将计数汇总到由另一列分组的多个新列中

Question

我试图减去两列，然后计算有多少具有相同的差异并将这些总和放入列中。总和的数量是按日期分组的-3或更多，-2，-1,0,1,2,3或更多的差异。

必须对DB2数据库执行查询。

数据...

------------------------------
| Date       | Num 1 | Num 2 |
------------------------------
| 2014-02-11 | 19872 | 19873 |
| 2014-02-11 | 19873 | 19873 |
| 2014-02-12 | 19875 | 19873 |
| 2014-02-13 | 19870 | 19873 |
| 2014-02-13 | 19872 | 19873 |
| 2014-02-14 | 19877 | 19869 |
| 2014-02-14 | 19873 | 19873 |

期望的输出......

-----------------------------------------------------------------------
| Date        | <= -3 |  -2   |  -1   |   0   |  +1   |  +2   | >= +3 |
-----------------------------------------------------------------------
| 2014-02-11  |     0 |   0   |   0   |   1   |   1   |   0   |    0  |
| 2014-02-12  |     0 |   1   |   0   |   0   |   0   |   0   |    0  |
| 2014-02-13  |     0 |   0   |   0   |   0   |   1   |   0   |    1  |
| 2014-02-14  |     1 |   0   |   0   |   1   |   9   |   0   |    0  |

Answer 1

试试这个：

select Date, 
sum(case when diff <= -3 then 1 else 0 ) AS [<=-3], 
sum(case when diff = -2 then 1 else 0 ) AS [-2], 
sum(case when diff = -1 then 1 else 0 ) AS [-1], 
sum(case when diff = 0 then 1 else 0 ) AS [0], 
sum(case when diff = 1 then 1 else 0 ) AS [+1], 
sum(case when diff = 2 then 1 else 0 ) AS [+2], 
sum(case when diff >= 3 then 1 else 0 ) AS [>=+3]
from 
(select Date, Num1, Num2, (Num1-Num2) diff from TableA)TableB
group by Date