我试图查询一些趋势统计数据，但是基准测试非常慢。查询执行时间约为134秒。
我有一个名为table_1的mysql表。
在create语句下面

CREATE TABLE `table_1` (
  `id` bigint(11) NOT NULL AUTO_INCREMENT,
  `original_id` bigint(11) DEFAULT NULL,
  `invoice_num` bigint(11) DEFAULT NULL,
  `registration` timestamp NULL DEFAULT NULL,
  `paid_amount` decimal(10,6) DEFAULT NULL,
  `cost_amount` decimal(10,6) DEFAULT NULL,
  `profit_amount` decimal(10,6) DEFAULT NULL,
  `net_amount` decimal(10,6) DEFAULT NULL,
  `customer_id` bigint(11) DEFAULT NULL,
  `recipient_id` text,
  `cashier_name` text,
  `sales_type` text,
  `sales_status` text,
  `sales_location` text,
  `invoice_duration` text,
  `store_id` double DEFAULT NULL,
  `is_cash` int(11) DEFAULT NULL,
  `is_card` int(11) DEFAULT NULL,
  `brandid` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`),
  KEY `idx_registration_compound` (`id`,`registration`)
) ENGINE=InnoDB AUTO_INCREMENT=47420958 DEFAULT CHARSET=latin1;

SELECT 

store_id,
            CONCAT('[',GROUP_CONCAT(tot SEPARATOR ','),']') timeline_transactions,
            SUM(tot) AS total_transactions,
            CONCAT('[',GROUP_CONCAT(totalRevenues SEPARATOR ','),']') timeline_revenues,
            SUM(totalRevenues) AS revenues,
            CONCAT('[',GROUP_CONCAT(totalProfit SEPARATOR ','),']') timeline_profit,
            SUM(totalProfit) AS profit,
            CONCAT('[',GROUP_CONCAT(totalCost SEPARATOR ','),']') timeline_costs,
            SUM(totalCost) AS costs



 FROM (select t1.md,
COALESCE(SUM(t1.amount+t2.revenues), 0) AS totalRevenues,
COALESCE(SUM(t1.amount+t2.profit), 0) AS totalProfit,
COALESCE(SUM(t1.amount+t2.costs), 0) AS totalCost,
COALESCE(SUM(t1.amount+t2.tot), 0) AS tot,
t1.store_id

from
(
 SELECT a.store_id,b.md,b.amount from ( SELECT DISTINCT store_id FROM  table_1) AS a
  CROSS JOIN 
 (
 SELECT
  DATE_FORMAT(a.DATE, "%m") as md,
  '0' as  amount
  from (
    select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) month as Date
    from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
    cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
    cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
  ) a
  where a.Date >='2019-01-01' and a.Date <= '2019-01-14'
  group by md) AS b 
)t1
left join
(
  SELECT
                COUNT(epl.invoice_num) AS tot,
                SUM(paid_amount) AS revenues,
                SUM(profit_amount) AS profit,
                SUM(cost_amount) AS costs,
                store_id,
                date_format(epl.registration, '%m') md
                FROM table_1 epl
                GROUP BY store_id, date_format(epl.registration, '%m')
)t2
ON   t2.md=t1.md AND t2.store_id=t1.store_id
group BY t1.md, t1.store_id) AS t3 GROUP BY store_id  ORDER BY total_transactions desc

最佳答案

大约90%的执行时间将用于执行GROUP BY store_id, date_format(epl.registration, '%m')。
不幸的是，您不能使用索引来group by派生值，因为这对您的报告至关重要，您需要预先计算它。您可以将该值添加到表中，例如使用生成的列：

alter table table_1 add md varchar(2) as (date_format(registration, '%m')) stored

alter table table_1 add md varchar(2) null;
create trigger tri_table_1 before insert on table_1
for each row set new.md = date_format(new.registration,'%m');
create trigger tru_table_1 before update on table_1
for each row set new.md = date_format(new.registration,'%m');

create index idx_table_1_storeid_md on table_1 
   (store_id, md, invoice_num, paid_amount, profit_amount, cost_amount)

      ...
            SUM(cost_amount) AS costs,
            store_id,
            md -- instead of date_format(epl.registration, '%m') md
            FROM table_1 epl
            GROUP BY store_id, md -- instead of date_format(epl.registration, '%m')
)t2   ...

  ...
        SUM(cost_amount) AS costs,
        store_id,
        md 
        FROM table_1 epl
        WHERE md = '01'
        GROUP BY store_id, md 
)t2   ...

  ...
        SUM(cost_amount) AS costs,
        store_id,
        md 
        FROM table_1 epl
        WHERE epl.registration >='2019-01-01' 
           and epl.registration <= '2019-01-14'
        GROUP BY store_id, md 
)t2   ...

关于mysql - MySQL查询太慢了，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/56139850/