Question

我有一个数据表，但我需要的结果是前几行的总和。每行代表一个日期的数据记录，我需要当前的总和。

表格：

+------+-------+--------+
| date | group | change |
+------+-------+--------+
|  0   |   a   |   1    |
+------+-------+--------+
|  0   |   b   |   2    |
+------+-------+--------+
|  1   |   a   |   3    |
+------+-------+--------+
|  1   |   b   |   4    |
+------+-------+--------+
|  2   |   a   |   5    |
+------+-------+--------+

结果：

+------+-------+-------+
| date | group |  sum  |
+------+-------+-------+
|  0   |   a   |   1   |
+------+-------+-------+
|  0   |   b   |   2   |
+------+-------+-------+
|  1   |   a   |   4   | = 1 + 3
+------+-------+-------+
|  1   |   b   |   6   | = 2 + 4
+------+-------+-------+
|  2   |   a   |   11  | = 1 + 3 + 5
+------+-------+-------+

我也需要不考虑分组的相同结果：

+------+-------+
| date |  sum  |
+------+-------+
|  0   |   3   |
+------+-------+
|  1   |   10  |
+------+-------+
|  2   |   15  |
+------+-------+

我的应用程序的另一个细节是我从每个日期/组的已知初始总和开始，即max(date)+1。我需要总和来考虑这些值。

我当前正在通过运行查询来完成此操作：

SELECT date, group, sum(change) FROM records GROUP BY date, group ORDER BY date DESC

然后在应用程序代码中，我从原始总和开始，向后遍历记录，再加上下一个日期，直到到达date_0。此解决方案对我有用，但是我想知道是否可以在SQL中完成所有这些操作？

Answer 1

您可以尝试以下-DEMO

select *,sum(change) over(partition by group order by date) as s from t
order by date

对于无组-

select date,sum(s)
from
(
select *,sum(change) over(partition by group order by date) as s from t
)A group by date

Answer 2

您似乎想要累积聚合。您可以按照以下方式进行操作：

select date, sum(sum(change)) over (order by date)
from t
group by date;

SQL查询汇总以前的结果

2 个答案: