我有下表：
date | shop_id | package_id | quantity
01/01/2017 | 1 | 1 | 1
01/01/2017 | 1 | 2 | 5
01/01/2017 | 2 | 1 | 10
01/01/2017 | 2 | 2 | 4
02/01/2017 | 1 | 1 | 14
02/01/2017 | 1 | 2 | 3
....
我想将具有相同商店和相同包装的所有先前记录的数量相加，如下所示
date | shop_id | package_id | sum_quantity
01/01/2017 | 1 | 1 | 1
01/01/2017 | 1 | 2 | 5
01/01/2017 | 2 | 1 | 10
01/01/2017 | 2 | 2 | 4
02/01/2017 | 1 | 1 | 15
02/01/2017 | 1 | 2 | 8
....
答案 0 :(得分：4)
这是基于shop_id和package_id的“运行总和”：
select "date", shop_id, package_id,
sum(quantity) over (partition by shop_id, package_id order by "date") as sum_quantity
from the_table
order by "date";
sum()用作window function（通过over()部分）。窗口定义中的order by使其成为“运行”总和，而不是“总体”总和。