I have a date(string) value in an XML file in this format:
Tue Apr 17 03:12:47 IST 2012
I want to use XSL transformation to convert the string/date into this format:
4/17/2012 03:12:47 AM
How can I do that in my XSL transform?
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Which xslt version (1.0, 2.0) can/would you use?hr_117– hr_1172013-06-03 08:05:34 +00:00Commented Jun 3, 2013 at 8:05
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I can use both in fact. Would prefer 1.0.Rg90– Rg902013-06-03 08:08:36 +00:00Commented Jun 3, 2013 at 8:08
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1 Answer
If you are using
XSLT 1.0 version, use EXSLT - date:format-date date extension
XSLT 2.0 version, use built-in: Formatting Dates and Times date extension
But my suggestion is to
Have a standard XSD datetime format on XML, on the code-behind (that is, on rendering time) you can format as you like.
Update:
Always XML to process through XSLT, dates should be in standard XSD format. Currently your input is not in standard format so that it throws error.
Example:
<xsl:variable name="dt" as="xs:dateTime" select="xs:dateTime('2012-10-21T22:10:15')"/>
<xsl:value-of select="format-dateTime($dt, '[Y0001]/[M01]/[D01]')"/>
OUTPUT:
2012/10/21
6 Comments
Rg90
I am using the format-dateTime function (2.0) and its giving me this error: 'Invalid dateTime value "Tue Apr 17 03:12:47 IST 2012" (Non-numeric year component)'
Siva Charan
@RohitGupta: Show your xslt
Rg90
<Date><xsl:value-of select="format-dateTime(CreatedDate, '[Y0001]-[M01]-[D01]T[H01]:[m01]:[s01]')" /></Date>
Artur Keyan
@SivaCharan I got an Error during XSLT transformation: An unknown XPath extension function was called. exaption, may be you know where is the problem
Siva Charan
@ArturKeyan: Its better you can post a new question with detailed error message and your XSLT. So that I can help you.
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