Re: [xsl] Date YYYYMMDD format XSLT 2.0

[Florent Georges <lists@xxxxxxxxxxxx>]

Pankaj Chaturvedi wrote:

  Hi

> <chapter version="Feb. 2000"/>

> The date could be optional can be taken as 01 by default. So the
> desired output required in "YYYYMMDD format".

  Depending on your precise input, the following can give you a
solution (or an idea to adapt):

    <xsl:variable name="months" as="element()+">
       <m name="Jan" num="01"/>
       <m name="Feb" num="02"/>
       <m name="Mar" num="03"/>
       <m name="Apr" num="04"/>
       <m name="May" num="05"/>
       <m name="Jun" num="06"/>
       <m name="Jul" num="07"/>
       <m name="Aug" num="08"/>
       <m name="Sep" num="09"/>
       <m name="Oct" num="10"/>
       <m name="Nov" num="11"/>
       <m name="Dec" num="12"/>
    </xsl:variable>

    <xsl:template name="test" match="/">
       <xsl:variable name="re" select="
           '^(([0-9]{{1,2}}) )?([A-Z][a-z][a-z]). ([0-9]{{4}})$'"/>
       <xsl:analyze-string select="'02 Feb. 2001'" regex="{ $re }">
          <xsl:matching-substring>
             <xsl:variable name="day" select="regex-group(2)"/>
             <xsl:sequence select="
                 concat(regex-group(4),
                        $months[@name eq regex-group(3)]/@num,
                        if ( string-length($day) eq 1 )
                          then '0' else '',
                        if ( $day ) then $day else '01')"/>
          </xsl:matching-substring>
       </xsl:analyze-string>
    </xsl:template>

  Regards,

--drkm