XML/XSLT stylesheet/format date

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December in German

File: Data.xml

File: Transform.xslt
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="2.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:xs="http://www.w3.org/2001/XMLSchema">
  <xsl:output method="text"/>
  <xsl:template match="/">
    
    <xsl:text>December 25, 1960 in German: </xsl:text>
    <xsl:value-of select="format-date(xs:date("1960-12-25"), "[D] [MNn,3-3] [Y0001]", "de", "AD", "DE")"/>
  </xsl:template>
</xsl:stylesheet>
Output:
December 25, 1960 in German: 25 Dez 1960



format-date(/orgchart/@date, "[D1] [MNn] [Y1]")

File: Data.xml
<?xml version="1.0"?>
<orgchart date="2004-03-31">
</orgchart>

File: Transform.xslt
<?xml version="1.0"?>
<html xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xsl:version="2.0">
  <head>
    <title>Management Structure</title>
  </head>
  <body>
    <p>
      <xsl:value-of
        select="format-date(/orgchart/@date, "[D1] [MNn] [Y1]")" />
    </p>
  </body>
</html>
Output:
<html>
   <head>
      <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
      <title>Management Structure</title>
   </head>
   <body>
      <p>31 March 2004</p>
   </body>
</html>



select="format-date(current-date(),"[dwo]")"

File: Data.xml

File: Transform.xslt
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="2.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:xs="http://www.w3.org/2001/XMLSchema">
  <xsl:output method="text"/>
  <xsl:template match="/">
    <xsl:text>&#xA;  Today is the </xsl:text>
    <xsl:value-of select="format-date(current-date(),"[dwo]")"/>
    <xsl:text> day of the year. </xsl:text>
    
  </xsl:template>
</xsl:stylesheet>
Output:

  Today is the three hundred and forty-fifth day of the year.
Источник — «http://wbex.ru/index.php?title=XML/XSLT_stylesheet/format_date&oldid=10345»