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I have SharePoint 2010 calendar and I am opening it as DataFormwebpart in SharePoint Designer and I noticed EndDate as follows: xsl:value-of select="@EndDate"/ and the result is 2012-03-15T14:30:00Z

How can I modify the xsl:value-of select="@EndDate"/ to Show DateTime as follows 2012-03-15 9:30 AM

Thanks,

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You'd have to parse the value with XSL and XPath. Check out these links, they should get you started:

XSL if: http://www.w3schools.com/xsl/xsl_if.asp

XPath transform(): http://www.w3schools.com/xpath/xpath_functions.asp

And some related SE posts:
https://stackoverflow.com/questions/268116/how-can-i-do-string-operations-in-xslt
https://stackoverflow.com/questions/1569523/can-an-xslt-parse-a-string-of-text

Hope that helps.

  • I can get the desired result with the following but just missing the time part, I want 9:00 AM instead of 14:00<xsl:value-of disable-output-escaping="no" select="ddwrt:FormatDateTime(string(@EventDate), 2057, 'yyyy/MM/dd HH:mm')" /> result is 12/03/2012 14:00 – 4uSharePoint Mar 15 '12 at 15:44
  • Try and use 1033 instead of 2057, since AM/PM probably is not used in your location (2057) it will not parse it with AM/PM. – Anders Aune Mar 15 '12 at 16:31
  • Try using this format string: 'yyyy/MM/dd hh:mm P' HH = hours (24 hour clock) hh = hours (12 hour clock) P = AM/PM marker Ref: w3.org/TR/xslt20/#function-format-dateTime (see section 16.5.1) – Wade Henderson Mar 15 '12 at 18:30
  • Thanks guys, the following is working, but AM/PM part is not working even I added P in the end but no luck. Can you plz let me know any alternative. select="ddwrt:FormatDateTime(string(@EventDate), 2057, 'yyyy/MM/dd HH:mm')" result is 12/03/2012 02:00 – 4uSharePoint Mar 16 '12 at 17:53

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