Question

我有以下查询，我正在尝试重写它以提高性能，我可以使用什么方法来重写它。

select 
    notes.id, notes.name, notes.parent_type, notes.contact_id from notes 
JOIN 
    ( 
    SELECT contact_id as id from accounts_contacts where account_id = 'acct1876' and deleted = '0' union 
    SELECT quote_id as id from quotes_accounts where account_id = 'acct1876' and deleted = '0' union 
    SELECT opportunity_id as id from accounts_opportunities where account_id = 'acct1876' and deleted = '0' union 
    SELECT leads.id as id from leads where account_id = 'acct1876' and deleted = '0' union 
    SELECT  project_id as id from projects_accounts where account_id = 'acct1876' and deleted = '0' union 
    select 'acct1876' as id 
    ) A 
    ON A.id = notes.parent_id and deleted = '0' OR contact_id in 
    ( SELECT contact_id from accounts_contacts where account_id = 'acct1876' and deleted = '0' ) and deleted = '0' 

    group by notes.id;

Answer 1

首先，你的最终OR是内连接开始的副本，否则毫无意义。

这部分

  ON A.id = notes.parent_id 
  and deleted = '0' 
  OR contact_id in ( SELECT contact_id 
                       from accounts_contacts 
                       where account_id = 'acct1876' 
                       and deleted = '0' ) 
  and deleted = '0'

可以

  ON A.id = notes.parent_id

接下来，您似乎正在尝试获取与给定帐户相关联的所有ID，包括相关帐户。我会确保每个表都有一个关于帐户ID和已删除列的索引。此外，对于此查询，我将它作为DISTINCT以防止重复项连接到notes表。然后我会交换订单（对我来说，在心理上查询你想要的ID，然后获得相关的注释）。下面是UNION查询的每个表的索引，以及加入的父ID列的notes表。

table                  index
accounts_contacts      (account_id, deleted, contact_id)
quotes_accounts        (account_id, deleted, quote_id )
accounts_opportunities (account_id, deleted, opportunity_id )
leads                  (account_id, deleted, id
projects_accounts      (account_id, deleted, project_id )
notes                  (parent_id)

现在，次要更新查询

select 
      notes.id, 
      notes.name, 
      notes.parent_type, 
      notes.contact_id 
   from 
         (SELECT DISTINCT contact_id as id 
             from accounts_contacts 
             where account_id = 'acct1876' and deleted = '0' 
          union 
          SELECT quote_id as id 
             from quotes_accounts 
             where account_id = 'acct1876' and deleted = '0' 
          union 
          SELECT opportunity_id as id 
             from accounts_opportunities 
             where account_id = 'acct1876' and deleted = '0' 
          union 
          SELECT leads.id as id 
             from leads 
             where account_id = 'acct1876' and deleted = '0' 
          union 
          SELECT project_id as id 
             from projects_accounts 
             where account_id = 'acct1876' and deleted = '0' 
          union 
          select 'acct1876' as id ) A 
         JOIN Notes
            ON A.id = notes.parent_id 
   group by 
     notes.id;

正如有人提到的那样，你有一个分组依据，但是没有总和或聚合的列，这将导致第一个条目被包括在内，并且由于出现了自动递增的ID列，所以无论帐户“ID”来自何处，都具有相同的价值。

Answer 2

为了回答你的问题并提升表现，我想建议一种看似奇怪的方法。

如果您有合理数量的记录，请使用MORE SELECTS，而不是单个（和重）记录。您可以在数组中收集ID，或者（最好）在字符串中收集ID，然后使用逗号等分隔ID列表进行另一个查询。这（虽然听起来完全是胡说八道）有许多优点，几乎没有任何缺点：单独的SELECT运行速度快，然后让数据库服务器呼吸。当您执行一个SELECT时，其他表将不会被锁定（!!!），而一个大的select将在整个查询时锁定所有涉及的表。因此，这样做更容易，更易读：

$idlist = fetch_idlist("select id from users where name like 'John%'");
$result = fetch_all("select * from mails where userid in ($idlist)");

比这个：

$result = fetch_all("select * from mails left join users on users.id=mails.userid ....")

我希望尽管功能不存在，但意义很明显。这只是原则。所以在你的情况下，也许你想选择联系人，构建id列表，然后对笔记等单独查询，并在php中编写最终结果。

同样，这只适用于低于百万的记录数，你不能以非常高的数量吃掉所有内存。但关键在于：在高负载时，分别耗费时间的查询和让其他进程适合介于两者之间（相对）长时间锁定一大堆表无比好。

很抱歉，如果不是100％的答案，但我认为值得解释。

如何重写MySQL查询

2 个答案: