SQL：选择n％带图片，（100-n）％不带图片

Question

我们有一个DB存储可能有图片的用户。

我在寻找一个优雅的方式在SQL中得到以下结果： 选择n个用户。在这n个用户中，例如60％应该有相关图片，40％不应该有图片。如果有少于60％的用户拥有图片，则结果应该由没有图片的用户填写。

SQL中有没有触发多个SELECT到数据库的优雅方式？

非常感谢。

Answer 1

丑陋的代码：

SELECT TOP @n * FROM 
(


     //-- We start selecting users who have a picture (ordered by HasPicture) 
     //-- If there is no more users with a picture, this query will fill the 
     //-- remaining rows with users without a picture 
     SELECT TOP 60 PERCENT * FROM tbUser 
     ORDER BY HasPicture DESC 

     UNION 

     //-- This is to make sure that we select at least 40% users without a picture 
     //-- AT LEAST because in the first query it is possible that users without a 
     //-- picture have been selected 
     SELECT TOP 40 PERCENT * FROM tblUser 
     WHERE HasPicture = 0 

     //-- We need to avoid duplicates because in the first select query we haven't 
     //-- specified HasPicture = 1 (and we didn't want to). 
     AND UserID not IN 
     (
     SELECT TOP 60 PERCENT UserID FROM tbUser 
     ORDER BY HavePicture DESC 
    ) 
) 

Answer 2

SELECT TOP(n) HasPicture --should be 0 or 1 to allow ORDER 
    FROM Users 
    ORDER BY 1 

Answer 3

所以你提供@n，是你想要的用户数。 您提供的@x是应该有图片的用户的百分比。

select top (@n) * 
from 
(
select top (@n * @x/100) * 
from users 
where picture is not null 

union all 

select top (@n) * 
from users 
where picture is null 
) u 
order by case when picture is not null then 1 else 2 end; 

所以...你想在谁拥有最多的图片@n * @x/100用户，其余的必须是没有照片的人谁。所以我在我的@ n * @ x/100图片人员和足够的其他人员之间做了一个'union all'来完成我的@n。然后我选择他们，订购我的TOP，以确保我保留有照片的人。

罗布

编辑：其实，这会更好：

select top (@n) * 
from 
(
select top (@n * @x/100) *, 0 as NoPicture 
from users 
where picture is not null 

union all 

select top (@n) *, 1 as NoPicture 
from users 
where picture is null 
) u 
order by NoPicture; 

...因为它消除了ORDER BY的影响。

Answer 4

对此类需求使用Select大小写。