结合这两个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   | 
+----------+---------+----------------------------+------------------+ 

这是所期望的结果：

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

这是我可以使用的连接，它似乎允许group_concat更正连接“admin，staff，other” - 但我无法解决如何将它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 

的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 

Answer 2

也许是这样的：

SELECT users.*, 
     SUM(overtime_list.shift_length) AS overtime_total, 
     (SELECT GROUP_CONCAT(users_roles.short_name) 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 

Answer 3

添加派生表，并加入回给用户。使用派生表是因为overtime_list上的聚合函数，因此数据不会重复。

SELECT users.*, 
    SUM(overtime_list.shift_length) AS overtime_total, 
    roles.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 >= '$totalovertimedays' 
LEFT JOIN 
(
    SELECT users_roles.user_ID, 
      GROUP_CONCAT(roles.short_name separator ', ') roles 
    from users_roles 
     INNER JOIN roles ON users_roles.role_ID = roles.role_ID 
    group by users_roles.user_ID 
) roles 
    ON users.user_ID = roles.user_ID 
WHERE availability_list.date = '$date' 
AND availability_list.type = '$type' 
GROUP BY users.user_ID 
ORDER BY overtime_total ASC 

Answer 4

我从来不喜欢使用的查询功能，用户数据库引擎将作为黑盒和着内部优化查询，这样就可以避免它们成为可能。随着跨应用你会得到相同的结果intendend：（没有测试所有查询作为我没有使用的所有对象）

SELECT users.*, 
     SUM(overtime_list.shift_length) AS overtime_total, 
     LEFT(ISNULL(roles.roles, ', '), LEN(ISNULL(roles.roles, ', ')) - 1) as roles 
    FROM availability_list 
    INNER JOIN users 
     ON users.user_ID = availability_list.user_ID 
    CROSS APPLY (
     SELECT short_name + ', ' 
     FROM roles 
     inner users_roles ON roles.role_id = users_roles.role_ID 
     WHERE users.user_ID = users_roles.user_ID 
     ORDER BY roles.role_id 
     FOR XML PATH('') 
    ) roles (roles)   
    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