[xquery-talk] Xquery : Sort and get only the first record
sudheshna iyer
sudheshnaiyer at yahoo.com
Sat Nov 21 16:42:18 PST 2015
Currently below works for me:
for $job in //joborder by xs:date(replace($job/StartDate, "(\d{2})-(\d{2})-(\d{4})", "$3-$1-$2")) descendingcount $positionwhere $position eq 1return $job
To this, I want to add:> Just need one more modification to this. In the input if the StartDate is
> blank, then I should consider that as the current job and return the job
> that has blank as the output.
Thank you for your response.
From: Christian Grün <christian.gruen at gmail.com>
To: sudheshna iyer <sudheshnaiyer at yahoo.com>
Cc: "talk at x-query.com" <talk at x-query.com>
Sent: Sunday, November 15, 2015 4:35 AM
Subject: Re: [xquery-talk] Xquery : Sort and get only the first record
How does your current XQuery expression look like?
On Sun, Nov 15, 2015 at 9:11 AM, sudheshna iyer <sudheshnaiyer at yahoo.com> wrote:
> Thank you all for the reply.
>
> Just need one more modification to this. In the input if the StartDate is
> blank, then I should consider that as the current job and return the job
> that has blank as the output.
>
> If there are no jobs which have startDate as blank, then it should return
> the recent startdate job (for which we already have the response as shown in
> email chain,,)
>
> For eg, in the below case, response should be the job element with id = 1:
> <?xml version="1.0" encoding="UTF-8"?>
> <Request>
> <SessionInfo>
> <uid>qq</uid>
> <pwd>qq</pwd>
> </SessionInfo>
> <Param>
> <CustomerInfo>
> <Contact>
> <Number>123</Number>
> <Name>aaa bbb</Name>
> <Jobs>
> <job>
> <Name>Analyst</Name>
> <Id>1</Id>
> <StartDate></StartDate>
> <EndDate>1</EndDate>
> </job>
> <job>
> <Name>Programmer</Name>
> <Id>2</Id>
> <StartDate>08-31-2015</StartDate>
> <EndDate>2</EndDate>
> </job>
> </Jobs>
> </Contact>
> </CustomerInfo>
> </Param>
> </Request>
>
>
>
>
> ________________________________
> From: Ghislain Fourny <g at 28.io>
> To: Ronald Bourret <rpbourret at rpbourret.com>
> Cc: "talk at x-query.com" <talk at x-query.com>; sudheshna iyer
> <sudheshnaiyer at yahoo.com>
> Sent: Tuesday, November 3, 2015 4:51 AM
> Subject: Re: [xquery-talk] Xquery : Sort and get only the first record
>
> Hi Ronald,
>
> You can also use a count clause if you have XQuery 3.0! It makes it easier
> to read. Like so:
>
> for $job in doc('x.xml')//job
> order by xs:date(replace($job/StartDate, "(\d{2})-(\d{2})-(\d{4})",
> "$3-$1-$2")) descending
> count $position
> where $position eq 1
> return $job
>
> I hope it helps.
>
> Kind regards,
> Ghislain
>
>
>
>
> On Mon, Nov 2, 2015 at 6:27 PM, Ronald Bourret <rpbourret at rpbourret.com>
> wrote:
>
> Can't use just use a predicate that returns the first element in the sorted
> sequence? Something like:
>
> let $seq = for $job in doc('x.xml')//job
> order by
> xs:date(replace($job/StartDate,
> "(\d{2})-(\d{2})-(\d{4})", "$3-$1-$2")
> ) descending
> return $job
> return $seq[1]
>
> (My apologies if the syntax isn't exact. I haven't written an XQuery for
> quite a while.)
>
> -- Ron
>
> On 11/2/2015 5:53 AM, sudheshna iyer wrote:
>
> Thank you very much, Christian. Let me try this..
>
>
>
> ------------------------------------------------------------------------
> *From:* Christian Grün <christian.gruen at gmail.com>
> *To:* sudheshna iyer <sudheshnaiyer at yahoo.com>
> *Cc:* "talk at x-query.com" <talk at x-query.com>
> *Sent:* Sunday, November 1, 2015 4:26 PM
> *Subject:* Re: [xquery-talk] Xquery : Sort and get only the first record
>
> If your XQuery processor supports XQuery Update, this would be one solution:
>
> copy $xml := doc('x.xml')
> modify (
> delete node subsequence(
> for $job in $xml//job
> order by xs:date(replace($job/StartDate,
> "(\d{2})-(\d{2})-(\d{4})", "$3-$1-$2")
> ) descending
> return $job
> , 2)
> )
> return $xml
>
> Best,
> Christian
>
>
>
>
> On Sun, Nov 1, 2015 at 8:51 PM, sudheshna iyer <sudheshnaiyer at yahoo.com
> <mailto:sudheshnaiyer at yahoo.com>> wrote:
> > Team,
> >
> > I have an xml which has multiple jobs elements. I want to sort the
> jobs and
> > want to output only the latest job. Note that <EndDate> can be empty
> > indicating that it is the current job.
> >
> > Basically I want to order by jobs/job/EndDate in descending fashion and
> > select only the first record..
> >
> > How do I do that using xquery?
> >
> > Input request:
> > <?xml version="1.0" encoding="UTF-8"?>
> > <Request>
> > <SessionInfo>
> > <uid>qq</uid>
> > <pwd>qq</pwd>
> > </SessionInfo>
> > <Param>
> > <CustomerInfo>
> > <Contact>
> > <Number>123</Number>
> > <Name>aaa bbb</Name>
> > <Jobs>
> > <job>
> > <Name>Analyst</Name>
> > <Id>1</Id>
> > <StartDate>01-01-2015</StartDate>
> > <EndDate>08-30-2015</EndDate>
> > </job>
> > <job>
> > <Name>Programmer</Name>
> > <Id>2</Id>
> > <StartDate>08-31-2015</StartDate>
> > <EndDate />
> > </job>
> > </Jobs>
> > </Contact>
> > </CustomerInfo>
> > </Param>
> > </Request>
> >
> > Expected output:
> >
> > <?xml version="1.0" encoding="UTF-8"?>
> > <Request>
> > <SessionInfo>
> > <uid>qq</uid>
> > <pwd>qq</pwd>
> > </SessionInfo>
> > <Param>
> > <CustomerInfo>
> > <Contact>
> > <Number>123</Number>
> > <Name>aaa bbb</Name>
> > <Jobs>
> > <job>
> > <Name>Programmer</Name>
> > <Id>2</Id>
> > <StartDate>08-31-2015</StartDate>
> > <EndDate/>
> > </job>
> > </Jobs>
> > </Contact>
> > </CustomerInfo>
> > </Param>
> > </Request>
> >
> > Your help is greatly appreciated.
>
> >
> >
> >
> >
> > _______________________________________________
> > talk at x-query.com <mailto:talk at x-query.com>
> > http://x-query.com/mailman/listinfo/talk
>
>
>
>
>
> _______________________________________________
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> http://x-query.com/mailman/listinfo/talk
>
>
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