在previous question关于“在正常表格表中连续N天计数”我在下面得到了这个答案，工作正常。 我意识到我需要跳过计算周末，例如：如果用户在周五和下周一开会，那么这应该算作连续2天而不是因为周末而停止。
以下是解释我现在所拥有的代码。
DROP TABLE IF EXISTS meetings;
CREATE TABLE IF NOT EXISTS meetings
( meeting_id int(10) unsigned NOT NULL AUTO_INCREMENT
, meeting_time datetime NOT NULL
, PRIMARY KEY (meeting_id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
DROP TABLE IF EXISTS meetings_users;
CREATE TABLE IF NOT EXISTS meetings_users
( user_id int(10) unsigned NOT NULL
, meeting_id int(10) unsigned NOT NULL
, PRIMARY KEY (meeting_id,user_id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
DROP TABLE IF EXISTS users;
CREATE TABLE IF NOT EXISTS users
( user_id int(10) unsigned NOT NULL AUTO_INCREMENT
, PRIMARY KEY (user_id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO users ( user_id ) VALUES (1),(2),(3),(4);
INSERT INTO meetings ( meeting_id, meeting_time ) VALUES
(1, '2013-01-14 10:00:00'),
(2, '2013-01-15 10:00:00'),
(3, '2013-01-16 10:00:00'),
(4, '2013-01-17 10:00:00'),
(5, '2013-01-18 10:00:00'),
(6, '2013-01-19 10:00:00'),
(7, '2013-01-20 10:00:00'),
(8, '2013-01-11 10:00:00');
INSERT INTO meetings_users (meeting_id, user_id ) VALUES
(1, 1),
(2, 1),
(2, 3),
(3, 1),
(3, 3),
(4, 2),
(4, 3),
(5, 2),
(6, 1),
(8, 1);
SET @dt = '2013-01-15';
SELECT user_id
, start
, DATEDIFF(@dt,start)+1 cons
FROM
(
SELECT a.user_id
, a.meeting_date Start
, MIN(c.meeting_date) End
, DATEDIFF(MIN(c.meeting_date),a.meeting_date) + 1 diff
FROM (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) a
LEFT
JOIN (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) b
ON b.user_id = a.user_id
AND a.meeting_date = b.meeting_date + INTERVAL 1 DAY
LEFT
JOIN (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) c
ON c.user_id = a.user_id
AND a.meeting_date <= c.meeting_date
LEFT
JOIN (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) d
ON d.user_id = a.user_id
AND c.meeting_date = d.meeting_date - INTERVAL 1 DAY
WHERE b.meeting_date IS NULL
AND c.meeting_date IS NOT NULL
AND d.meeting_date IS NULL
GROUP
BY a.user_id
, a.meeting_date
) x
WHERE @dt BETWEEN start AND end;
+---------+------------+------+
| user_id | start | cons |
+---------+------------+------+
| 1 | 2013-01-14 | 2 |
| 3 | 2013-01-15 | 1 |
+---------+------------+------+
我尝试创建自己的函数来替换原始的DATEDIFF，以便根据this回答（我将其命名为CustomDateDiff）跳过周末并且工作得非常好，但我认为{{1}上的问题仍然存在我猜我应该跳过a.meeting_date = b.meeting_date + INTERVAL 1 DAY。
答案 0 :(得分：2)
好的，从here抓取一些代码，我修改了之前对@dt中的因素的反应......
@dt = '2013-01-15';
SELECT *
, @dt target
, 5 * (DATEDIFF(@dt, start) DIV 7) + MID('0123444401233334012222340111123400001234000123440', 7 * WEEKDAY(start) + WEEKDAY(@dt) + 1, 1)+1 cons
FROM (
SELECT x.user_id
, MIN(x.start) start
, COALESCE(y.end,x.end) end
, MAX(COALESCE(y.diff,0)+x.diff) ttl
FROM
(
SELECT a.user_id
, a.meeting_date Start
, MIN(c.meeting_date) End
, DATEDIFF(MIN(c.meeting_date),a.meeting_date) + 1 diff
FROM (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) a
LEFT
JOIN (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) b
ON b.user_id = a.user_id
AND a.meeting_date = b.meeting_date + INTERVAL 1 DAY
LEFT
JOIN (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) c
ON c.user_id = a.user_id
AND a.meeting_date <= c.meeting_date
LEFT
JOIN (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) d
ON d.user_id = a.user_id
AND c.meeting_date = d.meeting_date - INTERVAL 1 DAY
WHERE b.meeting_date IS NULL
AND c.meeting_date IS NOT NULL
AND d.meeting_date IS NULL
GROUP
BY a.user_id
, a.meeting_date
)x
LEFT JOIN
(
SELECT a.user_id
, a.meeting_date Start
, MIN(c.meeting_date) End
, DATEDIFF(MIN(c.meeting_date),a.meeting_date) + 1 diff
FROM (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) a
LEFT
JOIN (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) b
ON b.user_id = a.user_id
AND a.meeting_date = b.meeting_date + INTERVAL 1 DAY
LEFT
JOIN (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) c
ON c.user_id = a.user_id
AND a.meeting_date <= c.meeting_date
LEFT
JOIN (SELECT DISTINCT mu.user_id,DATE(m.meeting_time) meeting_date FROM meetings_users mu JOIN meetings m ON m.meeting_id = mu.meeting_id) d
ON d.user_id = a.user_id
AND c.meeting_date = d.meeting_date - INTERVAL 1 DAY
WHERE b.meeting_date IS NULL
AND c.meeting_date IS NOT NULL
AND d.meeting_date IS NULL
GROUP
BY a.user_id
, a.meeting_date
)y
ON y.user_id = x.user_id
AND y.start = x.end+INTERVAL 3 DAY AND WEEKDAY(x.end) = 4
GROUP BY user_id,end
)j
WHERE @dt BETWEEN start AND end;