[xsl] Q on ISO TimeDate convertion

["Walter Torres" <walter@xxxxxxxxx>]

I have an XSLT Template that converts the ISO TimeDate format to an American
Standard display.

    2001-10-18T12:27:46

    10/18/2001 12:27:46

My problem is that I really don't know how to convert the Time part to an
AM/PM format, not just the 24 hour clock it is now.

Below is the code I am using to convert the date.

Can someone help with the Time conversion.

Also, is there a way to pass the ISO Date/Time and return the conversion,
just a function?

The Template I have does the conversion and the display all within the
Template.

Walter



   <!-- timestamp NODE Template -->
   <xsl:template match="timestamp">

      <!-- Pull the pieces apart -->
      <xsl:variable name="datetime" select="."/>
      <xsl:variable name="year"     select="substring( $datetime, 0 , 5 )"/>
      <xsl:variable name="month"    select="substring( $datetime, 6 , 2 )"/>
      <xsl:variable name="day"      select="substring( $datetime, 9 , 2 )"/>

      <td valign="top" class='data'>
         <!-- put the date together in American order -->
         <xsl:value-of select="concat($month, '/', $day, '/', $year )"/>
         <!-- Seperate the date and Time -->
         -
         <!-- Display the Time -->
         <xsl:value-of select="substring( $datetime, 12 , 5 )"/>
      </td>

   </xsl:template>


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