Question

我在MySQL 5.6.17中有表departments，employees和emails（对于Rails应用程序）。每个部门都有很多员工，部门和员工都有很多电子邮件。我想根据整个部门和部门内各个员工的电子邮件数量对departments进行排序。我的尝试：

SELECT departments.*, COUNT(DISTINCT employees.id) AS employees_count, COUNT(DISTINCT emails.id) AS emails_count
    FROM departments
    LEFT OUTER JOIN employees 
        ON employees.department_id = departments.id AND employees.is_employed = true
    LEFT OUTER JOIN emails 
        ON (emails.emailable_id = departments.id AND emails.emailable_type = 'department')
            OR (emails.emailable_id = employees.id AND emails.emailable_type = 'employee')
    GROUP BY departments.id 
    ORDER BY emails_count DESC 
    LIMIT 20;

不幸的是，此查询需要3分钟才能完成。由于此查询将在Web界面中使用，因此这不是可行的时间范围。 EXPLAIN给出了：

+----+-------------+-------------+-------+-------------------------------------------------+----------------------------------+---------+-------------------------------+-------+------------------------------------------------+
| id | select_type | table       | type  | possible_keys                                   | key                              | key_len | ref                           | rows  | Extra                                          |
+----+-------------+-------------+-------+-------------------------------------------------+----------------------------------+---------+-------------------------------+-------+------------------------------------------------+
|  1 | SIMPLE      | departments | index | PRIMARY                                         | PRIMARY                          | 4       | NULL                          | 37468 | Using where; Using temporary; Using filesort   |
|  1 | SIMPLE      | employees   | ref   | index_employees_on_department_id                | index_employees_on_department_id | 5       | development_db.departments.id |     5 | Using where                                    |
|  1 | SIMPLE      | emails      | ALL   | index_emails_on_emailable_id_and_emailable_type | NULL                             | NULL    | NULL                          | 10278 | Range checked for each record (index map: 0x2) |
+----+-------------+-------------+-------+-------------------------------------------------+----------------------------------+---------+-------------------------------+-------+------------------------------------------------+

然后，emails上的索引未被使用。当我将emails仅加入departments或仅加入employees时，此索引已成功使用，但不会同时加入{{1}}。

这是为什么？我该怎么办？有没有更有效的方法来查询所需的数据？

Answer 1

在连接之前首先进行聚合可能会有所帮助：

SELECT d.*, e.employees_count, em.emails_count
FROM d LEFT OUTER JOIN
     (SELECT e.department_id, count(*) as employees_count
      FROM employees e
      WHERE e.is_employed = true
      GROUP BY e.department_id
     ) e
     ON e.department_id = d.id LEFT OUTER JOIN
     (SELECT department_id, count(distinct id) as emails_count
      FROM (SELECT em.emailable_id as department_id, em.id
            FROM emails em
            WHERE em.emailable_type = 'department'
            UNION ALL
            SELECT e.department_id, em.id
            FROM emails em JOIN
                 employees e
                 ON em.emailable_id = e.id AND em.emailable_type = 'employee'
           ) ee
      GROUP BY department_id
     ) em
     ON em.department_id = d.id LEFT OUTER JOIN
ORDER BY emails_count DESC 
LIMIT 20;

您还需要emails(emailable_id, emailable_type, id)和emails(emailable_type, emailable_id, id)上的索引。

MySQL多态连接条件，OR不使用索引

1 个答案: