Question

这是我的查询，但是我无法成功显示pages的{{1}}高于列的平均值。

我在这里缺少什么？

percent_diff

Answer 1

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
)

Answer 2

有几种方法可以做到这一点。使用您的基本查询最快（写入）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

替代方案你可以获得avg然后使用if语句。你仍然会经历两次表（一次得到平均值，一次一次计算两列）

如何统计mySQL中高于平均水平的行？

2 个答案: