Question

我有一个数据表如下

id    status    conversation_id    message_id    date_created
1     1         1                  72            2012-01-01 00:00:00
2     2         1                  87            2012-03-03 00:00:00
3     2         2                  95            2012-05-05 00:00:00

我希望以date_created DESC顺序从表中获取所有行，但每conversation_id只能获得一行。因此，对于上面的示例数据，我希望得到id 2和3的行。

非常感谢任何建议。

Answer 1

SELECT t.id, t.status, t.conversation_id, t.message_id, t.date_created
    FROM YourTable t
        INNER JOIN (SELECT conversation_id, MAX(date_created) AS MaxDate
                        FROM YourTable
                        GROUP BY conversation_id) q
            ON t.conversation_id = q.conversation_id
                AND t.date_created = q.MaxDate
    ORDER BY t.date_created DESC;

Answer 2

请参阅SQL Fiddle

SELECT T.*
FROM T
WHERE NOT EXISTS (
  SELECT * 
  FROM T AS _T
  WHERE _T.conversation_id = T.conversation_id
  AND (
    _T.date_created > T.date_created
    OR
    _T.date_created = T.date_created AND _T.id > T.id) 
)
ORDER BY T.date_created DESC

得

ID      STATUS  CONVERSATION_ID   MESSAGE_ID    DATE_CREATED
3         2         2                95         May, 05 2012 
2         2         1                87         March, 03 2012

Answer 3

从@Incidently借用的创建和插入：

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');

假设id是主键，那么这解决了@ Joe的答案中@Incidentally指出的问题：

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 |
+------+--------+-----------------+------------+---------------------+

SQL - 从表中获取最新记录，其中字段是唯一的

3 个答案: