Question

我有一个我无法解决的故障，让我详细说明......

这些是我的MySQL表......

治疗师表

id  therapist_name
1   Therapist 1
2   Therapist 2

位置表

+-----+------------+--+
| id  |    name    |  |
+-----+------------+--+
|  1  | Location 1 |  |
|  2  | Location 2 |  |
|  3  | Location 3 |  |
+-----+------------+--+

Days_location表

+-----+-----------+--------------+-------------+--+
| id  |    day    | therapist_id | location_id |  |
+-----+-----------+--------------+-------------+--+
|  1  | monday    |            1 |           1 |  |
|  2  | monday    |            1 |           2 |  |
|   3 | wednesday |            1 |           3 |  |
|   4 | wednesday |            2 |           1 |  |
|   5 | tuesday   |            2 |           2 |  |
|   6 | friday    |            2 |           1 |  |
|   7 | friday    |            2 |           2 |  |
|   8 | friday    |            1 |           1 |  |
+-----+-----------+--------------+-------------+--+

现在我希望每个治疗师都能获得每天的位置，例如：

therapist_name =＆gt; Therapist 1，day_locations =＆gt;星期一（Location1，Location2），星期五（Location1）

我需要它作为一个选择变量，这是我的查询，但我卡在那里：

SELECT t.*,GROUP_CONCAT(
    SELECT CONCAT(dl2.day,GROUP_CONCAT(dl2.location_id)) as concated 
    FROM days_location dl2 
    WHERE therapist_id=85 
    GROUP BY dl2.day
) as day_location 
FROM therapists t  
LEFT JOIN days_location dl 
ON dl.therapist_id=t.id

这当然不起作用，我做错了什么......我应该尝试不同的方法还是让我的桌子与众不同？

Answer 1

我相信这正是您所寻找的，或者可以帮助您入手：

SELECT
    t.therapist_name,
    dl.day,
    GROUP_CONCAT(DISTINCT dl.name SEPARATOR ',') AS locations
FROM
    therapists t 
    LEFT JOIN days_location dl ON dl.therapist_id = t.id
    LEFT JOIN location l ON dl.location_id = l.id
GROUP BY t.therapist_name, dl.day

对于therapists.id = 1，这应该会给你结果：

+----------------+-----------+-----------------------+
| therapist_name |    day    |       locations       |
+----------------+-----------+-----------------------+
| Therapist 1    | monday    | Location 1,Location 2 |
| Therapist 1    | wednesday | Location 3            |
| Therapist 1    | friday    | Location 1            |
+----------------+-----------+-----------------------+

如果您需要将day与locations列连接起来，请使用简单的CONCAT()：

SELECT
    therapist_name,
    CONCAT(day, '(', locations, ')') AS locations
FROM (  
    SELECT
        t.therapist_name,
        dl.day,
        GROUP_CONCAT(DISTINCT dl.name SEPARATOR ',') AS locations
    FROM
        therapists t 
        LEFT JOIN days_location dl ON dl.therapist_id = t.id
        LEFT JOIN location l ON dl.location_id = l.id
    GROUP BY t.therapist_name, dl.day
    ) t
GROUP BY therapist_name, locations

输出应如下所示：

+----------------+-------------------------------+
| therapist_name |           locations           |
+----------------+-------------------------------+
| Therapist 1    | monday(Location 1,Location 2) |
| Therapist 1    | wednesday(Location 3)         |
| Therapist 1    | friday(Location 1)            |
+----------------+-------------------------------+

如果您需要将每个治疗师的所有组合成一行，那么您可以再次GROUP_CONCAT()。

在评论后修改：

SELECT
    therapist_name,
    GROUP_CONCAT( CONCAT(day, '(', locations, ')') SEPARATOR ',' ) AS locations
FROM (      
    SELECT
            t.therapist_name,
            dl.day,
            GROUP_CONCAT(DISTINCT dl.name SEPARATOR ',') AS locations
    FROM
            therapists t 
            LEFT JOIN days_location dl ON dl.therapist_id = t.id
            LEFT JOIN location l ON dl.location_id = l.id
    GROUP BY t.therapist_name, dl.day
    ) t
GROUP BY therapist_name

我没有测试过代码，所以可能会有一些小错误来调整。没办法测试它。

MySQL group_concat with select inside select

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