Question

我有这些表SQL Fiddle

项目表：

+----+----------+
| id |   name   |
+----+----------+
|  1 | Facebook |
|  2 | Twitter  |
|  3 | Amazon   |
+----+----------+

价格表：

+----+-----------+---------+-----------------------------+
| id |    buy    | item_id |             created_at      |
+----+-----------+---------+-----------------------------+
|  1 |   43000   |    1    | June, 18 2014 17:31:04+0000 |
|  2 |   44000   |    1    | June, 19 2014 17:31:04+0000 |
|  3 |   30000   |    2    | June, 20 2014 17:31:04+0000 |
|  4 |   33000   |    2    | June, 21 2014 17:31:04+0000 |
|  5 |   20000   |    3    | June, 22 2014 17:31:04+0000 |
|  6 |   21000   |    3    | June, 23 2014 17:31:04+0000 |
+----+-----------+---------+-----------------------------+

我希望根据价格日期获得每件商品的最后价格和最后价格购买价格之前的价格

期望的输出：

+----+---------+-----------------+---------+
| id |   buy   | last_before_buy | item_id |
+----+---------+-----------------+---------+
| 10 |  45000  |     43000       |    3    |
| 7  |  33000  |     31000       |    2    |
| 4  |  23000  |     23000       |    1    |
+----+---------+-----------------+---------+

Answer 1

您可以使用substring_index() / group_concat()技巧：

执行此操作

select max(id) as id,
       substring_index(group_concat(buy order by created_at desc), ',', 1) as buy,
       substring_index(substring_index(group_concat(buy  order by created_at desc), ',', 2), ',', -1) as lastbuy,
       item_id
from prices p
group by item_id;

Answer 2

这是另一种方法：

select a.id, a.buy, b.buy last_before_buy, a.item_id
from (select * from prices WHERE (created_at <= NOW() - INTERVAL 5 DAY) order by id desc) a
join (select * from prices order by id desc) b on a.item_id = b.item_id and a.id > b.id
group by a.item_id;

fiddle

按日期获取每个类别的最后记录

2 个答案: