Subject: Re: [xsl] Another grouping question From: Florent Georges <lists@xxxxxxxxxxxx> Date: Mon, 25 Feb 2008 15:07:02 +0100 (CET) |
James Cummings wrote: Hi > Is it because I've normalize-space()'d that I can't then traverse > up from the node to get the ancestor::tei:p? Because of normalize-space(), your sequence is a sequence of strings. If you want to get the original node in the grouping, group on the nodes directly. It is hard to tell without knowing your input, but I guess you want something like: <xsl:for-each-group select="$dMeetingC" group-by="normalize-space(lower-case(.))"> <!-- Here current-group() is element(tei:seg)+, so you might have to adapt your XPath expressions. --> ... <xsl:for-each select="current-group()"> <!-- Here the current item is element(tei:seg). --> ... </xsl:for-each> ... </xsl:for-each-group> Regards, --drkm _____________________________________________________________________________ Ne gardez plus qu'une seule adresse mail ! Copiez vos mails vers Yahoo! Mail http://mail.yahoo.fr
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