Subject | Re: R: R: [ib-support] SUM( TIME FIELD ) |
---|---|
Author | Ann Harrison |
Post date | 2001-02-10T21:28:44Z |
At 10:03 PM 2/10/2001 +0100, Arnaldo wrote:
is the number of seconds between the two times. The
answer InterBase returns is 14400 (4 * 60 * 60)
create table timecard (
worker integer,
start_time time,
end_time time);
insert into timecard (worker, start_time, end_time)
values ('23', '08:00', '18:00');
select cast (((end_time - start_time) / 60 /60) as numeric (9,3)) as
hours_worked
from timecard
where worker = 23;
select sum (cast (((end_time - start_time) / 60 /60) as numeric (9,3))) as
hours_worked
from timecard
where worker = 23;
Ann
www.ibphoenix.com
We have answers.
> >This isn't right, in fact. The result of the subtraction
> > > > >I do:
> > > > >Select SUM( TIMEFIELD ), but cannot find a result.
> >
> > Time is different.
> >
> > 07:30 + 3 = 10:00
> >
> > A time plus a number produces a time. The "3" is
> > a number representing a duration of three hours.
> >
> > 12:00 - 8:00 = 4
is the number of seconds between the two times. The
answer InterBase returns is 14400 (4 * 60 * 60)
>I have already done what You write, but with no success.Here, let me give you an example:
>I'll try another time, with more luck, I hope.
create table timecard (
worker integer,
start_time time,
end_time time);
insert into timecard (worker, start_time, end_time)
values ('23', '08:00', '18:00');
select cast (((end_time - start_time) / 60 /60) as numeric (9,3)) as
hours_worked
from timecard
where worker = 23;
select sum (cast (((end_time - start_time) / 60 /60) as numeric (9,3))) as
hours_worked
from timecard
where worker = 23;
Ann
www.ibphoenix.com
We have answers.