mysql - 如何通过数据透视表选择具有一定关系的行？

Question

示例数据库设计

  USERS            WORKDAYS                DAYS
[id] [name]      [user_id] [day_id]      [id] [name]
 1    john        1         2             1    sunday
 2    fred        1         3             2    monday
 3    bert        1         4             3    tuesday
 4    harry       1         5             4    wednesday
                  1         6             5    thursday
                  2         2             6    friday
                  4         1             7    saturday
                  4         2
                  4         3
                  4         4
                  4         5
                  4         6
                  4         7

如何查询

1. 像 John 这样的人，他完全一周的所有工作日都在工作:周一、周二、周三、周四、周五
2. 有人像哈利和约翰那样至少一直在工作吗？

我有一段时间使用 mysql，但目前找不到解决方案。通常我会为此使用位标志，但我尝试掌握 SQL 中更规范化的解决方案。

CREATE TABLE `users` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(100) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

CREATE TABLE `days` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(100) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=8 DEFAULT CHARSET=utf8;

CREATE TABLE `workdays` (
  `user_id` int(11) NOT NULL,
  `day_id` int(11) NOT NULL,
  KEY `workdays_users_FK` (`user_id`),
  KEY `workdays_days_FK` (`day_id`),
  CONSTRAINT `workdays_days_FK` FOREIGN KEY (`day_id`) REFERENCES `days` (`id`),
  CONSTRAINT `workdays_users_FK` FOREIGN KEY (`user_id`) REFERENCES `users` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;



INSERT INTO play.users
(id, name)
VALUES(1, 'john');
INSERT INTO play.users
(id, name)
VALUES(2, 'fred');
INSERT INTO play.users
(id, name)
VALUES(3, 'bert');
INSERT INTO play.users
(id, name)
VALUES(4, 'harry');

INSERT INTO play.days
(id, name)
VALUES(1, 'sunday');
INSERT INTO play.days
(id, name)
VALUES(2, 'monday');
INSERT INTO play.days
(id, name)
VALUES(3, 'tuesday');
INSERT INTO play.days
(id, name)
VALUES(4, 'wednesday');
INSERT INTO play.days
(id, name)
VALUES(5, 'thursday');
INSERT INTO play.days
(id, name)
VALUES(6, 'friday');
INSERT INTO play.days
(id, name)
VALUES(7, 'saturday');

INSERT INTO play.workdays
(user_id, day_id)
VALUES(1, 2);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(1, 3);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(1, 4);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(1, 5);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(1, 6);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(2, 2);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(4, 1);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(4, 2);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(4, 3);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(4, 4);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(4, 5);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(4, 6);
INSERT INTO play.workdays
(user_id, day_id)
VALUES(4, 7);

Answer 1

• 您可以根据用户 ID 和名称GROUP BY。
• 使用 HAVING 子句和 SUM() 聚合来过滤案例。
• 如果用户在某一天不工作，那一天的 SUM() 将为零，post joins。

对于第一种情况(在一周的所有工作日完全工作)，尝试:

SELECT u.id, u.name
FROM USERS AS u 
JOIN WORKDAYS AS wd ON wd.user_id = u.id 
JOIN DAYS AS d ON d.id = wd.day_id 
GROUP BY u.id, u.name 
HAVING SUM(d.name = 'monday') 
   AND SUM(d.name = 'tuesday') 
   AND SUM(d.name = 'wednesday') 
   AND SUM(d.name = 'thursday') 
   AND SUM(d.name = 'friday') 
   AND SUM(d.name = 'sunday') = 0 
   AND SUM(d.name = 'saturday') = 0

对于第二种情况，只需删除星期日 和星期六 的条件。尝试:

SELECT u.id, u.name
FROM USERS AS u 
JOIN WORKDAYS AS wd ON wd.user_id = u.id 
JOIN DAYS AS d ON d.id = wd.day_id 
GROUP BY u.id, u.name 
HAVING SUM(d.name = 'monday') 
   AND SUM(d.name = 'tuesday') 
   AND SUM(d.name = 'wednesday') 
   AND SUM(d.name = 'thursday') 
   AND SUM(d.name = 'friday')