Question

我有一个表，该表具有3 GB数据（此表将不断增加），并且我需要显示总销售额，排名靠前的产品和排名靠前的产品（该列中出现的次数最多）。以下是给我上述结果的查询：

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

以上查询大约需要2分钟25秒。我找不到任何优化方法。有人可以推荐其他方式吗？

示例表：

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

输出：

category  Top product   Total sales 
    C1         P1        27
    C2         P2        44
    C3         P1        44

Answer 1

您的查询可以这样写：

SELECT g1.category, g1.sum_sale, g2.product
FROM (
    SELECT category, SUM(sale) AS sum_sale
    FROM demo
    GROUP BY category
) AS g1
INNER JOIN (
    SELECT category, product, COUNT(*) AS product_count
    FROM demo
    GROUP BY category, product
) AS g2 ON g1.category = g2.category
INNER JOIN (
    SELECT category, MAX(product_count) AS product_count_max
    FROM (
        SELECT category, product, COUNT(*) AS product_count
        FROM demo
        GROUP BY category, product
    ) AS x
    GROUP BY category
) AS g3 ON g2.category = g3.category AND g2.product_count = g3.product_count_max

基本上，它尝试查找每个类别的最大计数（*），然后从中计算出乘积。可以从适当的索引中受益。

Answer 2

仅MySQL的骇客解决方案结合使用GROUP_CONCAT和嵌套的SUBSTRING_INDEX函数来获取有序逗号分隔字符串中的第一个元素。

这不是理想的方法；但它会减少所需的子查询数量，并且可能对您的特殊情况有效。

您还需要使用SET SESSION group_concat_max_len = @@max_allowed_packet;。

我们基本上确定产品和类别组合的销售额和发生次数。然后将此结果集用作Derived Table，我们使用Group_concat() hack来确定类别中具有最大数量的产品。

SET SESSION group_concat_max_len = @@max_allowed_packet;

SELECT 
  dt.category, 
  SUM(dt.sale_per_category_product) AS total_sales, 
  SUBSTRING_INDEX(
    SUBSTRING_INDEX(
      GROUP_CONCAT(dt.product ORDER BY dt.product_count_per_category DESC)
                    , ','
                    , 1
                   )
                 , ','
                 , -1
                ) AS top_product 
FROM 
(
  SELECT 
    category, 
    product, 
    SUM(sale) AS sale_per_category_product, 
    COUNT(*) AS product_count_per_category 
  FROM demo 
  GROUP BY category, product 
) AS dt 
GROUP BY dt.category

模式（MySQL v5.7）

| category | total_sales | top_product |
| -------- | ----------- | ------------|
| C1       | 27          | P1          |
| C2       | 44          | P2          |
| C3       | 44          | P1          |

View on DB Fiddle

如何优化我的sql查询，需要2分25秒才能运行

2 个答案: