我有下列表格

Customer (driver_id is UNQIUE)
+--------+------+-----------+
| cus_id | name | driver_id |
+--------+------+-----------+
|      1 | bob  |      2342 |
|      2 | sam  |      2463 |
+--------+------+-----------+

Items (manufacture_product_id is UNIQUE) 
+---------+-------+-------------------------+
| item_id | name  | manufacturer_product_id |
+---------+-------+-------------------------+
|       1 | shirt |                    2131 |
|       2 | jeans |                     383 |
|       3 | pants |                       2 |
|       4 | watch |                   34634 |
|       5 | belt  |                       5 |
+---------+-------+-------------------------+

Outfits
+-----------+--------+---------------------+
| outfit_id | cus_id |    creation_date    |
+-----------+--------+---------------------+
|         1 |      2 | 2020-01-28 12:31:00 |
|         2 |      2 | 2020-01-29 15:23:12 |
+-----------+--------+---------------------+

items_in_outfit
+----------------+-----------+---------+
| outfit_item_id | outfit_id | item_id |
+----------------+-----------+---------+
|              1 |         1 |       2 |
|              2 |         1 |       3 |
|              3 |         1 |       5 |
|              4 |         2 |       1 |
|              5 |         2 |       2 |
|              5 |         2 |       3 |
|              6 |         2 |       5 |
+----------------+-----------+---------+

总结一下。这个
customer sam 有两套装备
outfit_id's 1 以及 2 . 装备 1 包含 items_id's , 2,3,4 . 装备 2 包含 items_id's , 1,2,3,5 . item_id's 是无用的数据位，但我需要它来保持主键的完整性。我真正关心的是 manufacturer_product_id . 如何选择所有选项 cus_id=2's 连在一起的一串 manufacturer_product_id's . 所以我的结果是：

+-----------+--------------------------+---------------------+
| outfit_id | manufacturer_product_str |    creation_date    |
+-----------+--------------------------+---------------------+
|         1 | 2,383,34634              | 2020-01-28 12:31:00 |
|         2 | 2,5,383,2131             | 2020-01-29 15:23:12 |
+-----------+--------------------------+---------------------+

但是我得到的结果是

+-----------+--------------------------+---------------------+
| outfit_id | manufacturer_product_ids |    creation_date    |
+-----------+--------------------------+---------------------+
|         1 | 2,2,5,383,383,2131,34634 | 2020-01-28 12:31:00 |
+-----------+--------------------------+---------------------+

基本上是把所有的 outfit_id = 1 以及 outfit_id = 2 一起。我希望每件衣服都是分开的。

SELECT o.outfit_id, o.creation_date,
       GROUP_CONCAT(i.manufacturer_product_id ORDER BY i.manufacturer_product_id) as manufacturer_product_ids
FROM outfits o INNER JOIN
     items_in_outfit io
     ON o.outfit_id = io.outfit_id JOIN
     items i
     ON io.item_id = i.item_id JOIN
     customer c
     ON o.cus_id = c.cus_id 
WHERE c.driver_id = 2463
ORDER BY o.creation_date;