提高mysql中的查询性能

Question

我使用过这个查询

SELECT  COUNT(CASE WHEN C <= 1 THEN 1 END) AS nooffamiliesHavingcount1,
        COUNT(CASE WHEN C BETWEEN 2 AND 4 THEN 1 END) AS nooffamiliesHavingcountbetween2And4,
        COUNT(CASE WHEN C > 4 THEN 1 END) AS nooffamiliesHavingcountgreaterthan3
FROM    (   SELECT  COUNT(*) AS C
            FROM    user where user_id = (select user_id from location where location_id in(select location_id from country where state_name='STATE'))
            GROUP BY House_No
        ) t

这里的子查询返回大约10000条记录。 用户表有10,00,000条记录。 这花费了太多时间。然后它说错误是服务器消失了。 我正在使用mysql。

我从谷歌搜索过。但对我来说没有运气。

我需要对我的表进行哪些更改。如何通过提高查询性能来成功执行此查询。请提示我。谢谢提前....

Answer 1

试试这个查询

SELECT 
  COUNT(CASE WHEN C <= 1 THEN 1 END) AS nooffamiliesHavingcount1,
  COUNT(CASE WHEN C BETWEEN 2 AND 4 THEN 1 END) AS nooffamiliesHavingcountbetween2And4,
  COUNT(CASE WHEN C > 4 THEN 1 END) AS nooffamiliesHavingcountgreaterthan3
FROM  
  (SELECT 
    COUNT(*) AS C
  FROM 
    user u,  
    location l, 
    country c 
  where 
    l.state_name='STATE' AND 
    l.some_other_column_id= 4 AND  <------- Add your condition
    c.location_id = l.location_id AND 
    u.user_id = l.user_id 
  GROUP BY 
    u.House_No) t

使用正确的连接，因为它很容易理解..

SELECT 
  COUNT(CASE WHEN C <= 1 THEN 1 END) AS nooffamiliesHavingcount1,
  COUNT(CASE WHEN C BETWEEN 2 AND 4 THEN 1 END) AS nooffamiliesHavingcountbetween2And4,
  COUNT(CASE WHEN C > 4 THEN 1 END) AS nooffamiliesHavingcountgreaterthan3
FROM  
  (SELECT 
    COUNT(*) AS C
  FROM 
    user u
  INNER JOIN  
    location l
  ON 
    l.state_name='STATE' AND 
    l.some_other_column_id= 4   <------- Add your condition
    u.user_id = l.user_id 
  INNER JOIN
    country c 
  ON 
    c.location_id = l.location_id 
  GROUP BY 
    u.House_No) t

EDITED

在大多数情况下，JOIN比子查询更快，并且子查询的速度非常快。 我接受使用子查询更合乎逻辑且易于理解，但是当它涉及性能时，它不如连接那么好。 如果您正在使用连接，您的数据库将自行优化您的查询，这不是子查询的情况。 尝试对您的两个查询使用explain，您将清楚地了解查询是如何执行的。

希望这可以帮助...

Answer 2

你能尝试下面的：

SELECT COUNT（当COUNT（ ）<= 1那么1结束时的情况）作为nooffamiliesHavingcount1，COUNT（在 2和4之间COUNT（ ）的情况下1结束）作为nooffamiliesHavingcount Between2And4，COUNT（COE（*）> 4那么结束时的情况） AS nooffamiliesHavingcountgreaterthan3 FROM用户用户内部联接（从loc loc选择user_id，在loc.location_id = con.location_id，其中state_name ='STATE'）

在House_No上的user.user_id = temp.user_id组中