Question

我使用了这个查询

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

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

我是从谷歌搜索的。但对我来说没有运气。

我需要对我的表进行哪些更改。如何成功执行此查询提高查询性能。请建议我。提前谢谢......

Answer 1

尝试此查询

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

<强> 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的执行方式。

希望这会有所帮助......

Answer 2

你可以尝试下面的内容：

SELECT COUNT（COUNT（）＆lt; = 1 THEN 1 END时的情况）AS nooffamiliesHavingcount1， COUNT（在2和4之间计数（）的情况下，然后结束2）作为nooffamiliesHavecount Between2And4， COUNT（COUNT（*）＆gt; 4结束时的情况）作为nooffamiliesHavingcountgreaterthan3 来自用户用户内部联接（从位置loc中选择user_id Inner Join country con 在loc.location_id = con.location_id，其中state_name ='STATE'）为temp

在House_No

上的user.user_id = temp.user_id组

提高mysql中的查询性能

2 个答案: