作者：吴炳锡 来源：http://www.mysqlsupport.cn/ 联系方式： wubingxi#gmail.com 转载请注明作/译者和出处，并且不能用于商业用途，违者必究.
如下表：
CREATE TABLE `t1` (
`userid` int(11) DEFAULT NULL,
`atime` datetime DEFAULT NULL,
KEY `idx_userid` (`userid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8；
数据如下：
MySQL> select * from t1;
+--------+---------------------+
| userid | atime |
+--------+---------------------+
| 1 | 2013-08-12 11:05:25 |
| 2 | 2013-08-12 11:05:29 |
| 3 | 2013-08-12 11:05:32 |
| 5 | 2013-08-12 11:05:34 |
| 1 | 2013-08-12 11:05:40 |
| 2 | 2013-08-12 11:05:43 |
| 3 | 2013-08-12 11:05:48 |
| 5 | 2013-08-12 11:06:03 |
+--------+---------------------+
8 rows in set (0.00 sec)
其中userid不唯一，要求取表中每个userid对应的时间离现在最近的一条记录．初看到一个这条件一般都会想到借用临时表及添加主建借助于join操作之类的．
给一个简方法：
MySQL> select userid,substring_index(group_concat(atime order by atime desc),",",1) as atime from t1 group by userid;
+--------+---------------------+
| userid | atime |
+--------+---------------------+
| 1 | 2013-08-12 11:05:40 |
| 2 | 2013-08-12 11:05:43 |
| 3 | 2013-08-12 11:05:48 |
| 5 | 2013-08-12 11:06:03 |
+--------+---------------------+
4 rows in set (0.03 sec)
Good luck!
select userid,max(atime) from t1 group by userid having count(*)>1
这个例子不好, 因为可以直接用max()来完成…
(root@localhost) [test]> select max(atime) ,userid from t1 group by userid;
+———————+——–+
| max(atime) | userid |
+———————+——–+
| 2013-08-12 11:05:40 | 1 |
| 2013-08-12 11:05:43 | 2 |
| 2013-08-12 11:05:48 | 3 |
| 2013-08-12 11:06:03 | 5 |
+———————+——–+
@huarong, 顶．