Question

我有一张表 - 记录 - 看起来像这样

id | location_id | product | stock | date  
===|=============|=========|=======|======
1  | Bakery 1    |  cake   |  21   | 2
2  | Bakery 1    |  bread  |  23   | 2
3  | Bakery 2    |  cake   |  21   | 2
4  | Bakery 2    |  bread  |  21   | 2
5  | Bakery 1    |  cake   |  21   | 3
6  | Bakery 1    |  bread  |  23   | 3
7  | Bakery 2    |  cake   |  21   | 3
8  | Bakery 2    |  bread  |  21   | 3
9  | Bakery 1    |  cake   |  21   | 4
10 | Bakery 1    |  bread  |  23   | 4
11 | Bakery 2    |  bread  |  23   | 4

对于每个位置和每个产品，我想选择具有最高日期值的行。这看起来像是

id | location_id | product | stock | date  
===|=============|=========|=======|======
7  | Bakery 2    |  cake   |  21   | 3
9  | Bakery 1    |  cake   |  21   | 4
10 | Bakery 1    |  bread  |  23   | 4
11 | Bakery 2    |  bread  |  23   | 4

如何使用一个查询执行此操作？我可以循环遍历所有位置和所有产品并构建查询，但它耗费更多时间和内存？

Answer 1

select r1.*
from records r1
join
(
  select location_id, product, max(date) as date
  from records
  group by location_id, product
) r2 on r1.location_id = r2.location_id 
    and r1.product = r2.product
    and r1.date = r2.date

Answer 2

如果满足以下条件，您可以使用id列

非常有效地完成此项工作

您的id列是自动增量
date列代表当前日期：即您永远不会插入回溯记录。
您没有更新date列

如果这些条件为真，则表示唯一标识符id可用作每条记录的非唯一值date的代理。值date更高的行始终保证具有更高的id值。

如果你能做到这一点，你可以使你的查询表现得非常好。

首先，您创建一个子查询，以获取每个位置和产品组合的最新行的id。

SELECT MAX(id)
  FROM records
 GROUP BY location_id, date

然后使用该组id值从表中提取正确的记录

SELECT *
  FROM records
 WHERE id IN (
                SELECT MAX(id)
                 FROM records
                GROUP BY location_id, date
              )
ORDER BY location_id, product

Mysql从表中选择每个唯一location_id和产品的日期最大值

2 个答案: