For advanced computations, it’s sometimes easier to convert duration
items to milliseconds with our custom functions [1]:
let $ms-years := 1000 * 60 * 60 * 24 * 365
let $dur := xs:dayTimeDuration('P12345DT2M3S')
let $ms := convert:dayTime-to-integer($dur)
return $ms idiv $ms-years
[1]
I was going to ask if I could count on the difference always being in a
PxxxD format. I am happy to see that days-from-duration doesn't return 0. I
tried casting the xs:duration to many formats and kept getting 0.
Thanks for the tips
On Fri, Dec 6, 2019 at 11:38 AM Christian Grün
wrote:
> Hi
Hi France,
The function fn:years-from-duration is fairly basic; it only returns
the year component of your duration argument (as xs:dayTimeDuration
has none, in contrast to xs:yearMonthDuration). The background: A year
may have 365 or 366 days, and the duration is not sufficient to
compute the
Am 06.12.2019 um 11:32 schrieb Martin Honnen:
Am 06.12.2019 um 11:23 schrieb France Baril:
The following function returns P4511D: 0 but I expect P4511D: 12
Bug or bad usage?
let $d1 := xs:date('2007-07-31')
let $d2 := xs:date('2019-12-06')
let $duration := $d2 - $d1
let
Am 06.12.2019 um 11:23 schrieb France Baril:
The following function returns P4511D: 0 but I expect P4511D: 12
Bug or bad usage?
let $d1 := xs:date('2007-07-31')
let $d2 := xs:date('2019-12-06')
let $duration := $d2 - $d1
let $years := years-from-duration($duration)
return
Hi,
The following function returns P4511D: 0 but I expect P4511D: 12
Bug or bad usage?
let $d1 := xs:date('2007-07-31')
> let $d2 := xs:date('2019-12-06')
> let $duration := $d2 - $d1
> let $years := years-from-duration($duration)
> return $duration || ': ' || $years
--
France Baril
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