如何改善查询执行时间？

Question

我有一个相当复杂的查询，需要 1-2 分钟才能执行。有没有办法缩短执行时间？

这里是查询：

select  o.orders_id, o.customers_id, o.customers_name, s.orders_status_name,
        ot.text as order_total, ot.value, DATEDIFF(NOW(), payment_data_read_status) as numDaysLeft,
        ( SELECT  ifnull(sum(op.paid_amount), 0)
            from  orders_payment op
            where  op.orders_id=o.orders_id
              AND  op.confirm_payment='1'
        ) as paid_total
    from  orders o, orders_total ot, orders_status s
    where  o.orders_id = ot.orders_id
      and  ot.class = 'ot_total'
      and  o.orders_status = s.orders_status_id
      and  s.language_id = '1'
      AND  ROUND(ot.value,2) != ROUND(
                ( SELECT  ifnull(sum(op.paid_amount),0)
                    from  orders_payment op
                    where  op.orders_id=o.orders_id
                      AND  op.confirm_payment='1'
                ), 2)

查询说明

一些细节

number of records in orders = 7321

number of records in orders_total = 22167

number of records in orders_payment= 12038

number of records in orders_status= 9

orders_id 列在订单中自动递增 table。首先，我想在订单 table 中索引 orders_id 列，但由于它是主要的，所以我认为它不会起作用。

编辑错误

Answer 1

我发现嵌套查询不一定不好，但我尽量避免将它们放在 select 列表中。这是我的建议：

select
    o.orders_id,
    o.customers_id,
    o.customers_name,
    s.orders_status_name, 
    ot.text as order_total, 
    ot.value, 
    datediff(now(), payment_data_read_status) as numdaysleft, 
    ifnull(op.paid_total, 0) paid_total
from
    orders o
    join
    orders_total ot
    on o.orders_id = ot.orders_id 
    join
    orders_status s
    on o.orders_status = s.orders_status_id 
    left outer join
    (
        select 
            orders_id,
            sum(ifnull(paid_amount, 0)) as paid_total
        from
            orders_payment 
        where
            confirm_payment = '1'
        group by
            orders_id
    ) op 
    on
        op.orders_id = o.orders_id
where
    ot.class = 'ot_total' and 
    s.language_id = '1' and
    round(ot.value,2) != round(ifnull(op.paid_total, 0), 2);

我认为这将使优化器有更好的机会做好工作。

请注意，我在 "op" 的内部查询中放置了 "group by"。如果没有它，我认为您可能会欺骗优化器运行此查询针对每个结果行而不是一次。

有了这些卷，您应该不需要任何索引；他们可能会使事情变得更糟而不是更好，但请测试一下，看看会发生什么。

我无法测试我的建议，但如果您提供创建 table 脚本和一些数据，我会这样做。如果我在查询中有任何拼写错误，我们深表歉意。

Answer 2

类似于 Ron Ballard 的回答，但在子查询中进行舍入，并切换到显式连接语法：-

SELECT o.orders_id, 
        o.customers_id, 
        o.customers_name, 
        s.orders_status_name, 
        ot.text as order_total, 
        ot.value, 
        DATEDIFF(NOW(), payment_data_read_status) as numDaysLeft, 
        sub0.paid_amount_sum as paid_total 
FROM orders o
INNER JOIN orders_total ot ON o.orders_id = ot.orders_id
INNER JOIN orders_status s ON o.orders_status = s.orders_status_id 
INNER JOIN
(
    SELECT orders_id,
            COALESCE(SUM(op.paid_amount),0) AS paid_amount_sum,
            ROUND(COALESCE(SUM(op.paid_amount),0), 2) AS paid_amount_sum_rounded
    FROM orders_payment op 
    WHERE op.confirm_payment = '1'
    GROUP BY orders_id
) sub0
ON sub0.orders_id = o.orders_id 
WHERE ot.class = 'ot_total' 
AND s.language_id = '1' 
AND ROUND(ot.value,2) != sub0.paid_amount_sum_rounded

Answer 3

可能 233 个结果与 55 个结果的问题是测试 ... != ... 其中之一可能是 NULL。 returns NULL 被视为假，因此就好像 =.

解决这个问题的一种方法是添加

AND sub0.paid_amount_sum_rounded IS NOT NULL

需要索引：

o:  INDEX(orders_status, orders_id)  -- in this order
ot:  INDEX(class, orders_id)  -- in either order

从这个答案和其他答案中得出的结论：

使用JOIN ... ON ...语法。（这是为了清楚起见；它对性能没有影响。）
注意 LEFT JOIN 和关于可能 NULL 结果的子查询。
避免在 WHERE.
确保有 suitable 索引（以避免 table 扫描会降低性能）。
了解 'composite' 索引的好处。

如何改善查询执行时间？

How to improve query execution time?

mysql

sql

rdbms

query-optimization

查询说明

一些细节