Re: [xsl] Another grouping question

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

























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