Question

我正在使用 MySQL ..

我有一个简单的销售表如下：

o----o----------o-----------o
| id | store_id |  logDate  |
o----o----------o-----------o
| 1  |    1     | 2015-1-13 |
| 2  |    1     | 2015-1-14 |
| 3  |    2     | 2015-1-11 |
| 4  |    2     | 2015-1-18 |
o----o----------o-----------o

销售产品表

o----o----------o---------o------------o
| id | sale_id  |  qty    |    price   |
o----o----------o---------o------------o
| 1  |    1     |       1 |         10 |
| 2  |    2     |       1 |         10 |
| 3  |    2     |       1 |         10 |
| 4  |    3     |       1 |         10 |
| 5  |    3     |       1 |         10 |
| 6  |    3     |       1 |         10 |
| 7  |    4     |       1 |         10 |
| 8  |    4     |       1 |         10 |
o----o----------o---------o------------o

预期结果

o-- --------o----------------o---------------------o
| store_id  | SUM(price*qty) |  Highest Date On    | 
o-----------o----------------o---------------------o
|     1     |             20 |           2015-1-14 |
|     2     |             30 |           2015-1-11 |
O-----------o----------------o---------------------o

如何达到我的预期效果？我尝试过以下但是没有按预期工作：

SELECT store_id, MAX(total), highestSingleDateOn 
FROM (
   SELECT SUM(price * qty) AS total, 
      DATE(s.logDate) AS highestSingleDateOn, s.store_id AS store_id
   FROM sale_product sp JOIN sales s ON s.id = sp.sales_id
   GROUP BY DATE(s.logDate), s.store_id
   ORDER BY DATE(s.logDate) ASC
) AS result_for_highest_single_day
GROUP BY highestSingleDateOn, store_id

Answer 1

在MySQL中执行此操作的一种方法是使用多个聚合，然后使用连接。也许更简单的方法是使用变量：

SELECT sd.*
FROM (SELECT sd.*,
             (@rn := if(@s = store_id, @rn + 1,
                        if(@s := store_id, 1, 1)
                       )
             ) as rn 
      FROM (SELECT DATE(s.logDate) AS date, s.store_id, SUM(price * qty) AS total
            FROM sale_product sp JOIN sales s ON s.id = sp.sales_id
            GROUP BY DATE(s.logDate), s.store_id
            ORDER BY s.store_id, total desc
           ) sd cross join
           (SELECT @rn := 0, @s := -1) params
     ) sd
WHERE rn = 1;

Answer 2

SELECT store_id, MAX(total), highestSingleDateOn 
FROM (
   SELECT SUM(price * qty) AS total, 
      DATE(s.logDate) AS highestSingleDateOn, s.store_id AS store_id
   FROM sale_product sp JOIN sales s ON s.id = sp.sales_id
   GROUP BY DATE(s.logDate), s.store_id
   ORDER BY total DESC
) AS result_for_highest_single_day
GROUP BY  store_id

我刚修改了脚本ORDER BY DATE(s.logDate) ASC＆gt;＆gt; ORDER BY total DESC 和GROUP BY highestSingleDateOn, store_id＆gt;＆gt; GROUP BY store_id。

*在sql脚本之上，它使用了关于MYSQL的group by的不稳定特性。

*然后根据Mysql标准，我写了另一个版本的sql脚本。

select table1.*
from 
( SELECT SUM(price * qty) AS total, 
      DATE(s.logDate) AS highestSingleDateOn, s.store_id AS store_id
   FROM sale_product sp JOIN sales s ON s.id = sp.sale_id
   GROUP BY DATE(s.logDate), s.store_id) as table1
,
(select tmp.store_id,MAX(tmp.total) as max_total from 
        (SELECT SUM(price * qty) AS total, 
   DATE(s.logDate) AS highestSingleDateOn, s.store_id AS store_id
   FROM sale_product sp JOIN sales s ON s.id = sp.sale_id
   GROUP BY DATE(s.logDate), s.store_id ) as tmp group by tmp.store_id) as table2

where table1.store_id = table2.store_id and table1.total=table2.max_total

SQL - 从日期范围中查找最高单日销售额

2 个答案: