改进查询以根据没有联合的值连接其他表

Question

我有一个 table 包含以下数据：

+-----------------+--------------+----------------------+------------+--------------------+-----------+------+---------+---------+-----+------------------------+---------------------+
| notification_id | from_user_id | from_user_auth_level | to_user_id | to_user_auth_level | status_id | type | subject | message | url | timestamp_inserted_utc | timestamp_read_utc  |
+-----------------+--------------+----------------------+------------+--------------------+-----------+------+---------+---------+-----+------------------------+---------------------+
|               1 | NULL         | NULL                 |          1 |                  1 |         0 |    2 | test    | test    | url | 2010-10-10 00:00:00    | 2011-10-10 00:00:00 |
|               2 | 2            | 5                    |          1 |                  1 |         0 |    2 | test    | test    | url | 2010-10-10 00:00:00    | 2011-10-10 00:00:00 |
|               3 | 3            | 5                    |          1 |                  1 |         0 |    2 | test    | test    | url | 2010-10-10 00:00:00    | 2011-10-10 00:00:00 |
|               4 | 2295         | 4                    |          1 |                  1 |         0 |    2 | test    | test    | url | 2010-10-10 00:00:00    | 2011-10-10 00:00:00 |
|               5 | 10           | 1                    |          1 |                  1 |         0 |    2 | test    | test    | url | 2010-10-10 00:00:00    | 2011-10-10 00:00:00 |
+-----------------+--------------+----------------------+------------+--------------------+-----------+------+---------+---------+-----+------------------------+---------------------+

然后我还有其他一些 table，例如 'users'、'companies'、'organizations'、...等等

我需要能够获取每个通知的用户名、性别和图像（基于 from_user_id 和 from_user_auth_level）。

但问题在于，此信息位于不同的位置，具体取决于 user_auth_level 是什么。

例如：如果我的用户是 "regular" 用户，他的 auth_level 将是 1。图像将驻留在我的 "users" table 中，并且性别适用。但是如果用户有auth_level == 5，就说明他是一个组织。在这种情况下，性别不适用，图片位于 "organization" table，这需要通过用户链接到 user_roles，然后再链接到组织。

这适用于每种用户类型，他们都需要不同的加入。

我已经创建了一个有效的查询，但是它到处都使用 UNION，而且我了解到出于性能原因，它并不是最好的使用方式，所以我希望有人可以指导我改进这个查询的性能头脑：

                SELECT n.*, NULL as username, NULL as gender, NULL as picture
                FROM notification as n
                WHERE n.from_user_auth_level IS NULL
                AND n.to_user_id = $userid
                UNION
                SELECT n.*, u.username, u.gender as gender, u.profile_picture as picture
                FROM notification as n
                LEFT JOIN users AS u ON n.from_user_id = u.user_id
                WHERE n.from_user_auth_level = 1
                AND n.to_user_id = $userid
                UNION
                SELECT n.*, u.username, NULL as gender, c.logo as picture
                FROM notification as n
                LEFT JOIN users AS u ON n.from_user_id = u.user_id
                LEFT JOIN user_companies AS uc on u.user_id = uc.user_id
                LEFT JOIN company as c on uc.company_id = c.company_id
                WHERE n.from_user_auth_level = 4
                AND n.to_user_id = $userid
                UNION
                SELECT n.*, u.username, NULL as gender, o.logo as picture
                FROM notification as n
                LEFT JOIN users AS u ON n.from_user_id = u.user_id
                LEFT JOIN user_roles as ur on u.user_id = ur.user_id
                LEFT JOIN organization as o on ur.org_id = o.org_id
                WHERE n.from_user_auth_level = 5
                AND n.to_user_id = $userid
                UNION
                SELECT n.*, u.username, NULL as gender, o.logo as picture
                FROM notification as n
                LEFT JOIN users AS u ON n.from_user_id = u.user_id
                LEFT JOIN user_roles as ur on u.user_id = ur.user_id
                LEFT JOIN organization as o on ur.org_id = o.org_id
                WHERE n.from_user_auth_level = 7
                AND n.to_user_id = $userid
                UNION
                SELECT n.*, u.username, NULL as gender, NULL as picture
                FROM notification as n
                LEFT JOIN users AS u ON n.from_user_id = u.user_id
                WHERE n.from_user_auth_level = 9
                AND n.to_user_id = $userid"

得到这个结果后，我使用 PHP 根据 timestamp_inserted_utc 对结果进行排序，因为使用 UNION 不可能得到正确的结果。

Answer 1

我将使用通知 table 作为基础并使用条件外连接作为：

select 
  n.*,t1.gender, t2.orgNo 
from 
  notifications n
left outer join table1 t1 on (n.auth=1 and more join)
left outer join table2 t2 on (n.auth=2 and more..)

您将有更多的列，但它们的名称是有意义的，您可以在应用程序级别合并。

Answer 2

建议索引（大部分为"covering"）：

n:  (from_user_auth_level, to_user_id, from_user_id)
u:  (user_id, username, profile_picture, gender)
o:  (org_id, logo)
ur:  (org_id, user_id)
c:  (company_id, logo)
uc:  (company_id, user_id)

（他们甚至可以加快 Teson 的回答速度。）

改进查询以根据没有联合的值连接其他表

Improve query to join other tables depending on value without union

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