如何统计mySQL中高于平均水平的行？

Question

这是我的查询，但是我无法成功显示其高于列平均值的pages的数量。

我在这里错过了什么？

SELECT count(page) from 
(
select * FROM pages_diff where company_id = 3 
group by page 
having percent_diff > avg(percent_diff) 
) nested 

Answer 1

有几种方法可以做到这一点。与您的基本查询最快（写）将是这样的：

SELECT above_avg,below_avg from 
(
select count(page) as above_avg FROM pages_diff where company_id = 3 
group by page 
having percent_diff > avg(percent_diff) 
) nested_above 
, 
(
select count(page) as below_avg FROM pages_diff where company_id = 3 
group by page 
having percent_diff < avg(percent_diff) 
) nested_below 

替代你可以只是得到平均然后使用if语句。你还是会去通过表两次（一次得到平均，一次做对两列算一次）

Answer 2

我认为这将是解决方案：

SELECT count(page) FROM pages_diff 
WHERE company_id = 3 AND percent_diff > (
    SELECT avg(percent_diff) FROM pages_diff 
    WHERE company_id = 3 
)