Subject: RE: [xsl] sorting parameter ? From: "Ricaud Matthieu" <matthieu.ricaud@xxxxxxx> Date: Mon, 8 Mar 2004 13:50:35 +0100 |
I did not thought about this simple way of doing ! and indeed i sort my datats according to an attributes node, i just did : <xsl:sort select="@*[name()=$sortparam]"/>. One more time Thanks David! Matthieu. -----Message d'origine----- De : owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]De la part de David Carlisle Envoyé : vendredi 5 mars 2004 16:32 À : xsl-list@xxxxxxxxxxxxxxxxxxxxxx Objet : Re: [xsl] sorting parameter ? It work fine but I'm just wondering if there's not another way of doing it using standard xsl. the usual case, where the sortparam is not an arbitrary xpath but just the name of a child element can be done by <xsl:sort select="*[name()=$sortparam]"/> Slightly more complicated expressions can easily be added to this mechanism, essentially anything that involves a fixed length Xpath without arbitrary predicates. But if you really want to be able to pass in an arbitrary Xpath and sort on the result of that an extension function is realistically the only option. David -- http://www.dcarlisle.demon.co.uk/matthew ________________________________________________________________________ This e-mail has been scanned for all viruses by Star Internet. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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