时间:2021-07-01 10:21:17 帮助过:13人阅读
CREATE TABLE `t1` ( `userid` int(11) DEFAULT NULL, `atime` datetime DEFAULT NULL, KEY `idx_userid` (`userid`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; 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) 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) 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!