GROUP BY 未来几个月超过 N 年

Question

我需要汇总 5 年内 12 个下个月按“水平”分组的金额：
假设我们是 2015-08-15

SUM amount from  0 to 12 next months (from 2015-08-16 to 2016-08-15)
SUM amount from 12 to 24 next months (from 2016-08-16 to 2017-08-15)
SUM amount from 24 to 36 next months ...
SUM amount from 36 to 48 next months
SUM amount from 48 to 60 next months

这是一个fiddled 数据集示例：

+----+------------+--------+
| id | date       | amount |
+----+------------+--------+
|  1 | 2015-09-01 |     10 |
|  2 | 2015-10-01 |     10 |
|  3 | 2016-10-01 |     10 |
|  4 | 2017-06-01 |     10 |
|  5 | 2018-06-01 |     10 |
|  6 | 2019-05-01 |     10 |
|  7 | 2019-04-01 |     10 |
|  8 | 2020-04-01 |     10 |
+----+------------+--------+

这是预期的结果：

+---------+--------+
| horizon | amount |
+---------+--------+
|       1 |     20 |
|       2 |     20 |
|       3 |     10 |
|       4 |     20 |
|       5 |     10 |
+---------+--------+

我怎样才能得到这 12 个月的分组“视野”？

---

我标记了 PostgreSQL，但我实际上使用的是 ORM，所以只是为了找到这个想法。 （顺便说一下，我无法使用日期格式化功能）

Answer 1

我会按 12 个月的时间范围划分并按此分组：

SELECT
  FLOOR(
      (EXTRACT(EPOCH FROM date) - EXTRACT(EPOCH FROM now()))
        / EXTRACT(EPOCH FROM INTERVAL '12 month')
    ) + 1 AS "horizon",
  SUM(amount) AS "amount"
FROM dataset 
GROUP BY horizon
ORDER BY horizon;

SQL Fiddle

灵感来自：Postgresql SQL GROUP BY time interval with arbitrary accuracy (down to milli seconds)

Answer 2

很简单：

SELECT horizon, sum(amount) AS amount
FROM generate_series(1, 5) AS s(horizon)
JOIN dataset ON "date" >= current_date + (horizon - 1) * interval '1 year'
             AND "date" < current_date + horizon * interval '1 year'
GROUP BY horizon
ORDER BY horizon;

Answer 3

也许是 CTE？

WITH RECURSIVE grps AS
(
  SELECT 1 AS Horizon, (date '2015-08-15') + interval '1' day AS FromDate, (date '2015-08-15') + interval '1' year AS ToDate
  UNION ALL
  SELECT Horizon + 1, ToDate + interval '1' day AS FromDate, ToDate + interval '1' year
  FROM grps WHERE Horizon < 5
)
SELECT 
  Horizon, 
  (SELECT SUM(amount) FROM dataset WHERE date BETWEEN g.FromDate AND g.ToDate) AS SumOfAmount
FROM 
  grps g

SQL fiddle

Answer 4

假设您需要从当前日期到明年这一天的间隔等等，我会这样查询：

SELECT 1 AS horizon, SUM(amount) FROM dataset
WHERE date > now()
AND date < (now() + '12 months'::INTERVAL)
UNION
SELECT 2 AS horizon, SUM(amount) FROM dataset
WHERE date > (now() + '12 months'::INTERVAL)
AND date < (now() + '24 months'::INTERVAL) 
UNION
SELECT 3 AS horizon, SUM(amount) FROM dataset
WHERE date > (now() + '24 months'::INTERVAL)
AND date < (now() + '36 months'::INTERVAL)
UNION
SELECT 4 AS horizon, SUM(amount) FROM dataset
WHERE date > (now() + '36 months'::INTERVAL)
AND date < (now() + '48 months'::INTERVAL)
UNION
SELECT 5 AS horizon, SUM(amount) FROM dataset
WHERE date > (now() + '48 months'::INTERVAL)
AND date < (now() + '60 months'::INTERVAL)
ORDER BY horizon;

您可以对其进行概括并使用附加变量进行类似的操作：

SELECT number AS horizon, SUM(amount) FROM dataset
WHERE date > (now() + ((number - 1) * '12 months'::INTERVAL))
AND date < (now() + (number * '12 months'::INTERVAL));

其中number 是[1,5] 范围内的整数

这是我从小提琴中得到的：

| horizon | sum |
|---------|-----|
|       1 |  20 |
|       2 |  20 |
|       3 |  10 |
|       4 |  20 |
|       5 |  10 |

Answer 5

试试这个

select 
id,
sum(case when date>=current_date and date<current_date+interval 1 year then amount else 0 end) as year1,
sum(case when date>=current_date+interval 1 year and date<current_date+interval 2 year then amount else 0 end) as year2,
sum(case when date>=current_date+interval 2 year and date<current_date+interval 3 year then amount else 0 end) as year3,
sum(case when date>=current_date+interval 3 year and date<current_date+interval 4 year then amount else 0 end) as year4,
sum(case when date>=current_date+interval 4 year and date<current_date+interval 5 year then amount else 0 end) as year5
from table
group by id

Answer 6

你需要一个联合和一个聚合函数：

select 1 as horizon, 
       sum(amount) amount
from the_table
where date >= current_date 
  and date < current_date + interval '12' month
union all
select 2 as horizon, 
       sum(amount) amount
where date >= current_date + interval '12' month
  and date < current_date + interval '24' month
union all 
select 3 as horizon, 
       sum(amount) amount
where date >= current_date + interval '24' month
  and date < current_date + interval '36' month
... and so on ...

但我不知道如何使用混淆层（又名 ORM）来做到这一点，但我确信它支持（或者它应该）聚合和联合。

这可以很容易地封装到一个 PL/PgSQL 函数中，您可以在其中传递“地平线”，并且 SQL 是动态构建的，因此您需要调用的只是：select * from sum_horizon(5) 其中5 表示年。

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^{顺便说一句：date 是一个可怕的列名称。一方面是因为它是一个保留字，但更重要的是因为它没有记录该列的含义。这是“发布日期”吗？ “截止日期”？ “订单日期”？}