Re: Time problem again?

On Tuesday 30 September 2003 09:50, Bjørn T Johansen wrote:
> On Tue, 2003-09-30 at 10:42, Richard Huxton wrote:
> > So - your rule is something like:
> >
> > For some block of times...
> > IF max(my_time) - min(my_time) > 12 hours
> > THEN sort "through midnight"
> > ELSE sort "naturally"
> >
> > Which is what Tom said (I was wondering where he got his 12 hours from).
> > Have I got that right?
>
> Yes, that sounds about right.. :)
> But how do I code this in an Select statement?

Well, I'd write a function (notice the double-quoting):

-- sort_times(TARGET-TIME, DIFFERENCE)
-- Takes a target time and the difference max(t)-min(t) in its group
-- Returns a timestamp you can sort on
--
CREATE FUNCTION sort_times(time, interval)
RETURNS timestamptz AS '
SELECT
CASE
WHEN $2 > ''12:00''::interval AND $1<=''12:00:00''::time
THEN ''1970-01-02 00:00:00+00''::timestamptz + $1
ELSE ''1970-01-01 00:00:00+00''::timestamptz + $1
END
' LANGUAGE 'SQL' IMMUTABLE;

Then you have the wrong way:

SELECT
id, grp, ts
FROM
timetest
ORDER BY
grp, ts
;

id | grp | ts
----+-----+----------
1 | a | 11:00:00
2 | a | 14:00:00
3 | a | 17:00:00
4 | a | 20:00:00
7 | b | 01:00:00 ***
8 | b | 04:00:00 *** Oops - these are not
5 | b | 20:00:00 *** what we wanted
6 | b | 22:00:00 ***
9 | c | 03:00:00
10 | c | 06:00:00
11 | c | 08:00:00
(11 rows)

And the right way:
SELECT
t.id, t.grp, t.ts
FROM
timetest t,
(SELECT grp, max(ts)-min(ts) as tdiff FROM timetest GROUP BY grp) AS diffs
WHERE
t.grp = diffs.grp
ORDER BY
t.grp, sort_times(t.ts, diffs.tdiff)
;

id | grp | ts
----+-----+----------
1 | a | 11:00:00
2 | a | 14:00:00
3 | a | 17:00:00
4 | a | 20:00:00
5 | b | 20:00:00 ***
6 | b | 22:00:00 *** Ah - better!
7 | b | 01:00:00 ***
8 | b | 04:00:00 ***
9 | c | 03:00:00
10 | c | 06:00:00
11 | c | 08:00:00
(11 rows)

I'm not sure how PG will optimise the correctly sorted one - you'll have to
try it on your real data and see.

--
Richard Huxton
Archonet Ltd