如何优化这个SQL查询,涉及200万行数据。

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英文:

How to optimize this sql query on + 2 million rows

问题

我有一个包含200多万行数据并且在快速增长的SQL数据库。列不多,只有code, price, date和stationID

目标是按code和stationID获取最新的价格。
查询功能正常,但需要超过10秒。

有没有优化查询的方法?

$statement = $this->pdo->prepare(
    'WITH cte AS 
    (
        SELECT stationID AS ind, code, CAST(price AS DOUBLE) AS price, date,
        ROW_NUMBER() OVER (
            PARTITION BY code, stationID
            ORDER BY date DESC
        ) AS latest
        FROM price
    )
    SELECT *
    FROM cte
    WHERE latest = 1'
);

$statement->execute();
$results = $statement->fetchAll(PDO::FETCH_GROUP | PDO::FETCH_ASSOC);

编辑:
第一列上有一个名为'id'的索引。我不知道这是否有帮助。

数据库(InnoDB)如下所示:

  • id primary - int
  • stationID - int
  • code - int
  • price - decimal(10,5)
  • date - datetime

编辑2:
结果需要按stationID分组,每个stationID需要显示多行。每个code都显示最新的日期,如下所示:

22456:
    code: 1
    price: 3
    date: 2023-06-21

    code: 2
    price: 2
    date: 2023-06-21

    code: 3
    price: 5
    date: 2023-06-21

22457:
    code: 1
    price: 10
    date: 2023-06-21

    code: 2
    price: 1
    date: 2023-06-21

    code: 3
    price: 33
    date: 2023-06-21

JSON输出应如下所示:

{"1000001":[{"code":1,"price":1.661,"date":"2023-06-06 12:46:32","latest":1},{"code":2,"price":1.867,"date":"2023-06-06 12:46:32","latest":1},{"code":3,"price":1.05,"date":"2023-06-06 12:46:32","latest":1},{"code":5,"price":1.818,"date":"2023-06-06 12:46:32","latest":1},{"code":6,"price":1.879,"date":"2023-06-06 12:46:32","latest":1}],
"1000002":[{"code":1,"price":1.65,"date":"2023-06-03 08:53:26","latest":1},{"code":2,"price":1.868,"date":"2023-06-03 08:53:26","latest":1},{"code":6,"price":1.889,"date":"2023-06-03 08:53:27","latest":1}],...}
英文:

I have a Sql database with +2 million rows and growing fast. There aren't many columns, only code, price, date and stationID.

The aim is to get the latest price by code and stationID.
The query works well, but takes more than 10s.

Is there a way to optimize the query?

 $statement = $this->pdo->prepare(

        'WITH cte AS 
        (

        SELECT stationID AS ind, code, CAST(price AS DOUBLE ) AS price, date

        ,ROW_NUMBER() OVER(
             PARTITION BY code, stationID
             ORDER BY date DESC
         ) AS latest

        FROM price
        )
        
        SELECT *
        FROM cte
        WHERE latest  = 1
        '
    );

    $statement->execute();
    $results = $statement->fetchAll(PDO::FETCH_GROUP | PDO::FETCH_ASSOC);

Edit:
There is an index on the first column called 'id'. I don't know if this helps.

The database (InnoDB) look like this:

id primary - int
stationID - int
code - int
price - decimal(10,5)
date - datetime

Edit 2:

The result need to be grouped by stationID and each stationID need to show many line. One line for every code with latest date.
Like this:

22456:
      code:  1
      price: 3
      date:  2023-06-21

      code:  2
      price: 2
      date:  2023-06-21

      code:  3
      price: 5
      date:  2023-06-21

22457:
      code:  1
      price: 10
      date:  2023-06-21

      code:  2
      price: 1
      date:  2023-06-21

      code:  3
      price: 33
      date:  2023-06-21

The json output should be like this:

{"1000001":[{"code":1,"price":1.661,"date":"2023-06-06 12:46:32","latest":1},{"code":2,"price":1.867,"date":"2023-06-06 12:46:32","latest":1},{"code":3,"price":1.05,"date":"2023-06-06 12:46:32","latest":1},{"code":5,"price":1.818,"date":"2023-06-06 12:46:32","latest":1},{"code":6,"price":1.879,"date":"2023-06-06 12:46:32","latest":1}],"1000002":[{"code":1,"price":1.65,"date":"2023-06-03 08:53:26","latest":1},{"code":2,"price":1.868,"date":"2023-06-03 08:53:26","latest":1},{"code":6,"price":1.889,"date":"2023-06-03 08:53:27","latest":1}],… 

答案1

得分: 1

只要在相同的 code, stationID 对中不能有相同日期的两行,使用窗口函数就有点像用大锤来砸坚果。

select p.stationID, p.code, p.price, p.date
from (
    select code, stationID, max(date) as max_date
    from price
    group by code, stationID
) max
join price p
    on max.code = p.code
   and max.stationID = p.stationID
   and max.max_date = p.date;

需要以下索引:

alter table price add index (code, stationID, date desc);

此查询应该在不到1毫秒内完成,因为派生表可以从索引中构建,然后只需从表中读取所需的行。

或者,如果您知道每个 code, stationID 对都会在特定时间段内(1小时、1天、1周)收到更新的价格,那么您可以通过向窗口函数添加一个 where 子句来显著减少窗口函数所需的工作量:

with cte as 
(
    select stationID as ind, code, price, date, row_number() over(partition by code, stationID order by date desc) as latest
    from price
    where date >= now() - interval 1 week
)
select * from cte where latest  = 1;
英文:

As long as you cannot have two rows with the same datetime for the same code, stationID pair, using the window function is a bit like using a sledgehammer to crack a nut.

select p.stationID, p.code, p.price, p.date
from (
    select code, stationID, max(date) as max_date
    from price
    group by code, stationID
) max
join price p
    on max.code = p.code
   and max.stationID = p.stationID
   and max.max_date = p.date;

It requires the following index:

alter table price add index (code, stationID, date desc);

This query should take less than 1 ms, as the derived table can be built from the index, and then it is reading just the required rows from the table.

Alternatively, if you know that every code, stationID pair will have received an updated price within a specific time period (1 hr, 1 day, 1 week), then you could significantly reduce the work required by the window function by adding a where clause to it:

with cte as 
(
    select stationID as ind, code, price, date, row_number() over(partition by code, stationID order by date desc) as latest
    from price
    where date >= now() - interval 1 week
)
select * from cte where latest  = 1;

答案2

得分: 0

我会想象你需要以下索引,以使你的查询性能良好(你只需要在数据库设计的一部分中执行一次)。

CREATE INDEX IX ON price
  (code, stationID, date DESC, price)

前两列可以以任何顺序排列。

英文:

I would imagine you need the following index for your query to perform well (you only need to do this once, as part of your database design).

CREATE INDEX IX ON price
  (code, stationID, date DESC, price)

The first two columns can be in either order.

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  • 本文由 发表于 2023年6月6日 03:56:02
  • 转载请务必保留本文链接:https://go.coder-hub.com/76409615.html
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