Re: Interval Question

Is is a "bare bones" copy of the function. The travel time is hard coded for
the example.

By the way, I'm on version 7.4.

On Tuesday 11 January 2005 10:10 am, Timothy Perrigo saith:
> Can you post the code for the function you are having trouble with?
> The following psql query works as expected (returns a negative
> interval):
>
> select '3 days 4 hours 17 mins'::interval - '3 days 6 hours 17
> mins'::interval;
> ?column?
> -----------
> -02:00:00
>
> Inserting the difference into a table with an interval column also
> seems to work:
>
> azrael=# create temp table interval_test(dif interval);
> CREATE TABLE
> azrael=# insert into interval_test select '3 days 4 hours 17
> mins'::interval - '3 days 6 hours 17 mins'::interval;
> INSERT 13615943 1
> azrael=# select * from interval_test;
> dif
> -----------
> -02:00:00
> (1 row)
>
> On Jan 11, 2005, at 8:34 AM, Terry Lee Tucker wrote:
> > Greetings:
> >
> > I am working on a function which returns an interval value. The work
> > of the
> > function is to calculate the difference between the appointment
> > timestamp and
> > and the current timestamp, represented as an interval, and the the time
> > required to travel from point A to B, represented as an interval.
> >
> > Appoint time: 01/14/2004 15:30
> > Current time: 01/11/2004 10:43
> > Appt Interval: @ 3 days 4 hours 17 mins
> >
> > Travel Time: 78 hours 17 minutes
> > Travel Interval: @ 3 days 6 hours 17 mins
> >
> > As you can see, this truck is going to be 2 hours late. The return
> > value I'm
> > looking for is the difference between Appt. Interval and Travel
> > Interval, as
> > in: return (appt_interval - travel_interval). This value will be
> > stored in a
> > column of type interval. I would like for late values to be shown as
> > negative. @ -2 hours. I thought that subtracting the larger interval
> > from
> > the small would return this but it is always the absolute value. The
> > documentation states:
> >
> > interval values can be written with the following syntax:
> >
> > [(at)] quantity unit [quantity unit...] [direction]
> >
> > Where: quantity is a number (possibly signed);
> >
> > Considering the above statement I believed that I could show this
> > difference
> > as a negative value but I haven't been able to figure out how to do
> > it. Can I
> > do this, and if so, how?
> >
> > Thanks...
> >
> > Work: 1-336-372-6812
> > Cell: 1-336-363-4719
> > email: terry(at)esc1(dot)com
> >
> > ---------------------------(end of
> > broadcast)---------------------------
> > TIP 8: explain analyze is your friend
>
> ---------------------------(end of broadcast)---------------------------
> TIP 7: don't forget to increase your free space map settings

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Work: 1-336-372-6812
Cell: 1-336-363-4719
email: terry(at)esc1(dot)com