sql-server - SQL: SELECT n% 有图片，(100-n)% 没有图片

Question

我们有一个数据库来存储可能有图片的用户。

我正在寻找一种优雅的 SQL 方法来获得以下结果：选择 n 个用户。在这 n 个用户中，例如 60% 的用户应该有相关的图片，而 40% 的用户不应该有图片。如果少于 60% 的用户有图片，则结果应该由没有图片的用户填充。

在 SQL 中是否有一些优雅的方式，而无需向数据库触发多个 SELECT？

非常感谢你。

Answer 1

因此，您提供@n，即您想要的用户数。您提供的@x 是那些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...您最多需要@n * @x / 100 个有图片的用户，其余的必须是没有图片的人。所以我在我的@n*@x/100 图片人和足够多的其他人之间做一个“联合”来完成我的@n。然后我选择他们回来，订购我的 T​​OP 以确保我保留有照片的人。

抢

编辑：实际上，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 2

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

Answer 3

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 4

对此类需求使用 Select 案例。