I have an XML where I get this attribute:
Trans_dtmBookingStamp="Fri, 15 Nov, 2013 @ 3:20pm"
I want to convert the time stamp, with XSLT, to an ISO8601 Format like this:
2013-11-13T15:20:00+02:00
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Daniel Haley
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user2261714
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What version of XSLT? Can you post an example of what you've tried? – Daniel Haley Dec 19 '13 at 07:01
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possible duplicate of [How to parse string to date in xslt 2.0](http://stackoverflow.com/questions/16851726/how-to-parse-string-to-date-in-xslt-2-0) – ceving Dec 19 '13 at 11:09
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http://stackoverflow.com/questions/16892344/convert-a-string-to-date-format-in-xslt – ceving Dec 19 '13 at 11:10
1 Answers
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See if this can get you started:
<xsl:template name="convertDate">
<xsl:param name="datestring" />
<xsl:param name="d" select="substring-before(substring-after($datestring, ' '), ' ')"/>
<xsl:param name="mmm" select="substring(substring-after(substring-after($datestring, ' '),' '), 1, 3)"/>
<xsl:param name="m" select="string-length(substring-before('JanFebMarAprMayJunJulAugSepOctNovDec', $mmm)) div 3 + 1"/>
<xsl:param name="y" select="substring(substring-after($datestring, $mmm), 3 , 4)"/>
<xsl:param name="ymmdd" select="10000*$y+100*$m+$d"/>
<xsl:param name="mm" select="substring($ymmdd, 5, 2)"/>
<xsl:param name="dd" select="substring($ymmdd, 7, 2)"/>
<xsl:value-of select="concat ($y, '-', $mm, '-', $dd, 'T')" />
</xsl:template>
Calling this template with a datestring parameter of "Fri, 15 Nov, 2013 @ 3:20pm" will return a value of "2013-11-15T". The time portion is left as an exercise for the reader.
michael.hor257k
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Seriously disturbed people vote me down - but this is good code nevertheless. – michael.hor257k Oct 21 '15 at 22:30