Playing with DateTime is not possible with XSLT 1.0 alone .. In a similar situations I took help of scripting .. (C#)
Sample XML:
<?xml version="1.0" encoding="utf-8"?>
<root>
<Apple>2011-12-01T16:33:33Z</Apple>
</root>
Sample XSLT:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl" xmlns:cs="urn:cs">
<xsl:output method="xml" indent="yes"/>
<msxsl:script language="C#" implements-prefix="cs">
<![CDATA[
public string datenow()
{
return(DateTime.Now.ToString("yyyy'-'MM'-'dd'T'HH':'mm':'ss'Z'"));
}
]]>
</msxsl:script>
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Apple">
<xsl:copy>
<xsl:value-of select="cs:datenow()"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Resulting Output:
<?xml version="1.0" encoding="utf-8"?>
<root>
<Apple>2012-02-22T18:03:12Z</Apple>
</root>
The script may reside in a same file (like I have it in my sample XSLT code) or if the code triggering XSLTransformation is C# then move the same code in the calling place :)
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