获取MySQL的表中每个userid最后一条记录的方法

2024-07-24 12:46:07

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如下表：

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!