Re: [xsl] Another grouping question

["James Cummings" <cummings.james@xxxxxxxxx>]

On Mon, Feb 25, 2008 at 2:07 PM, Florent Georges <lists@xxxxxxxxxxxx> wrote:
> James Cummings wrote:
>   Hi
>   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(.))">

Yup, this is exactly what I tried for myself moments after posting
(*doh* why don't I try these things *before* posting!)

What I did was remove the normalize-space(lower-case(.)) from the
earlier variable, and put it there on the for-each-group, and then had
the following as my date variable:
=====
<xsl:variable name="date">
   <list type="date">
      <xsl:for-each select="current-group()">
         <xsl:variable name="thisDate"
             select="./ancestor::tei:dDay//tei:date[1]/@when"/>
         <item><ref target="{concat('http://www.example.com/',
            substring-before($thisDate, '-'), '.xml#p',
$thisDate)}"><xsl:value-of select="$thisDate"/></ref>
         </item>
       </xsl:for-each>
    </list>
</xsl:variable>
=====

With then an <xsl:copy-of select="$date"/> further down when creating
the output.

Thanks for you help, I should have persevered before posting this time. *sigh*

-James