[xsl]xpath for getting file version

[Antony Suryadinata <antony dot suryadinata at spike dot com dot au>]

Hi All,

The XML:

<?xml version="1.0"?>
<files>
   <file name="a">
     <versions>
       <version day="01" month="03" year="2001' name="test - B"/>
       <version day="03" month="12" year="2000' name="test - A"/>
       <version day="31" month="03" year="2001' name="test - C"/>
     </versions>
    </file>
</files>

What is the xpath for getting the latest version? I have tried:

<xsl:value-of select="versions/version[number(concat(@year, @month, 
@day)) > number(concat(../version/@year, ../version/@month,
../version/@day))]/@name"/>

but it doesn't seem to give the right result

I need it to be an xpath because I want to use it in a <xsl:key>. i.e:

<xsl:key name="file-version' match="file" use="the correct xpath here"/>

Thank you,

Antony



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