AW: Finding node having maximum value for an attribute

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:  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.





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