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?
t:\ftemp>type iyer.xq
dateTime(xs:date('2010-06-14'),xs:time('00:00:00'))
t:\ftemp>xquery iyer.xq
<?xml version="1.0" encoding="UTF-8"?>2010-06-14T00:00:00
t:\ftemp>type iyer.xsl
<?xml version="1.0" encoding="US-ASCII"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns:xsd="http://www.w3.org/2001/XMLSchema";
version="2.0">
<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
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