各种类别的百分比计算，以满足的条件与SQL Server

Question

我有一个关于城市看起来像这样之间的航班信息的表：

    origin_city dest_city   time    
    Dothan AL   Atlanta GA    171    
    Dothan AL   Atlanta GA    171    
    Dothan AL   Elsewhere AL    2    
    Dothan AL   Elsewhere AL    2    
    Dothan AL   Elsewhere AL    2    
    Boston MA   New York NY     5    
    Boston MA   City MA         1    
    New York NY Boston MA       5    
    New York NY Boston MA       5    
    New York NY Boston MA       5    
    New York NY Poughkipsie NY  2

我想找到，每个起点城市，是不到3个小时长途飞行的百分比。所以结果会是这样：

    Dothan AL    60
    Boston MA    50
    New York NY  25

我以为会工作的代码如下所示：

     SELECT F.origin_city as origin_city,    
       ((SELECT COUNT(*) FROM Flights as F2
       WHERE F2.actual_time < 3) / (SELECT COUNT(*) FROM Flights as  F3)) * 100
     AS percentage
     FROM Flights as F
     GROUP BY F.origin_city
     ORDER BY percentage;
     GO

当我运行它，我得到原产地城市的列表和百分比列，符合市场预期，但该比例始终为0。我仍然很困惑子查询（因为你可以看到）。

Answer 1

我会用AVG()作为窗口函数做到这一点：

SELECT F.origin_city as origin_city, 
       AVG( CASE WHEN F2.actual_time < 3 THEN 100.0 ELSE 0 END) as percentage
FROM Flights F
GROUP BY F.origin_city
ORDER BY percentage;

这假定该时间以小时测定。据谷歌地图，可以在68小时内从Dothan核心步行到亚特兰大，所以171是可疑的。

Answer 2

您的百​​分比超过了整个表，而不是由起点城市群。尝试是这样的：

 SELECT F.origin_city as origin_city,    
   (SUM(CASE WHEN F.actual_time < 3 THEN 1 ELSE 0 END) / COUNT(*) )  * 100 AS percentage
 FROM Flights as F
 GROUP BY F.origin_city
 ORDER BY percentage;
 GO

FWIW与您当前的子查询的问题是，你有没有你的当前行和子查询中的数据之间的连接。你也许可以把它改写为这样的：

 SELECT F.origin_city as origin_city,    
 ((SELECT COUNT(*) FROM Flights as F2
 WHERE F2.origin_city = F.origin_city and F2.actual_time < 3) / (SELECT COUNT(*) FROM Flights as  F3 where F3.origin_city = F.origin_city)) * 100
 AS percentage
 FROM Flights as F
 GROUP BY F.origin_city
 ORDER BY percentage;
 GO

但它是一种不必要的重新查询表时，你已经有足够的数据做你的计算，如上图所示的每一行。