Question

我想返回数据库中“持续时间”列中平均值的20％以内的值。

我想建立在下面的代码上，但是我不想返回其中Duration小于持续时间平均值的值，而是返回所有在AVG（Duration）值的20％之内的值。

Select * From table
Where Duration < (Select AVG(Duration) from table)

Answer 1

这是一种方法...

Select * From table
Where Duration between (Select AVG(Duration)*0.8 from table)
and (Select AVG(Duration)*1.2 from table)

也许这样可以避免重复扫描：

with cte as ( Select AVG(Duration) as AvgDuration from table )
Select * From table
Where Duration between (Select AvgDuration*0.8 from cte)
and (Select AvgDuration*1.2 from cte)

或

Select table.* From table
cross join ( Select AVG(Duration) as AvgDuration from table ) cj
Where Duration between cj.AvgDuration*0.8 and cj.AvgDuration*1.2

或使用窗口功能：

Select d.* 
from (
     SELECT table.*
        , AVG(Duration) OVER() as AvgDuration
     From table
     ) d
Where d.Duration between d.AvgDuration*0.8 and d.AvgDuration*1.2

最后一个可能是最有效的方法。

在SQL中找到平均值的百分比（％）范围

1 个答案: