="ln3"/> This worked properly in XT: <xsl:for-each select="PERSON[following-sibling::PERSON/@lastname = ./@lastname and following-sibling::PERSON/@firstname = ./@firstname and not(preceding-sibling::PERSON/@lastname = ./@lastname and preceding-sibling::PERSON/@firstname = ./@firstname)] "> <xsl:copy-of select="."/> </xsl:for-each> It always gets the oldest brother :) Nikita Ogievetsky http://www.cogx.com XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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