优化SELECT中的子查询答案

【问题标题】：Optimize subquery in SELECT优化SELECT中的子查询
【发布时间】：2020-02-27 23:34:58
【问题描述】：

我的表schema如下：

索引：

products.id 主键
products.description 唯一
expenses.id 主键
expenses.product_id product.id 的外键

我的目标是加载

当月每种产品的成本（AS cost_november）
上个月每种产品的成本（AS cost_october）
当月成本与上月相比的变化
（当月成本 - 上月成本）（AS 成本）
当月费用与上一月相比的百分比变化
（上月费用 * 100 / 当月费用）(AS percent_diff)

我已经编写了完全符合此功能的 SQL：

SELECT description, (SUM(cost) - IFNULL(
(
    SELECT SUM(cost)
    FROM expenses
    WHERE month = 9 AND year = 2019 AND product_id = e.product_id
    GROUP BY product_id
), 0)) AS costs,

SUM(cost) * 100 / 
(
    SELECT SUM(cost)
    FROM expenses
    WHERE month = 9 AND year = 2019 AND product_id = e.product_id
    GROUP BY product_id
) AS percent_diff,

SUM(cost) AS costs_october,

IFNULL(
(
    SELECT SUM(cost)
    FROM expenses
    WHERE month = 9 AND year = 2019 AND product_id = e.product_id
    GROUP BY product_id
), 0) AS costs_september

FROM expenses e
JOIN products p ON (e.product_id = p.id)
WHERE month = 10 AND year = 2019
GROUP BY product_id
ORDER BY product_id;

但是将相同的子查询复制粘贴三次真的是解决方案吗？理论上，每个产品需要运行四个查询。有没有更优雅的方式？

感谢您的帮助！

【问题讨论】：

您从未有过月内成本变化吗？
请列出您目前拥有的索引。
@P.Salmon 费用/成本以后不会改变。

标签： mysql sql join query-optimization innodb

【解决方案1】：

您可以一次计算所有月份和所有产品：

SELECT year, month,
       SUM(costs) as curr_month_costs,
       LAG(SUM(costs)) OVER (ORDER BY year, month) as prev_month_costs,
       (SUM(costs) -
        LAG(SUM(costs)) OVER (ORDER BY year, month) 
       ) as diff,
       LAG(SUM(costs)) OVER (ORDER BY year, month) * 100 / SUM(costs)
FROM expenses e JOIN
     products p
     ON e.product_id = p.id
GROUP BY product_id, year, month
ORDER BY year, month, product_id;

如果您只想选择当前月份，可以使用子查询。

【讨论】：

我想我不能使用 LAG() 因为我还在使用 MySQL 5.7.26

【解决方案2】：

我会通过条件聚合来解决这个问题：

select 
    p.description,
    sum(case when e.month = 11 then e.cost else 0 end) costs_november,
    sum(case when e.month = 10 then e.cost else 0 end) costs_october,
    sum(case when e.month = 11 then e.cost else -1 * e.cost end) costs,
    sum(case when e.month = 10 then e.cost else 0 end)
        * 100
        / nullif(
            sum(case when e.month = 11 then e.cost else 0 end),
            0
        ) percent_diff
from expenses e
inner join products p on p.id = e.product_id
where e.year = 2019 and e.month in (10, 11)
goup by e.product_id

您可以通过使用子查询来避免重复相同的条件求和（您的 RDBMS 可能无论如何都会对其进行优化，但这往往会使查询更具可读性）：

select 
    description,
    costs_november,
    costs_october,
    costs_november - costs_october costs,
    costs_october * 100 / nullif(costs_november, 0) percent_diff
from (
    select 
        p.description,
        sum(case when e.month = 11 then e.cost else 0 end) costs_november,
        sum(case when e.month = 10 then e.cost else 0 end) costs_october
    from expenses e
    inner join products p on p.id = e.product_id
    where e.year = 2019 and e.month in (10, 11)
    goup by e.product_id
) t

【讨论】：