xquery order by date Time - datetime

Question

Say I have the following documents in my database: a_doc1, a_doc2, b_doc1, and b_doc2
All these documents are of the following format
<doc>
....
<updatedTime>2011-02-07T14:41:02.133-05:00</updatedTime>
....
</doc
>
The value of the "updatedTime" element is inserted when the document is created using the fn:current-dateTime()
Now I am trying to do the following:
find all documents whose name starts with "a_"
order these documents by their <updatedTime> element in descending order
Return the first document name from the descending order
I tried the following:
for $doc_name in db:list()
where fn:starts-with($doc_name, 'a_')
order by xs:dateTime(doc($doc_name)/updatedTime) descending
return $doc_name
Say "a_doc1" is created at "2011-02-07T14:40:00.78-05:00" and "a_doc2" is created at "2011-02-07T14:41:02.133-05:00", the desired output is a_doc2. In short the document name(starting with a_) of the most recent document created must be returned.
When I try my sample code, the output returned is : [a_doc1, a_doc2].
The expected output is: [a_doc2, a_doc1].
Thanks,
Sony

Answer 1

order by dateTime(doc($doc_name)/updatedTime/text())
Instead of this use:
order by xs:dateTime(doc($doc_name)/updatedTime)

Answer 2

you should use (xs: http://www.w3.org/2001/XMLSchema):
xs:dateTime($arg as xs:anyAtomicType?) as xs:dateTime?
Example:
xs:dateTime("1999-12-31T12:00:00")
instead of (fn: http://www.w3.org/2005/xpath-functions)
fn:dateTime($arg1 as xs:date?, $arg2 as xs:time?) as xs:dateTime?
Example:
fn:dateTime(xs:date("1999-12-31"), xs:time("12:00:00"))