Question

我有一张这样的桌子：

CREATE TABLE IF NOT EXISTS `logging` (
  `id` int(6) unsigned NOT NULL,
  `status` varchar(150) NOT NULL,
  `timestamp` DATETIME NOT NULL,
  PRIMARY KEY (`id`)
) DEFAULT CHARSET=utf8;
INSERT INTO `logging` (`id`, `status`, `timestamp`) VALUES
  ('1', 'logout', '2021-01-01 05:01:00'),
  ('2', 'login', '2021-01-01 06:02:00'),
  ('3', 'online', '2021-01-01 06:03:00'),
  ('4', 'away', '2021-01-01 06:04:00'),
  ('5', 'online', '2021-01-01 06:05:00'),
  ('6', 'logout', '2021-01-02 04:00:00'),
  ('7', 'login', '2021-01-02 04:05:00'),
  ('8', 'online', '2021-01-02 04:07:00'),
  ('9', 'logout', '2021-01-02 04:55:00');

<头>

id	状态	时间戳
1	注销	2021-01-01 05:01:00
2	登录	2021-01-01 06:02:00
3	在线	2021-01-01 06:03:00
4	离开	2021-01-01 06:04:00
5	在线	2021-01-01 06:05:00
6	注销	2021-01-02 04:00:00
7	登录	2021-01-02 04:05:00
8	在线	2021-01-02 04:07:00
9	注销	2021-01-02 04:55:00

我想要一个输出：

<头>

日期	A	B
2021-01-01	2021-01-01 06:03:00	2021-01-02 04:00:00
2021-01-02	...	...
2021-01-03	...	...

规则是，1 个日志日是从 5:00:00 -（第二天）04:49:49。 'A'是从第一次在线（登录后）开始的时间戳，'B'是从第一次退出（在线后）开始的时间戳。

目前我正在使用此查询来应用日志日规则：

select dt, timestamp from ( 
    select id, left(timestamp, 10) dt, timestamp, status
    from logging
    where TIME(timestamp)  >= '05:00:00' and status = 'online'
    union all 
    select id, left((timestamp-interval 1 day), 10) dt ,timestamp, status
    from logging
    where TIME(timestamp)  < '05:00:00' and status = 'online'
    ) x group by dt;

但是我在应用第二条规则来获取时间戳时遇到问题..

更新：这是测试数据和表格：https://www.db-fiddle.com/f/tDJL2JHwbtSNEurrLEj11r/0 或http://sqlfiddle.com/#!9/aadb80/2

Answer 1

SELECT t1.ts A, t3.ts B
FROM test t1
JOIN test t2 ON t1.ts < t2.ts
JOIN test t3 ON t2.ts < t3.ts
WHERE t1.status = 'login'
  AND t2.status = 'online'
  AND t3.status = 'logout'
  AND NOT EXISTS ( SELECT NULL
                   FROM test t4
                   WHERE t1.ts < t4.ts
                     AND t4.ts < t2.ts
                     AND t4.status IN ('login', 'online', 'logout') )
  AND NOT EXISTS ( SELECT NULL
                   FROM test t5
                   WHERE t2.ts < t5.ts
                     AND t5.ts < t3.ts
                     AND t5.status IN ('login', 'logout') )
  AND DATE(t1.ts - INTERVAL 5 HOUR) = DATE(t3.ts - INTERVAL '04:49:49' HOUR_SECOND);

fiddle

从同一个表中选择每个条件后的行 myql

1 个答案: