Hi Bart,
If you store a date in an Object as a string, 4D interprets that date as a UTF
date.
So “2019-02-24” (giving only the date but not time) is interpreted as midnight
on the 24th. That is at the stroke of midnight when the 24th day **starts**.
When you retrieve the value around 6:30 PST on the 24th, 4D is converting to
your local time — which is several hours **before** the 24th begins. Hence you
see the 23rd.
If you pass a 4D date, the problem won’t happen. That is replace the string
("2019-02-24”) with a 4D date (#2019-02-24#).
There’s also a compatibility setting that where you can specify that string
dates are to be treated as 4D dates. That is, no time associated with the date.
Then your expression:
> OB SET($obj;"apiDate";"2019-02-24")
would always return the 24th
Best regards,
Jeremy
> On Feb 24, 2019, at 9:24 PM, Bart Davis via 4D_Tech <4d_tech@lists.4d.com>
> wrote:
>
> If you store a 4D date value in an object and retrieve it the value is
> correct. However, if you retrieve a date that is stored in the format
> YYYY-MM-DD, the date returned is one day less that the date stored.
>
> The code below illustrates the issue. I have only tested this on Mac OS
> using 4D v17.1 (Note: I tested this at 6:30 pm PST and am not sure if GMT
> offset has any effect)
>
> Has anyone else experienced this issue?
>
> Thanks,
> Bart
>
> C_DATE($4dDate;$apiDate)
> C_OBJECT($obj)
>
> $obj:=New object
> $today:=Current date
>
> OB SET($obj;"4Ddate";$today)
> OB SET($obj;"apiDate";"2019-02-24")
>
> $4dDate:=ob get($obj;"4Ddate";Is date) // date returned is correct 2019-02-24
> $apiDate:=ob get($obj;"apiDate";Is date) // ERROR: date returned is 2019-02-23
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