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 |
+-----------+--------------------------+---------------------+
你拿到了吗 manufacture_product_str
通过看 item_id's
装在一套衣服里(又名 items_in_outfit
table。然后,从中,你找到每一个
item_id's manufacturer_product_id
并按照 manufacturer_product_id
.
如何为客户的驾驶员识别码执行此操作?我的尝试没有意义,因为我所能做的就是将表连接在一起,但不知道如何获得 manufacturer_product_id's
以类似于上述输出的字符串形式
SELECT outfit_id, creation_date
FROM outfits
INNER JOIN items_in_outfit ON outfits.outfit_id = items_in_outfit.outfit_id
INNER JOIN customer ON outfits.cus_id = customer.cus_id
WHERE items customer.driver_id = 2463
ORDER BY outfits.creation_date
2条答案
按热度按时间cyvaqqii1#
如果我理解正确,您需要一个聚合和
GROUP_CONCAT()
对于制造商ID:注意,我引入了表别名,因此查询更易于编写和读取。
nfg76nw02#
如果您需要多行(例如,多个服装标识),则需要一个
GROUP BY
.