Subject: Re: XSL Decending Order Query
From: Trevor Nash <tcn@xxxxxxxxxxxxx>
Date: Sun, 03 Nov 2002 16:57:10 +0000
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>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
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