Re: problem with xsl:sort
> <xsl:for-each select="."> > <xsl:sort select="@type" data-type="text" order="ascending"/> > > <xsl:for-each select="."> processes a node-set that always contains > exactly one node. So there isn't very much point in sorting it. What's the problem? Have I done something wrong ? Not necessarily wrong, just pointless. If you sort a list of length 1 then clearly the result will be the same as the original list so there is no point in sorting. "." matches just the current node so <xsl:for-each select="."> will execute over a current node list which is just the current node so <xsl:for-each select="."> <xsl:sort select="@type" data-type="text" order="ascending"/> is exactly the same thing, whataver sort options you specify. David ________________________________________________________________________ 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
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