当我使用group by … with rollup运行查询时：

select a, b, sum(c) 
from <table> 
group by a, b with rollup;

我在(我认为是)查询的PK(即group-by列)中获得了重复的行：

+------+------+--------+
| a    | b    | sum(c) |
+------+------+--------+
| NULL | NULL |     13 |
| NULL |    1 |      4 |
| NULL |    3 |      8 |
| NULL |    4 |      9 |
| NULL | NULL |     34 |
|    1 |    3 |     17 |
|    1 |    4 |   NULL |
|    1 |   17 |      2 |
|    1 | NULL |     19 |
|    2 | NULL |      6 |
|    2 |    1 |     17 |
|    2 |    3 |     17 |
|    2 | NULL |     40 |
|    4 |   17 |      2 |
|    4 | NULL |      2 |
|    5 | NULL |     11 |
|    5 |    6 |      7 |
|    5 | NULL |     18 |
|   13 |    4 |      2 |
|   13 | NULL |      2 |
|   14 |   41 |      3 |
|   14 | NULL |      3 |
|   18 |    1 |      2 |
|   18 | NULL |      2 |
|   41 |    2 |     17 |
|   41 | NULL |     17 |

…更多行跟随…

我如何区分(NULL,NULL,13)(NULL,NULL,34)？也就是说,我如何区分由于底层数据而具有空值的行和具有空值的行,因为它是由汇总添加的？ (请注意,还有更多示例 – (2,NULL,6)和(2,NULL,40))

解决方法:

好问题.我能想到的一个选择就是这样做：

select COALESCE(a, -1) AS a, COALESCE(b, -1) AS b, sum(c) 
from <table> 
group by COALESCE(a, -1), COALESCE(b, -1) with rollup;