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.
>
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