Subject: AW: [xsl] Finding node having maximum value for an attribute From: "Pfitzner, Jan" <Jan.Pfitzner@xxxxxx> Date: Fri, 31 May 2002 10:36:35 +0200 |
Yes, you can. You can sort the Archive-nodes using the ID with xsl:sort and extract the first or last node of the sorted tree. <xsl:variable name="max"> <xsl:sort data-type="number" select="//Archive/@ID" order="descending"/> <xsl:value-of select="//Archive[1]/@ID"/> </xsl:variable> JP -----Ursprüngliche Nachricht----- Von: Bagchi Ratul [mailto:bratul@xxxxxxxx] Gesendet: Freitag, 31. Mai 2002 10:24 An: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Betreff: [xsl] Finding node having maximum value for an attribute Hi, I have a XML which has the following structure : <Archives> <Archive ID="1"> <somenode>somevalue</somenode> <someothernode>someothervalue</someothernode> </Archive> <Archive ID="2"> <somenode>somevalue</somenode> <someothernode>someothervalue</someothernode> </Archive> </Archives> Now my question is, is it possible to wite a XPath query which will return me the highest ID attribute if Archive node in the whole file??? Regards, Ratul. 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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