Question

我有表entrances，用于记录用户进入房间并走出房间的时间。这样的事情：

user   |   action    |    time
-------------------------------------------
Ivan   |   in        |  2016-08-28 12:00:00 
John   |   in        |  2016-08-28 12:00:01
Ann    |   in        |  2016-08-28 12:00:02
Ivan   |   out       |  2016-08-28 12:00:03
Ivan   |   in        |  2016-08-28 12:00:04
Ann    |   out       |  2016-08-28 12:00:05
Ivan   |   out       |  2016-08-28 12:00:06
Mike   |   in        |  2016-08-28 12:00:07
John   |   out       |  2016-08-28 12:00:08
Ann    |   out       |  2016-08-18 12:00:09
John   |   in        |  2016-08-18 12:00:10
John   |   out       |  2016-08-18 12:00:11
Ann    |   in        |  2016-08-18 12:00:12

用户操作是独立的。只知道第一个操作总是in，用户无法in两次out（并反转）。

我的目标是找到目前在场的所有用户。

我有两个想法：

选择最新out

in

选择计数in的计数为out

如何在mysql上实现这个？还是其他任何想法？

用于测试的SQL：

CREATE TABLE `entrances` (
  `id` int(11) NOT NULL,
  `user` varchar(10) COLLATE utf8_bin NOT NULL,
  `action` varchar(3) COLLATE utf8_bin NOT NULL,
  `time` datetime NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_bin;
INSERT INTO `entrances` (`id`, `user`, `action`, `time`) VALUES
(1, 'Ivan', 'in', '2016-08-28 12:00:00'),
(2, 'John', 'in', '2016-08-28 12:00:01'),
(3, 'Ann', 'in', '2016-08-28 12:00:02'),
(4, 'Ivan', 'out', '2016-08-28 12:00:03'),
(5, 'Ivan', 'in', '2016-08-28 12:00:04'),
(6, 'Ann', 'out', '2016-08-28 12:00:05'),
(7, 'Ivan', 'out', '2016-08-28 12:00:06'),
(8, 'Mike', 'in', '2016-08-28 12:00:07'),
(9, 'John', 'out', '2016-08-28 12:00:08'),
(10, 'Ann', 'out', '2016-08-28 12:00:09'),
(11, 'John', 'in', '2016-08-28 12:00:10'),
(12, 'John', 'out', '2016-08-28 12:00:11'),
(13, 'Ann', 'in', '2016-08-28 12:00:12');
ALTER TABLE `entrances` ADD PRIMARY KEY (`id`);

Answer 1

首先从子查询

time

user

然后加入该子查询，只包含每个user
然后只记录那些action = in带where子句

喜欢这个

select e.*
from entrances e
join
(
   select user, max(time) as mtime
   from entrances
   group by user
) t on t.user = e.user 
   and t.mtime = e.time
where e.action = 'in'

Answer 2

使用操作<script src="//cdnjs.cloudflare.com/ajax/libs/moment.js/2.14.1/moment.min.js"></script>加入所有行，并选择没有out操作或out操作更新

的行

in

Answer 3

我会尝试这样的事情，但是测试真实数据会更好。

select user from entrances e inner join 
(select user, count(*) cnt from entrances where action='in') as e_in 
on e.user=e_in.user left join 
(select user, count(*) cnt from entrances where action='out') as e_out 
on e.user=e_out.user
where e_out.user is null or (e_in.cnt - e_out.cnt) = 1

Answer 4

我的想法1的Mysql实现：

SELECT e.* FROM entrances e
WHERE NOT EXISTS (
    SELECT * FROM entrances WHERE e.user=user AND action="out" AND time > (
        SELECT MAX(time) FROM entrances WHERE e.user=user AND action="in"
    )
)

这按照字面意思书写：找到在out之后没有in的用户。利用NOT EXISTS功能。优点是可读性，没有魔法，简单的algorythm。

Mysql：如何找到目前在房间里的用户？

4 个答案: