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At 2010-09-26 11:52 -0700, sudheshna iyer wrote:
I wanted to convert "2010-06-14" to xs:datetime ->2010-06-14T00:00:00Z using Xquery. Are there any functions to achieve this without writing java code?
<xsl:output method='text'/> <xsl:template match="/"> <xsl:value-of select="dateTime(xsd:date('2010-06-14'),xsd:time('00:00:00'))"/> </xsl:template> </xsl:stylesheet> t:\ftemp>xslt2 iyer.xsl iyer.xsl 2010-06-14T00:00:00 t:\ftemp> If this is not the right forum, can you suggest the correct mailing list to post this question? This is not the correct forum for XQuery questions, but since the answer is based on XPath, I've included an XSLT solution above, under the XQuery solution. The XQuery list I monitor is http://www.x-query.com/mailman/listinfo/talk I hope this helps. . . . . . . . . . Ken -- XSLT/XQuery training: after http://XMLPrague.cz 2011-03-28/04-01 Vote for your XML training: http://www.CraneSoftwrights.com/s/i/ Crane Softwrights Ltd. http://www.CraneSoftwrights.com/s/ G. Ken Holman mailto:gkholman@xxxxxxxxxxxxxxxxxxxx Male Cancer Awareness Nov'07 http://www.CraneSoftwrights.com/s/bc Legal business disclaimers: http://www.CraneSoftwrights.com/legal
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