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
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