Re: [xsl] Date conversion

[Steven dot C dot Kienle at am dot pnu dot com]

  I responded once, but to give a more "complex" solution:
  
  <?xml version='1.0' encoding='utf-8' ?>
  <xsl:stylesheet version="1.0" 
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
  xmlns:months="months.uri">
  
  <xsl:output method="text" />
  
  
  <months:month abbr="Jan" num="01"/>
  <months:month abbr="Feb" num="02"/>
  <months:month abbr="Mar" num="03"/>
  <months:month abbr="Apr" num="04"/>
  <months:month abbr="May" num="05"/>
  <months:month abbr="Jun" num="06"/>
  <months:month abbr="Jul" num="07"/>
  <months:month abbr="Aug" num="08"/>
  <months:month abbr="Sep" num="09"/>
  <months:month abbr="Oct" num="10"/>
  <months:month abbr="Nov" num="11"/>
  <months:month abbr="Dec" num="12"/>
  
  <xsl:variable name="the-months" select="document('')/*/months:month" 
  />
  
  <xsl:template match="//date">
  
  <xsl:value-of select="$the-months[@abbr = substring(current(), 4, 
  3)]/@num" />
  <xsl:text>/</xsl:text>
  <xsl:value-of select="substring(text(), 1, 2)" />
  <xsl:text>/</xsl:text>
  <xsl:value-of select="substring(text(), 8, 4)" />
  </xsl:template>
  
  </xsl:stylesheet>
  
  This uses the document function to pull in the lookup table from the 
  stylesheet itself.  Then the select state find the correct month 
  number using a xpath query.
  

______________________________ Reply Separator _________________________________
Subject: [xsl] Date conversion
Author:  xsl-list@lists.mulberrytech.com at Internet-America
Date:    01-08-2001 9:45 AM


The format of the date returned in my XML is 17-Dec-2000 07:23:31 I need to 
display the date as 12/17/2000.

How do I do this using xsl?

Thanks,
Katie

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