[xsl]xpath for getting file version

Subject: [xsl]xpath for getting file version
From: Antony Suryadinata <antony.suryadinata@xxxxxxxxxxxx>
Date: Wed, 15 Aug 2001 09:42:05 +1000
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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