Re: [xsl] XSL Decending Order Query

Subject: Re: [xsl] XSL Decending Order Query
From: Trevor Nash <tcn@xxxxxxxxxxxxx>
Date: Sun, 03 Nov 2002 16:57:10 +0000
>My query is that of a basic one, but with trying to many times, i have 
>failed in getting my xml data churned out in descending order, rather than 
>ascending, which is the de facto standard.
>
>     <xsl:for-each select="archive/news">
>     <xsl:sort select="archive/news" order="descending">

The 'select' attribute specifies the key on which to sort, as an XPath
expression to be evaluated in the context of each selected node.
Given a 'news' element, 'archive/news' will always result in nothing,
because 'news' elements do not have 'archive' elements as children.
So the sort comes out in document order.

select="date" might do what you want, providing your dates are in
year-month-day format such as "2002/11/03".

If you simply want reverse document order, then select="position()"
data-type="number" will do the trick.

(in both cases, you still want order="descending").

Trevor Nash
--
Traditional training & distance learning,
Consultancy by email

Melvaig Software Engineering Limited
voice:     +44 (0) 1445 771 271 
email:     tcn@xxxxxxxxxxxxx

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


Current Thread