創建和插入件從@Incidently借:
CREATE TABLE T
(id int, status int, conversation_id int, message_id int, date_created datetime);
insert into T (id, status, conversation_id, message_id, date_created)
values(1, 1, 1, 72, '2012-01-01 00:00:00');
insert into T (id, status, conversation_id, message_id, date_created)
values(2, 2, 1, 87, '2012-03-03 00:00:00');
insert into T (id, status, conversation_id, message_id, date_created)
values(3, 2 , 2 , 95 , '2012-05-05 00:00:00');
假設的ID是主鍵然後這個SOLV ES的問題,通過@Incidentally中指出@喬的回答是:
select T.*
from
T
inner join (
select conversation_id, max(id) as id
from T
group by conversation_id
) s on s.id = T.id
order by T.date_created desc, T.conversation_id
;
+------+--------+-----------------+------------+---------------------+
| id | status | conversation_id | message_id | date_created |
+------+--------+-----------------+------------+---------------------+
| 3 | 2 | 2 | 95 | 2012-05-05 00:00:00 |
| 2 | 2 | 1 | 87 | 2012-03-03 00:00:00 |
+------+--------+-----------------+------------+---------------------+
人常常使用這個連接與 - GROUP_BY和-MAX模式,但不幸的是,這是錯誤的。如果有幾個具有相同conversation_id和date_created的記錄 - 它將全部返回。請參閱http://sqlfiddle.com/#!2/6f3cb/3 – Incidently