Question

我正在尝试创建一个在我当前查询中添加... elif (score == "quit") ...作为新列的查询，首先是我的表格：

GROUP_CONCAT

基本上，我想要一个查询，它获取所有用户（显示在表中）的自定义列，该列显示附加到该用户的所有移位（如果没有移位附加到显然有空结果）

这是我到目前为止所做的查询：

Users table
+----+----------+--------------+
| id | username | date_created |
+----+----------+--------------+
|  1 | user1    | 2000-03-16   |
|  2 | user2    | 2001-05-14   |
|  3 | user3    | 2002-01-13   |
|  4 | user4    | 2003-03-14   |
+----+----------+--------------+

Shifts table
+----+------------+--------------+
| id | shift_name | date_created |
+----+------------+--------------+
|  1 | shift1     | 2002-05-10   |
|  2 | shift2     | 2002-07-11   |
|  3 | shift3     | 2002-09-23   |
+----+------------+--------------+

Accounts table
+----+--------------+--------------+
| id | account_name | date_created |
+----+--------------+--------------+
|  1 | account1     | 2001-05-01   |
|  2 | account2     | 2001-05-02   |
|  3 | account3     | 2001-05-03   |
+----+--------------+--------------+

Shift Mapping table
+----+---------+----------+------------+
| id | user_id | shift_id | account_id |
+----+---------+----------+------------+
|  1 |       1 |        1 |          1 |
|  2 |       1 |        2 |          2 |
|  3 |       3 |        1 |          1 |
+----+---------+----------+------------+

现在使用转移附加的用户没有问题，我的问题是没有转移附加的用户只返回由SELECT users.id AS user_id, users.username, users.date_created, GROUP_CONCAT(DISTINCT (t.shift_name)) AS shifts FROM (`users`) LEFT JOIN (SELECT s.shift_name, sm.user_id FROM shift_map sm LEFT JOIN shifts s ON sm.shift_id = s.id) t ON users.id = t.user_id GROUP BY user_id ORDER BY `users`.`date_created` DESC;引起的1结果我如何排除用户GROUP BY没有转换，或者如何返回所有附加轮班且没有附加班次的用户？感谢。

更新

以下是我想看到的示例结果：

GROUP BY user_id

我的查询中的问题是它只显示了一个空转的用户。

Answer 1

    SELECT users.id AS user_id, users.username, users.date_created,
GROUP_CONCAT(DISTINCT (t.shift_name)) AS shifts
FROM (`users`)
LEFT JOIN 
    (SELECT s.shift_name, sm.user_id FROM mapping sm
    LEFT JOIN shifts s ON sm.shift_id = s.id) t 
ON users.id = t.user_id
GROUP BY username
ORDER BY user_id

+---------+----------+--------------+---------------+
| user_id | username | date_created | shifts        |
+---------+----------+--------------+---------------+
|       1 | user1    | 2000-03-16   | shift1,shift2 |
|       2 | user2    | 2001-03-16   | NULL          |
|       3 | user3    | 2002-03-16   | shift1        |
|       4 | user4    | 2003-03-16   | NULL          |
+---------+----------+--------------+---------------+

Answer 2

我的坏，我可以使用简单的LEFT JOIN：

SELECT users.id AS user_id, users.username, users.date_created,
GROUP_CONCAT(DISTINCT (shift.shift_name)) AS shifts
FROM (`users`)
LEFT JOIN shifts_map ON users.id = shifts_map.user_id
LEFT JOIN shifts ON shifts_map.shift_id = shift.id
GROUP BY user_id
ORDER BY `users`.`date_created` DESC;

我只是使查询复杂化，忘了简单的LEFT JOIN可以解决问题。感谢。

左连接子查询排除GROUP BY，结果为null

2 个答案: