如何統計mySQL中高於平均水平的行？

Question

這是我的查詢，但是我無法成功顯示其高於列平均值的pages的數量。

我在這裏錯過了什麼？

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

Answer 1

有幾種方法可以做到這一點。與您的基本查詢最快（寫）將是這樣的：

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後使用if語句。你還是會去通過表兩次（一次得到平均，一次做對兩列算一次）

Answer 2

我認爲這將是解決方案：

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