Subject: Re: [xsl] Problem sorting Elements when using the document() function From: David Carlisle <davidc@xxxxxxxxx> Date: Wed, 3 Dec 2003 15:21:25 GMT |
<xsl:variable name="taxfile"> <xsl:value-of select="@file"></xsl:value-of> </xsl:variable> That makes a result tree fragment, it's a lot less typing and more efficient for the system to store the attribute directly: <xsl:variable name="taxfile" select="@file"/> Although you don't really need a variable at all here concat($taxfile,'') just coerces the result tree fragment to a tring and then appends '' which is a no-op. You can just do <xsl:apply-templates select="document(@file)//family"/> or <xsl:apply-templates select="document(string(@file))//family"/> depending on whether you want the relative URI to be based on the URI of the stylesheet or your master doc. <xsl:sort data-type="text" select="/name"/> /name will evaluate to the same thing (probably the empty string unless your documents have a top level element called name) so you are sorting with a constant key, so nothing gets sorted. You want a relative path so different items get a different sort key. You didn't show your input, but if name is a child of family you want <xsl:sort data-type="text" select="name"/> 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
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