Question

我有一个表，可以存储特定用户/过滤器/类型的每天每小时的是/否值。每一天都是自己的一排。因此，对于任何给定的用户/过滤器/类型组合，总会有7行。

我想要完成的是每个用户/过滤器/类型组合的一个结果，其中包含一周中每天的所有小时数。我认为我在这里需要的方法是自己加入别名和组，但我尝试的一切都失败了。我在fiddle

设置了一个基本的小提琴

如果在数据库端有一个更简单和/或更优选的方法来处理这个问题，我也能够为此更改数据库模式（我的直觉说有）。也许是由filter_id链接的一周中每一天的表？

INSERT INTO filters
 (`filter_id`, `user_id`, `filter`, `type`, `day`, `12a`, `1a`... and so on for each hour)
VALUES
(1, 1, 'filter1', 1, 1, 1, 1),
(2, 1, 'filter1', 1, 2, 1, 1),
(3, 1, 'filter1', 1, 3, 1, 1),
(4, 1, 'filter1', 1, 4, 1, 1),
(5, 1, 'filter1', 1, 5, 1, 1),
(6, 1, 'filter1', 1, 6, 1, 1),
(7, 1, 'filter1', 1, 7, 1, 1),
(8, 1, 'filter2', 1, 1, 0, 0),
(9, 1, 'filter2', 1, 2, 0, 0),
(10, 1, 'filter2', 1, 3, 0, 0),
(11, 1, 'filter2', 1, 4, 0, 0),
(12, 1, 'filter2', 1, 5, 0, 0),
(13, 1, 'filter2', 1, 6, 0, 0),
(14, 1, 'filter2', 1, 7, 0, 0),
(15, 1, 'filter3', 1, 1, 0, 0),
(16, 1, 'filter3', 1, 2, 0, 0),
(17, 1, 'filter3', 1, 3, 0, 0),
(18, 1, 'filter3', 1, 4, 0, 0),
(19, 1, 'filter3', 1, 5, 0, 0),
(20, 1, 'filter3', 1, 6, 0, 0),
(21, 1, 'filter3', 1, 7, 0, 0)
;

编辑：

我在这方面取得了一些进展，它在一个结果中显示了过滤器的全天工作...但是......当用户有多个过滤器时，它不起作用。我似乎无法使分组正确和/或其他东西，因此结果只显示唯一的用户/类型/过滤器组合......目前它只显示每个用户的一个过滤结果。

这只是星期一，星期二，星期三加入...必须有一个更简单的方法来做到这一点。就像我说的那样，我完全愿意为此更改数据库的模式，但不确定除此之外最好的方法是什么。我当然不能在一个表中列出整个星期的所有小时数（小时数为168列，每行为172列）。

$stmt = $db->prepare("
    SELECT users.user_id, users.username, c.computer_name, filters.user_id, filters.filter, filters.type,
    monday.12a as m12a,
    monday.1a as m1a,
    monday.2a as m2a,
    monday.3a as m3a,
    monday.4a as m4a,
    monday.5a as m5a,
    monday.6a as m6a,
    monday.7a as m7a,
    monday.8a as m8a,
    monday.9a as m9a,
    monday.10a as m10a,
    monday.11a as m11a,
    monday.12p as m12p,
    monday.1p as m1p,
    monday.2p as m2p,
    monday.3p as m3p,
    monday.4p as m4p,
    monday.5p as m5p,
    monday.6p as m6p,
    monday.7p as m8p,
    monday.9p as m9p,
    monday.10p as m10p,
    monday.11p as m11p,
    tuesday.12a as t12a,
    tuesday.1a as t1a,
    tuesday.2a as t2a,
    tuesday.3a as t3a,
    tuesday.4a as t4a,
    tuesday.5a as t5a,
    tuesday.6a as t6a,
    tuesday.7a as t7a,
    tuesday.8a as t8a,
    tuesday.9a as t9a,
    tuesday.10a as t10a,
    tuesday.11a as t11a,
    tuesday.12p as t12p,
    tuesday.1p as t1p,
    tuesday.2p as t2p,
    tuesday.3p as t3p,
    tuesday.4p as t4p,
    tuesday.5p as t5p,
    tuesday.6p as t6p,
    tuesday.7p as t8p,
    tuesday.9p as t9p,
    tuesday.10p as t10p,
    tuesday.11p as t11p,
    wednesday.12a as w12a,
    wednesday.1a as w1a,
    wednesday.2a as w2a,
    wednesday.3a as w3a,
    wednesday.4a as w4a,
    wednesday.5a as w5a,
    wednesday.6a as w6a,
    wednesday.7a as w7a,
    wednesday.8a as w8a,
    wednesday.9a as w9a,
    wednesday.10a as w10a,
    wednesday.11a as w11a,
    wednesday.12p as w12p,
    wednesday.1p as w1p,
    wednesday.2p as w2p,
    wednesday.3p as w3p,
    wednesday.4p as w4p,
    wednesday.5p as w5p,
    wednesday.6p as w6p,
    wednesday.7p as w8p,
    wednesday.9p as w9p,
    wednesday.10p as w10p,
    wednesday.11p as w11p   
    FROM
        ( SELECT account_id, computer_id, computer_name
            FROM computers
            WHERE account_id = 1
            ORDER BY computer_id ASC LIMIT 0, 5
        ) as c
        LEFT JOIN users
            on users.computer_id = c.computer_id

        LEFT JOIN filters
            on filters.user_id = users.user_id

        LEFT JOIN filters as monday
            on monday.user_id = filters.user_id and monday.filter = filters.filter and monday.day = 1

        LEFT JOIN filters as tuesday
            on tuesday.user_id = filters.user_id and tuesday.filter = filters.filter and tuesday.day = 2

        LEFT JOIN filters as wednesday
            on wednesday.user_id = filters.user_id and wednesday.filter = filters.filter and wednesday.day = 3


    WHERE filters.type = 1
    GROUP BY users.user_id
");

Answer 1

只要两个表中的列之间存在匹配，INNER JOIN关键字就会选择两个表中的所有行。

SELECT column_name(s)
FROM table1
INNER JOIN table2
ON table1.column_name=table2.column_name;

这是官方的mysql页面。 http://dev.mysql.com/doc/refman/5.0/en/join.html

组合同一个表中的多个行（一周中的几天）

1 个答案: