MySQL select 在特定规则的行之后

Question

我有一个 table 这样的：

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', 'away', '2021-01-02 04:08:00'),
  ('10', 'break', '2021-01-02 04:10:00'),
  ('11', 'online', '2021-01-02 04:15:00'),
  ('12', 'logout', '2021-01-02 04:55:00'),
  ('13', 'login', '2021-01-02 05:04:00'),
  ('14', 'online', '2021-01-02 05:05:00'),
  ('15', 'away', '2021-01-03 05:01:00'),
  ('16', 'logout', '2021-01-03 05:02:00'),
  ('17', 'login', '2021-01-03 05:04:00'),
  ('18', 'online', '2021-01-03 05:05:00'),
  ('19', 'logout', '2021-01-04 03:05:00'),
  ('20', 'login', '2021-01-04 05:07:00'),
  ('21', 'online', '2021-01-04 06:00:00'),
  ('22', 'logout', '2021-01-05 10:00:00'),
  ('23', 'login', '2021-01-05 11:00:00'),
  ('24', 'away', '2021-01-05 11:01:00'),
  ('25', 'online', '2021-01-06 06:01:00'),
  ('26', 'login', '2021-01-07 06:01:00'),
  ('26', 'logout', '2021-01-07 07:01:00');

id	status	timestamp
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

......

在上面的插入查询中提供的数据。

我想要一个输出：

date	A (online)	B (Logout)
2021-01-01	2021-01-02 04:07:00	2021-01-02 04:55:00
2021-01-02	2021-01-02 05:05:00	2021-01-03 04:59:59
2021-01-03	2021-01-03 05:05:00	2021-01-04 03:05:00
2021-01-04	2021-01-04 06:00:00	2021-01-04 04:59:59
2021-01-05	2021-01-04 05:00:00	2021-01-05 10:00:00
2021-01-06	2021-01-04 11:00:00	2021-01-06 04:59:59

规则是，1 个记录日是从 5:00:00 -（次日）04:59:59。 'A'是从上次上线开始的时间戳（登录后，如果有的话），'B'是从上次注销开始的时间戳（如果没有注销，B设置为04:59:59。另一个规则是，如果最后一天在'online'之后没有'logout'，则计算到下一个日志日（下一个日志日设置为05.00.00，如果最后一条记录来自最后记录日是 'online')

目前我正在使用此查询来应用记录日规则：

SELECT date(t1.timestamp) dt, (t2.timestamp) A, (t3.timestamp) B, t1.status, t2.status, t3.status
FROM logging t1
JOIN logging t2 ON t1.timestamp < t2.timestamp
JOIN logging t3 ON t2.timestamp < t3.timestamp
WHERE 
t1.status = 'login'
AND t2.status = 'online'
AND t3.status = 'logout'
AND NOT EXISTS ( SELECT NULL
               FROM loggingt4
               WHERE t1.timestamp < t4.timestamp
                 AND t4.timestamp < t2.timestamp
                 AND t4.status IN ('login', 'online', 'logout') )
AND NOT EXISTS ( SELECT NULL
               FROM logging t5
               WHERE t2.timestamp < t5.timestamp
                 AND t5.timestamp < t3.timestamp
                 AND t5.status IN ('login', 'logout'))
AND DATE(t1.timestamp - INTERVAL 6 HOUR) = DATE(t3.timestamp - INTERVAL '05:59:59' HOUR_SECOND);

Answer 1

检查这个：

WITH cte AS (
SELECT DATE(t1.`timestamp` - INTERVAL 5 HOUR) `date`,
       MAX(t1.`timestamp`) login, 
       MAX(t2.`timestamp`) online, 
       MAX(t3.`timestamp`) logout
FROM logging t1
JOIN logging t2 ON t1.`timestamp` < t2.`timestamp`
JOIN logging t3 ON t2.`timestamp` < t3.`timestamp`
WHERE t1.status = 'login'
  AND t2.status = 'online'
  AND t3.status = 'logout'
  AND NOT EXISTS ( SELECT NULL
                   FROM logging t4
                   WHERE t1.`timestamp` < t4.`timestamp`
                     AND t4.`timestamp` < t2.`timestamp`
                     AND t4.status IN ('login', 'online', 'logout') )
  AND NOT EXISTS ( SELECT NULL
                   FROM logging t5
                   WHERE t2.`timestamp` < t5.`timestamp`
                     AND t5.`timestamp` < t3.`timestamp`
                     AND t5.status IN ('login', 'logout') )
GROUP BY `date`
)
SELECT `date`,
       online,
       CASE WHEN DATE(online - INTERVAL 5 HOUR) = DATE(logout - INTERVAL '04:59:59' HOUR_SECOND)
            THEN logout
            ELSE DATE(online + INTERVAL 19 HOUR) + INTERVAL '04:59:59' HOUR_SECOND
            END logout
FROM cte

fiddle

CTE 仅用于可见性 - 您可以将所有内容合并到一个查询中。

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