Question

我有下表：

category  product    sale 
C1         P1        10
C2         P2        12
C3         P1        14
C1         P2        15
C1         P1        16
C2         P2        10
C2         P3        22
C3         P1        01
C3         P2        27
C3         P3        02

我想按类别分组并获得销售总额，以及每个类别中的顶级产品。（顶级产品实际上是产品中出现的最多产品，而不是通过销售）

预期输出：

category  Top product   Total sales 
    C1         P1        38
    C2         P2        44
    C3         P1        44

我编写了以下MySQL查询，但出现错误（产品的列名称无效）：

SELECT SUM(product) AS pro FROM
(SELECT category, count(sale)
FROM `table`
GROUP BY `table`.`category`
) as t2
ORDER BY pro DESC LIMIT 1

我正在努力编写正确的MySQL查询，有人可以指出问题吗？

Answer 1

要获得每个类别的销售总额以及顶级产品，您可以使用相关/相关子查询，如下所示：

select t.category, 
       sum(t.sale) sales,
        (select product 
        from demo 
        where  category = t.category
        group by product
        order by count(*) desc
        limit 1) top_product
from demo t
group by t.category

Demo

Answer 2

您可以使用Correlated Subquery来获得某个类别中销量最高（销量最高）的产品。请尝试以下操作：

SELECT 
  t.category, 
  SUM(t.sale) AS total_sales, 
  (SELECT t2.product 
   FROM `table` AS t2 
   WHERE t2.category = t.category 
   ORDER BY t2.sale DESC LIMIT 1) AS top_product 
FROM
  `table` AS t 
GROUP BY t.category

DB Fiddle DEMO

Answer 3

仅MySQL解决方案，因此当您需要代码中的数据库可移植性时不要使用此解决方案。

结合使用GROUP_CONCAT和嵌套的SUBSTRING_INDEX函数来获取第一个元素/项目，也可以用来获取额外的列，以保留正确的分组信息。

注意：SET SESSION group_concat_max_len = @@max_allowed_packet;的使用非常重要，否则GROUP_CONCAT仅显示1024字节。

查询

SET SESSION group_concat_max_len = @@max_allowed_packet;

SELECT 
   Table1.category
 , SUM(Table1.sale) sum_per_category
 , SUBSTRING_INDEX(
     SUBSTRING_INDEX(
         GROUP_CONCAT(product ORDER BY sale DESC)
       , ','
       , 1
     )
     , ','
     , -1
   ) AS most_sold_product_in_category
FROM 
 Table1 
GROUP BY 
 Table1.category

结果

| category | sum_per_category | most_sold_product_in_category |
|----------|------------------|-------------------------------|
|       C1 |               26 |                            P3 |
|       C2 |               12 |                            P2 |
|       C3 |               14 |                            P1 |
|       C4 |               15 |                            P2 |

参见演示http://sqlfiddle.com/#!9/0eb073/13

Answer 4

这可能有效：

SELECT t2.category, t2.sales,(select product        
                              from `table` 
                              where sale=(select max(b.sale) 
                                          from `table` b
                                          where b.category=t2.category)  limit 1)
FROM
(SELECT category, sum(sale) as sales
FROM `table`
GROUP BY `table`.`category`
) as t2

Answer 5

Select q.*,SUM(sale) from

(SELECT *
FROM `categories` order by sale desc limit 50 
) as q group by q.category

查询以显示MySQL中每个类别的顶级产品

5 个答案:

DB Fiddle DEMO