A few days ago I responded to:
>>if a date field is created with legacy tools, can it be used it as a datetime?
>>And an hour field?
>Apparently so:
>
>C_DATE($dOutput)
>C_TIME($dtOutput)
>
> // Misc is a table created using 4D legacy tool (structure editor)
>
>$sKey:="DateTime Test"
>Begin SQL
> insert into Misc (sKey, dDate) VALUES (:$sKey, CURRENT_TIMESTAMP());
>
> select dDate from Misc where sKey = :$sKey into :$dOutput -- yields the
> date part: 08/31/16 End SQL
>Begin SQL
> select dDate from Misc where sKey = :$sKey into :$dtOutput -- yields the
> time part: 07:24:00 End SQL
>So my statement about needing to create the field with SQL commands is not
>correct. If you fill a date field using SQL time >stamp 4D will store both the
>date and time parts either of which may be retrieved using SQL, no matter how
>the date field >was created.
I neglected to mention that once you add the time value to a 4D field (as
described above) you can no longer use the equality comparator (=) in queries
when you are querying with only a date.
If you run the above code to fill the date field with a datetime value the
following query will fail:
QUERY([Misc];[Misc]dDate=CURRENT DATE)
That is because 4D is including both the date and the time in the query and
CURRENT date does include the time part.
So be careful if you start including time in your 4D date fields.
Tom Benedict
Optum Inc
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