mysql - 组合这两个 mySQL 查询

Question

此查询产生正确的答案:

SELECT users.*,
        SUM(overtime_list.shift_length) AS overtime_total,
        (SELECT GROUP_CONCAT(users_roles.role_ID) FROM users_roles WHERE users.user_ID = users_roles.user_ID) AS roles
    FROM availability_list
    INNER JOIN users
        ON users.user_ID = availability_list.user_ID
    INNER JOIN stations
        ON users.station_ID = stations.station_ID
    INNER JOIN overtime_list
        ON overtime_list.user_ID = users.user_ID
        AND overtime_list.date >= '$totalovertimedays'
    WHERE availability_list.date = '$date'
    AND availability_list.type = '$type'
    GROUP BY users.user_ID
    ORDER BY overtime_total ASC

输出:

 +----------+---------+----------------------------+------------------+
 |  user_ID | user    |   roles                    |  overtime_total  |
 +----------+---------+----------------------------+------------------+
 |        1 |   Smith |    1,2                     |        12        |
 +----------+---------+----------------------------+------------------+
 |        2 |   Jones |    1,2,3                   |        7         |
 +----------+---------+----------------------------+------------------+

这是期望的结果:

 +----------+---------+----------------------------+------------------+
 |  user_ID | user    |   roles                    |  overtime_total  |
 +----------+---------+----------------------------+------------------+
 |        1 |   Smith |    Admin, Staff            |        12        |
 +----------+---------+----------------------------+------------------+
 |        2 |   Jones |    Admin, Staff, Other     |        7         |
 +----------+---------+----------------------------+------------------+

这是我可以使用的联接，它似乎允许 group_concat 正确联接“admin、staff、other” - 但我无法弄清楚如何将其合并到上面的主查询中？

SELECT users.user_id, GROUP_CONCAT(roles.short_name  separator ', ') roles
FROM users
JOIN users_roles ON users.user_ID = users_roles.user_ID
JOIN roles ON users_roles.role_ID= users_roles.role_ID
GROUP BY users.user_ID

users_roles 表:

+----------+---------+
|  user_ID | role_ID |
+----------+---------+
|        1 |   1     |
+----------+---------+
|        2 |   1     |
+----------+---------+
|        2 |   2     |
+----------+---------+
|        2 |   3     |
+----------+---------+
|        1 |   3     |
+----------+---------+

角色表:

+----------+------------+
|  role_ID | short_name |
+----------+------------+
|        1 |   Admin    |
+----------+------------+
|        2 |   Super    |
+----------+------------+
|        3 |   Other    |
+----------+------------+

Answer 1

你可以尝试:

SELECT users.*,
        SUM(overtime_list.shift_length) AS overtime_total,
        (SELECT GROUP_CONCAT(roles.short_name) FROM users_roles 
         INNER JOIN roles ON user_roles.role_ID = roles.role_ID
         WHERE users.user_ID = users_roles.user_ID) AS roles
    FROM availability_list
    INNER JOIN users
        ON users.user_ID = availability_list.user_ID
    INNER JOIN stations
        ON users.station_ID = stations.station_ID
    INNER JOIN overtime_list
        ON overtime_list.user_ID = users.user_ID
        AND overtime_list.date >= '$totalovertimedays'
    WHERE availability_list.date = '$date'
    AND availability_list.type = '$type'
    GROUP BY users.user_ID
    ORDER BY overtime_total ASC