当表之间没有物理连接时优化 SQL 查询

Question

当 table 之间没有逻辑联系时，我需要帮助编写更优化的 SQL 查询。我写了一些子查询，但我对它不满意，因为我重复了两次几乎相同的子查询。
方案是这样的：

create table rate (
  rate_date date,
  value decimal(10,2),
  multiplier integer,
  name varchar(3)
  );

create table amount (
  amount decimal(10,2),
  amount_year date,
  rate_name varchar(3)
  );

我希望 AMOUNT table 中的每个金额都使用 RATE table 中的乘数和比率值来计算，就像这样 amount / multiplier * rate。 table 的共同点是名称（一些代码标识符）和日期。这些应该用于收集数据。仅考虑 amount_year 的最新汇率。
如果您愿意提供帮助，可以查看此 fiddle。这是我的尝试。

select am.amount / 
    (
      select ra.multiplier 
      from rate ra
      where ra.rate_date = (select max(rate_date) 
                            from rate 
                            where year(rate_date) = year(am.amount_year) and name = am.rate_name) 
        and ra.name = am.rate_name
    ) * (
      select ra.value 
      from rate ra
      where ra.rate_date = (select max(rate_date) 
                            from rate 
                            where year(rate_date) = year(am.amount_year) and name = am.rate_name) 
        and ra.name = am.rate_name
    ) as calculated_amount
    from amount am;

Answer 1

你的子查询是一样的，所以你可以合二为一：

select am.amount / 
       (select ra.multiplier * ra.value
        from rate ra
        where ra.rate_date = (select max(rate_date) 
                              from rate 
                              where year(rate_date) = year(am.amount_year) and name = am.rate_name)  and ra.name = am.rate_name
                             )
       ) as calculated_amount
from amount am;

或者这可以表述为：

select am.amount / 
       (select ra.multiplier * ra.value
        from rate ra
        where year(ra.rate_date) = year(am.amount_year) and name = am.rate_name
        order by ra.rate_date desc
        limit 1
       ) as calculated_amount
from amount am;

如果您愿意，可以将此作为带有 where 子句的 join 执行：

select (am.amount / ra.multiplier * ra.value) as calculated_amount
from amount am join
     rate ra
     on ra.name = am.name and
        year(rate_date) = year(am.amount_year)
where ra.date = (select max(r2.rate_date)
                 from rate r2
                 where year(r2.rate_date) = year(r.rate_date) and
                       r2.name = ra.name
                );

为了性能，从 rate(name, rate_date, multiplier, value) 上的索引开始。对于任一查询。

为了获得更好的性能，您可能需要将 "year" 作为单独的列存储在 table 中。

Answer 2

您可以使用单个子查询获取每年的最新乘数和值。然后将其与 amount table 一起进行除法。

SELECT am.amount / (r2.multiplier * r2.value) AS calculated_amount
FROM amount AS am
JOIN (SELECT YEAR(rate_date) AS year, MAX(rate_date) AS maxdate
      FROM rate
      GROUP BY year) AS r1 ON YEAR(am.rate_date) = r1.year
JOIN rate AS r2 ON r1.maxdate = r2.rate_date

当表之间没有物理连接时优化 SQL 查询

Optimized SQL query when there's no physical connection between tables

mysql

sql

join

subquery

sql-optimization