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 ---------------------------(end of broadcast)--------------------------- TIP 2: you can get off all lists at once with the unregister command (send "unregister YourEmailAddressHere" to [EMAIL PROTECTED])