Question

下面是我桌子的复制品。如下所示，该行按ID和日期排序。

id     |        date         |    points   
1          Jan 1, 2014             110
2          Feb 12, 2014            20
3          Mar 23, 2014            40
4          Apr 10, 2014            80
5          May 20, 2014            30
6          June 3, 2014            60
7          Jan 15, 2014            10
8          Mar 5, 2014             70
9          Feb 8, 2014             50
10         June 10, 2014            5

我的目标是按50到10之间的点（从最高到最低）订购行，并且到目前为止也是最短的。我怎样才能做到这一点？

必须包含所有行作为最终结果

Answer 1

类似的东西：

SELECT *, CASE WHEN (points >= 10 AND points <= 50) THEN 1 ELSE 0 END AS order_points     
    FROM points
ORDER BY
    order_points DESC, 
    CASE 
        WHEN order_points = 1 THEN points
    END DESC,  DATEDIFF( date, NOW() )

Answer 2

有点猜测：

select a_date,points 
from t 
order by case when points between 10 and 50 then -1 else 1 end * points
   , abs(DATEDIFF( a_date, now()));

对于您的数据的子集：

+------------+--------+
| a_date     | points |
+------------+--------+
| 2014-03-23 |     40 |
| 2014-02-12 |     20 |
| 2014-06-10 |      5 |
| 2014-04-10 |     80 |
| 2014-01-01 |    110 |
+------------+--------+

Answer 3

SELECT *
FROM MyTable
order by CASE WHEN Points BETWEEN 10 AND 50 THEN 0 ELSE 1 END, TheDate desc;

ID  TheDate    Points
--  ---------  ------
 5  20-MAY-14      30
 3  23-MAR-14      40
 2  12-FEB-14      20
 9  08-FEB-14      50
 7  15-JAN-14      10
10  10-JUN-14       5
 6  03-JUN-14      60
 4  10-APR-14      80
 8  05-MAR-14      70
 1  01-JAN-14     110

按点和日期排序表

3 个答案: