Thread: query problem
Hi suppose I have the following situation: city date ---------+----------------------- London | 2002-08-08 07:05:16+00 London | 2002-07-30 13:08:22+00 London | 2002-07-30 07:39:15+00 London | 2002-07-29 17:51:47+00 London | 2002-07-29 17:45:49+00 London | 2002-07-29 17:45:47+00 Paris | 2002-04-08 15:04:28+00 Paris | 2002-03-29 17:22:18+00 Paris | 2002-02-15 12:50:32+00 Paris | 2002-01-22 11:40:22+00 Paris | 2002-01-07 17:41:23+00 Paris | 2001-11-12 16:37:37+00 Paris | 2001-11-05 15:28:23+00 Paris | 2001-11-05 08:21:19+00 Oslo | 2002-07-19 15:42:20+00 Oslo | 2002-07-19 15:42:18+00 Oslo | 2002-07-18 10:03:58+00 Oslo | 2002-07-18 08:56:30+00 Oslo | 2002-07-17 17:17:27+00 Oslo | 2002-07-17 16:11:38+00 For each city I have a couple of dates in DESC order. For each city i need to get the first record which date comes after a given date. If the given date was for example "2002-07-19 15:39:15+00", I would get the following records: London | 2002-07-29 17:45:47+00 Oslo | 2002-07-19 15:42:18+00 Is there a way to obtain this records by performing one single query and not by making for each city something like "SELECT city,date FROM table WHERE city='London' AND date>'2002-07-19 15:39:15+00' ORDER BY date ASC LIMIT 1;"? Thanks.
Marco,
> Is there a way to obtain this records by performing one
> single query and not by making for each city something like
> "SELECT city,date FROM table WHERE city='London' AND date>'2002-07-19
> 15:39:15+00' ORDER BY date ASC LIMIT 1;"?
Close. Try:
SELECT city, MAX("date") as last_date
FROM table
WHERE "date" > $date
GROUP BY city
ORDER BY city
Though as an aggregate query, this will be slow on large tables.
--
-Josh BerkusAglio Database SolutionsSan Francisco
How about: select city, min(date) from thetable where date > '2002-07-19 15:39:15+00' group by city; JLL Marco Muratori wrote: > > Hi > suppose I have the following situation: > > city date > ---------+----------------------- > London | 2002-08-08 07:05:16+00 > London | 2002-07-30 13:08:22+00 > London | 2002-07-30 07:39:15+00 > London | 2002-07-29 17:51:47+00 > London | 2002-07-29 17:45:49+00 > London | 2002-07-29 17:45:47+00 > Paris | 2002-04-08 15:04:28+00 > Paris | 2002-03-29 17:22:18+00 > Paris | 2002-02-15 12:50:32+00 > Paris | 2002-01-22 11:40:22+00 > Paris | 2002-01-07 17:41:23+00 > Paris | 2001-11-12 16:37:37+00 > Paris | 2001-11-05 15:28:23+00 > Paris | 2001-11-05 08:21:19+00 > Oslo | 2002-07-19 15:42:20+00 > Oslo | 2002-07-19 15:42:18+00 > Oslo | 2002-07-18 10:03:58+00 > Oslo | 2002-07-18 08:56:30+00 > Oslo | 2002-07-17 17:17:27+00 > Oslo | 2002-07-17 16:11:38+00 > > For each city I have a couple of dates in DESC order. > For each city i need to get the first record which date > comes after a given date. If the given date was for example > "2002-07-19 15:39:15+00", I would get the following > records: > > London | 2002-07-29 17:45:47+00 > Oslo | 2002-07-19 15:42:18+00 > > Is there a way to obtain this records by performing one > single query and not by making for each city something like > "SELECT city,date FROM table WHERE city='London' AND date>'2002-07-19 > 15:39:15+00' ORDER BY date ASC LIMIT 1;"? > Thanks. > > ---------------------------(end of broadcast)--------------------------- > TIP 6: Have you searched our list archives? > > http://archives.postgresql.org
I think you meant min(date)...
Josh Berkus wrote:
>
> Marco,
>
> > Is there a way to obtain this records by performing one
> > single query and not by making for each city something like
> > "SELECT city,date FROM table WHERE city='London' AND date>'2002-07-19
> > 15:39:15+00' ORDER BY date ASC LIMIT 1;"?
>
> Close. Try:
>
> SELECT city, MAX("date") as last_date
> FROM table
> WHERE "date" > $date
> GROUP BY city
> ORDER BY city
>
> Though as an aggregate query, this will be slow on large tables.
>
> --
> -Josh Berkus
> Aglio Database Solutions
> San Francisco
>
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