计算每个月的运行总数

Question

假设我有这张桌子：

id; date; units
1; Jan 1; 1
2; Jan 2; 4
3; Feb 9; 6
4; Mar 1; 1
5; Mar 4; 2

我现在如何相应地计算“accumulated_month”列？计算应该在每个新的月份重新开始。

id; date; units; accumulated_month
1; Jan 1; 1; 1
2; Jan 2; 4; 5
3; Feb 9; 6; 6
4; Mar 1; 1; 1
5; Mar 4; 2; 3

我得到的只是这个：

id; date; units; accumulated_month
1; Jan 1; 1; 5
2; Jan 2; 4; 5
3; Feb 9; 6; 6
4; Mar 1; 1; 3
5; Mar 4; 2; 3

Answer 1

create table t (id int, date date, units int);
insert into t (id, date, units) values
(1, '2016-01-01', 1),
(2, '2016-01-02', 4),
(3, '2016-02-09', 6),
(4, '2016-03-01', 1),
(5, '2016-03-04', 2);

不清楚您是想要月份总计还是月份内的运行总计。当月总计：

select
    id, date, units,
    sum(units) over (partition by date_trunc('month', date)) as acumm
from t
order by 1,2
;
 id |    date    | units | acumm 
----+------------+-------+-------
  1 | 2016-01-01 |     1 |     5
  2 | 2016-01-02 |     4 |     5
  3 | 2016-02-09 |     6 |     6
  4 | 2016-03-01 |     1 |     3
  5 | 2016-03-04 |     2 |     3

如果您想要一个月内的运行总计，则将 order by 添加到窗口函数：

select
    id, date, units,
    sum(units) over (
        partition by date_trunc('month', date)
        order by id
    ) as acumm
from t
order by 1,2
;
 id |    date    | units | acumm 
----+------------+-------+-------
  1 | 2016-01-01 |     1 |     1
  2 | 2016-01-02 |     4 |     5
  3 | 2016-02-09 |     6 |     6
  4 | 2016-03-01 |     1 |     1
  5 | 2016-03-04 |     2 |     3

Answer 2

运行这个：

SELECT
    t1.id, 
    t1.date, 
    t1.units,
    SUM(t2.units) accumulated_month
FROM t t1 
    JOIN t t2 
        ON date_trunc('month', t1.date) = date_trunc('month', t2.date)
            AND t2.date <= t1.date
GROUP BY t1.id, t1.date, t1.units
ORDER BY t1.id

编辑：

简化 SQL：

SELECT
    t1.*, SUM(t2.units) accumulated_month
FROM t t1 
    JOIN t t2 
        ON date_trunc('month', t1.date) = date_trunc('month', t2.date)
            AND t2.date <= t1.date
GROUP BY t1.id
ORDER BY t1.id

逻辑

背后的逻辑是JOIN 与自己的表。然后对于t1JOIN中的每一行t2中满足两者的所有行（AND）：

1. 同年/同月
2. t2.date

有了这些ON 约束，t1 中的每一行将是 JOIN，到目前为止该月累计。所以如果没有分组，你会得到类似的东西：

╔════╦════════════╦═══════╦════════════╦══════════╗
║ id ║    date    ║ units ║  t2_date   ║ t2_units ║
╠════╬════════════╬═══════╬════════════╬══════════╣
║  1 ║ 2016-01-01 ║     1 ║ 2016-01-01 ║        1 ║
║  2 ║ 2016-01-02 ║     4 ║ 2016-01-01 ║        1 ║
║  2 ║ 2016-01-02 ║     4 ║ 2016-01-02 ║        4 ║
║  3 ║ 2016-02-09 ║     6 ║ 2016-02-09 ║        6 ║
║  4 ║ 2016-03-01 ║     1 ║ 2016-03-01 ║        1 ║
║  5 ║ 2016-03-04 ║     2 ║ 2016-03-01 ║        1 ║
║  5 ║ 2016-03-04 ║     2 ║ 2016-03-04 ║        2 ║
╚════╩════════════╩═══════╩════════════╩══════════╝

在GROUP BY t1.id 之后，您可以通过SUM(t2.units) 获得您所期望的。