Thank you for the quick responses. After reviewing those approaches and
the not getting the expected results for my use case. I came up with the
following function. For my use case I needed to ignore days and calculate
the duration based solely from the year and month of each date.
declare function yearMonthDuration($date1 as xs:dateTime, $date2 as
xs:dateTime) {
let $negative := if ( $date1 lt $date2 ) then ( "-" ) else ()
let $year1 := fn:year-from-dateTime($date1)
let $year2 := fn:year-from-dateTime($date2)
let $month1 := fn:month-from-dateTime($date1)
let $month2 := fn:month-from-dateTime($date2)
let $totalMonths := fn:abs( (($year1 - $year2 ) * 12) + ($month1 -
$month2) )
let $years := $totalMonths idiv 12
let $months := $totalMonths mod 12
return (
xs:yearMonthDuration( $negative || "P" || $years || "Y" || $months
|| "M" )
)
};
-Will
On Thu, Feb 19, 2015 at 6:10 PM, Danny Sokolsky <
danny.sokol...@marklogic.com> wrote:
> I'm not sure if this helps, but here is something similar using a
> technique like the functx function does:
>
> http://www.xqueryfunctions.com/xq/functx_yearmonthduration.html
>
>
> xquery version "1.0-ml";
>
> let $date1 := xs:dateTime(xs:date("2015-03-27"))
> let $date2 := xs:dateTime(xs:date("2024-03-27"))
> let $test := xs:duration($date2 - $date1)
> let $years := fn:round(fn:days-from-duration($test) div 365)
> let $months := fn:round((fn:days-from-duration($test) div 365) div 12)
> return
> ( xs:yearMonthDuration('P1M') * $months +
> xs:yearMonthDuration('P1Y') * $years)
>
> But duration math is strange so I might be doing something wrong.
>
> -Danny
>
>
> -----Original Message-----
> From: general-boun...@developer.marklogic.com [mailto:
> general-boun...@developer.marklogic.com] On Behalf Of Florent Georges
> Sent: Thursday, February 19, 2015 4:26 PM
> To: MarkLogic Developer Discussion
> Subject: Re: [MarkLogic Dev General] xs:yearMonthDuration in 1.0-ml
>
> That function never found its way from the working drafts to the
> recommendation. You can achieve something similar by using the following
> (yes, by using the average number of days in a month, which is fragile and
> works fine only for large numbers of days, over several
> years):
>
> let $days := xs:dayTimeDuration("-P3278D") div
> xs:dayTimeDuration('P1D')
> let $months := $days idiv 30.43684991666667
> return
> $months * xs:yearMonthDuration('P1M')
>
> Regards,
>
> --
> Florent Georges
> http://fgeorges.org/
> http://h2oconsulting.be/
>
>
> On 20 February 2015 at 01:02, William Sawyer wrote:
> > I have two dates I am trying to subtract and get a
> xs:yearMonthDuration. In
> > "0.9-ml" there is a function you can call but in "1.0-ml" the
> documentation
> > says to subtract the dates. But how do I specify that I want a
> > xs:yearMonthDuration instead of a xs:dayTimeDuration.
> > MarkLogic Version: 7.0-2.3
> >
> > Thanks,
> > -Will
> >
> >
> > xquery version "1.0-ml";
> >
> > let $date1 := xs:dateTime(xs:date("2015-03-27"))
> > let $date2 := xs:dateTime(xs:date("2024-03-27"))
> > let $test := $date1 - $date2
> > return $test
> > ***************
> > xs:dayTimeDuration("-P3288D")
> >
> >
> > xquery version "0.9-ml"
> >
> > let $date1 := xs:dateTime(xs:date("2015-03-27"))
> > let $date2 := xs:dateTime(xs:date("2024-03-27"))
> > let $test := fn:subtract-dateTimes-yielding-yearMonthDuration($date1,
> > $date2)
> > return $test
> > **************
> > xs:yearMonthDuration("-P8Y11M")
> >
> > _______________________________________________
> > General mailing list
> > General@developer.marklogic.com
> > http://developer.marklogic.com/mailman/listinfo/general
> >
>
> --
> Florent Georges
> http://fgeorges.org/
> http://h2oconsulting.be/
> _______________________________________________
> General mailing list
> General@developer.marklogic.com
> http://developer.marklogic.com/mailman/listinfo/general
> _______________________________________________
> General mailing list
> General@developer.marklogic.com
> http://developer.marklogic.com/mailman/listinfo/general
>
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