Ann. This is what i need.

THANKS, for Your patience.

Ciao.

Arnaldo

----- Original Message -----
From: Ann Harrison <aharrison@...>
To: <ib-support@yahoogroups.com>
Sent: Saturday, February 10, 2001 10:28 PM
Subject: Re: R: R: [ib-support] SUM( TIME FIELD )


> At 10:03 PM 2/10/2001 +0100, Arnaldo wrote:
>
> > >
> > > > > >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
>
> This isn't right, in fact. The result of the subtraction
> 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.
> >I'll try another time, with more luck, I hope.
>
> Here, let me give you an example:
>
> 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.
>
>
>
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>