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获取MySQL的表中每个userid最后一条记录的方法

来自:种豆 时间:2018-01-12 阅读:646次 原文链接

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!

 
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