Question

下面是我的代码的副本，我的主要目标是根据其photo_id获取某些图像的等级。下面的代码工作得很好，但问题是，如果有总关系点，那么这两个图像得到相同的排名。我真正想要的是，如果这些图像具有相同的总分，我想按日期对它们进行排名。我怎么能做到这一点？

SELECT  uo.*, 
    ( SELECT  COUNT(DISTINCT ui.total_points)
    FROM    photo_list ui
            WHERE   ui.total_points>= uo.total_points
            ) AS rank
    FROM    photo_list uo
    WHERE photo_id = :photo_id

以下是我的表格：

id  | photo_id | total_points | date
1      432cfc        0          1403076412
2      fsd324        1          1403076413
3      gd43ds        0          1403076411

这是我的愿望结果：

   rank  | photo_id | total_points | date
    1      fsd324        1          1403076413
    2      gd43ds        0          1403076411
    3      432cfc        0          1403076412

Answer 1

select * from (
    SELECT  uo.*, 
    (SELECT  COUNT(DISTINCT ui.total_points)
    FROM    photo_list ui
    WHERE   ui.total_points>= uo.total_points
    ) AS rank
    FROM    photo_list uo
    WHERE photo_id = :photo_id
    ) as alias
order by  total_points,date;

Answer 2

这将是查询。

SELECT   
    ( SELECT  COUNT(DISTINCT ui.total_points)
    FROM    photo_list ui
        WHERE   ui.total_points>= uo.total_points
        ) AS rank,uo.photo_id,uo.total_points,uo.date1
    FROM    photo_list uo
    ORDER BY rank ASC, date1 ASC

排名不重复

2 个答案: