Subject: Re: [xsl] XSLT 2 - Sorting data with two elements named the same From: "Andrew Welch" <andrew.j.welch@xxxxxxxxx> Date: Thu, 12 Jun 2008 10:28:58 +0100 |
2008/6/12 Chris Hughes <chris_hughes22@xxxxxxxxxxx>: > XTTE1020: A sequence of more than one item is not allowed as the @select attribute of xsl:sort. > <Person name="Chris"> > <Rating type="Adjusted" value="100"> > <Rating type="Actual" value="99.6"> > </Person> > <Person name="John"> > <Rating type="Adjusted" value="95"> > <Rating type="Actual" value="97.6"> > </Person> > <Person name="Dave"> > <Rating type="Adjusted" value="90"> > <Rating type="Actual" value="81.6"> > </Person> > <xsl:variable name="sortedratings"> > <xsl:for-each select="Person[Rating/@type = 'Adjusted']"> > <xsl:sort select="Rating/@value" data-type="number" order="descending"/> > <xsl:copy-of select="."/> > </xsl:for-each> > </xsl:variable> > So my question is how can I sort this data just by the one adjusted Rating element, and still have access to the partent node. Isn't it just: <xsl:sort select="Rating[@type = 'Adjusted']/@value" .... ? -- Andrew Welch http://andrewjwelch.com Kernow: http://kernowforsaxon.sf.net/
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