Sql使用两个聚合函数进行透视

Question

我在下面提到的sql，主表中的问题下陷入困境

|PLACE| DATE_1  |ID | FQ | PLAN|
---------------------------------
Alpha |01-03-16 | A | 100| YES |
Alpha |01-03-16 | A | 50 | YES |
Alpha |02-03-16 | A | 10 | YES |
BETA  |02-03-16 | C | 80 | YES |
BETA  |02-03-16 | D | 9  | YES |
BETA  |03-03-16 | D | 11 | YES |
BETA  |03-03-16 | E | 13 | YES |
--------------------------------

但需要以下格式输出

        |Sum of FQ | Plan      |Sum of FQ  | Plan      |
| Place |ID |01-03-16  | 01-03-16  | 02-03-16  | 02-03-16  | 
| Alaph |A  |100       | YES       | 10        | NO        |
| Alaph |B  |50        | YES       |           |           |
| Beta  |C  |          |           | 80        | NO        |

Answer 1

根据输入数据，您的输出没有意义。我猜你想要条件聚合：

select place, id,
       max(case when date_1 = '2016-03-01' then fq end) as fq_20160301,
       max(case when date_1 = '2016-03-01' then plan end) as plan_20160301,
       max(case when date_1 = '2016-03-02' then fq end) as fq_20160302,
       max(case when date_1 = '2016-03-02' then plan end) as plan_20160302
from t
group by place, id;

请注意，这使用ISO标准日期常量。如果您将日期存储为字符串（这是一个非常糟糕的主意），那么您需要调整数据的逻辑。