由isoweek分组 - postgresql

Question

这是基于前一个问题的子问题：

Split values from an interval and group by isoweek - Postgresql

问题是如何通过isoweek

对此进行分组

CREATE TABLE task
    (id int4, start date, stop date, hr int4);    
INSERT INTO task
    (id, start, stop, hr)
VALUES
    (1, '2017-01-01','2017-01-31', 80),
    (2, '2017-01-01','2017-02-28', 120);

根据Patricks的回答，我找到了这个解决方案：

SELECT id,to_char(iso, 'iyyy-iw'),(hr/weeks)::numeric (5,2) as hr_week 
FROM (SELECT id,hr,generate_series(start,stop,interval '1 week') as iso, 
(stop - start)/7 as weeks FROM task) as sub

http://sqlfiddle.com/#!15/93ee1/78

下一步是“分组”，如下所示：

2016-52 35
2017-01 35
2017-02 35
2017-03 35
2017-04 35
2017-05 15
2017-06 15
2017-07 15
2017-08 15

我无法弄清楚如何做到这一点。任何帮助表示赞赏。

TIA，

Answer 1

SQL Fiddle

PostgreSQL 9.3架构设置：

CREATE TABLE task
    (id int4, start date, stop date, hr int4);    
INSERT INTO task
    (id, start, stop, hr)
VALUES
    (1, '2017-01-01','2017-01-31', 80),
    (2, '2017-01-01','2017-02-28', 120);

查询1 ：

SELECT
     to_char(iso, 'iyyy-iw')  as YYY_WK
   , max(weeks) as weeks
   , sum((hr/weeks)::numeric (5,2)) as hr_week 
FROM (
    SELECT
       id
     , hr
     , generate_series(start,stop,interval '1 week') as iso
     , (stop - start)/7 as weeks 
    FROM task
    ) as sub
group by
     to_char(iso, 'iyyy-iw')

<强> Results ：

|  yyy_wk | weeks | hr_week |
|---------|-------|---------|
| 2017-08 |     8 |      15 |
| 2017-06 |     8 |      15 |
| 2017-02 |     8 |      35 |
| 2017-03 |     8 |      35 |
| 2017-07 |     8 |      15 |
| 2016-52 |     8 |      35 |
| 2017-05 |     8 |      15 |
| 2017-01 |     8 |      35 |
| 2017-04 |     8 |      35 |

Answer 2

同时我找到了另一种基于cte的解决方案。然而，这不包括空周，但也是一个解决方案。我会认为@ used-by-already是一个更可靠的答案。

UrlLibrary sportsurl= new UrlLibrary(sports);

http://sqlfiddle.com/#!15/93ee1/113