gpt4 book ai didi

mysql - 相同用户和相同状态的连续实例 - MYSQL

转载 作者:行者123 更新时间:2023-11-30 01:07:08 25 4
gpt4 key购买 nike

需要一个 MYSQL 查询来查找数据库中连续且具有相同 user_id 和 status_id 的所有事件。如何在避免循环的同时找到连续事件?

我遇到的问题是我不知道如何检查时间之间的休息时间。因此,如果某人在上午 10:00 至 10:30 之间有空,然后在上午 11:00 至中午 12:30 之间有空,如何防止系统只显示上午 10:00 至下午 12:30。

我们使用 PHP 和 Javascript 来创建日历,但希望有一个可以用于许多不同查询的 TableView 。

如果您有任何疑问,请告诉我。

CREATE TABLE `events` (
`event_id` int(11) NOT NULL auto_increment COMMENT 'ID',
`user_id` int(11) NOT NULL,
`date` date NOT NULL,
`time` time NOT NULL,
`status_id` int(11) default NULL,
PRIMARY KEY (`event_id`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=1568 ;


INSERT INTO `events` VALUES(1, 101, '2013-08-14', '23:00:00', 2);
INSERT INTO `events` VALUES(2, 101, '2013-08-14', '23:15:00', 2);
INSERT INTO `events` VALUES(3, 101, '2013-08-14', '23:30:00', 2);
INSERT INTO `events` VALUES(4, 101, '2013-08-14', '23:45:00', 2);
INSERT INTO `events` VALUES(5, 101, '2013-08-15', '00:00:00', 2);
INSERT INTO `events` VALUES(6, 101, '2013-08-15', '00:15:00', 1);
INSERT INTO `events` VALUES(7, 500, '2013-08-14', '23:45:00', 1);
INSERT INTO `events` VALUES(8, 500, '2013-08-15', '00:00:00', 1);
INSERT INTO `events` VALUES(9, 500, '2013-08-15', '00:15:00', 2);
INSERT INTO `events` VALUES(10, 500, '2013-08-15', '00:30:00', 2);
INSERT INTO `events` VALUES(11, 500, '2013-08-15', '00:45:00', 1);
INSERT INTO `events` VALUES(12, 101, '2013-08-15', '01:15:00', 1);
INSERT INTO `events` VALUES(13, 101, '2013-08-15', '01:30:00', 1);
INSERT INTO `events` VALUES(14, 101, '2013-08-15', '01:45:00', 1);
INSERT INTO `events` VALUES(15, 101, '2013-08-15', '02:00:00', 1);

期望的输出

row |user_id | date_start | time_start | date_end   | time_end | status_id | duration
1 |101 |'2013-08-14'| '23:00:00' |'2013-08-15'|'00:15:00'| 2 | 5
2 |101 |'2013-08-15'| '00:15:00' |'2013-08-15'|'00:30:00'| 1 | 1
3 |500 |'2013-08-14'| '23:45:00' |'2013-08-15'|'00:15:00'| 1 | 2
4 |500 |'2013-08-15'| '00:15:00' |'2013-08-15'|'00:45:00'| 2 | 2
5 |500 |'2013-08-15'| '00:45:00' |'2013-08-15'|'01:00:00'| 2 | 1
6 |101 |'2013-08-15'| '01:15:00' |'2013-08-15'|'02:15:00'| 1 | 4

最佳答案

所以我能够创建一个解决方案......大概。请看一下并找到可以优化的方法!

SELECT user1, user2, status_id, min(Timestamp1), Timestamp2, max(duration) FROM (
SELECT x.user1, x.user2, x.status_id, x.Timestamp1, x.Timestamp2, min(x.difference) as duration
FROM (

SELECT e1.user_id AS user1, e2.user_id AS user2, e1.status_id, CONCAT(e1.date, " ", e1.time) AS Timestamp1, CONCAT(e2.date, " ", e2.time)AS Timestamp2,
CAST(TIMEDIFF(CONCAT(e2.date, " ", e2.time),CONCAT(e1.date, " ", e1.time)) AS SIGNED) AS 'difference'
FROM `events` e1, `events` e2
WHERE e1.user_id = e2.user_id AND e1.status_id = e2.status_id
AND CAST(TIMEDIFF(CONCAT(e2.date, " ", e2.time),CONCAT(e1.date, " ", e1.time)) AS SIGNED) >= 0
AND CAST(TIMEDIFF(CONCAT(e2.date, " ", e2.time),CONCAT(e1.date, " ", e1.time)) AS SIGNED) < 240000
ORDER BY e1.user_id, e1.status_id, Timestamp1, Timestamp2) x,
(

SELECT e1.user_id AS user1, e2.user_id AS user2, e1.status_id, CONCAT(e1.date, " ", e1.time) AS Timestamp1, CONCAT(e2.date, " ", e2.time)AS Timestamp2,
CAST(TIMEDIFF(CONCAT(e2.date, " ", e2.time),CONCAT(e1.date, " ", e1.time)) AS SIGNED) AS 'difference'
FROM `events` e1, `events` e2
WHERE e1.user_id = e2.user_id AND e1.status_id = e2.status_id
AND CAST(TIMEDIFF(CONCAT(e2.date, " ", e2.time),CONCAT(e1.date, " ", e1.time)) AS SIGNED) >= 0
ORDER BY e1.user_id, e1.status_id, Timestamp1, Timestamp2) y

WHERE x.user1 = y.user1 AND x.user2 = y.user2 AND x.Timestamp1 = y.Timestamp1
AND ADDTIME(x.Timestamp2, '00:15:00') NOT IN (
SELECT CONCAT(e2.date, " ", e2.time)AS Timestamp2
FROM `events` e1, `events` e2
WHERE e1.user_id = e2.user_id AND e1.status_id = e2.status_id
AND x.user1 = e1.user_id AND x.status_id = e1.status_id
AND x.Timestamp1 = CONCAT(e1.date, " ", e1.time)
AND CAST(TIMEDIFF(CONCAT(e2.date, " ", e2.time),CONCAT(e1.date, " ", e1.time)) AS SIGNED) >= 0)

GROUP BY x.user1, x.user2, x.status_id, x.Timestamp1
) z
GROUP BY user1, user2, status_id, Timestamp2

以下是它提供的结果:

用户1用户2状态最小值(时间戳1)时间戳2最大值(持续时间)

101 101 1 2013-08-15 00:15:00 2013-08-15 00:15:00 0

101 101 1 2013-08-15 01:15:00 2013-08-15 02:00:00 4500

101 101 2 2013-08-14 23:00:00 2013-08-15 00:00:00 10000

500 500 1 2013-08-14 23:45:00 2013-08-15 00:00:00 1500

500 500 1 2013-08-15 00:45:00 2013-08-15 00:45:00 0

500 500 2 2013-08-15 00:15:00 2013-08-15 00:30:00 1500

关于mysql - 相同用户和相同状态的连续实例 - MYSQL,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19669573/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com