集团通过isoweek - PostgreSQL的

Question

这是一个基于前一个问题一个subquestion：

Split values from an interval and group by isoweek - Postgresql

的问题是如何分组这个由isoweek

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); 

基于帕特里克答案我发现这解决方案：

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

下一个步骤是“组群”是这样的：

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。任何帮助赞赏。

TIA，

Answer 1

SQL Fiddle

的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') 

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，我发现了另一个解决方案，基于CTE。然而，这不包括空的周，但也是一个解决方案。我会认为@已被用作更可靠的答案。

WITH list as (
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) 

SELECT DISTINCT ON (week) week, sum(hr_week) 
FROM list 
GROUP BY 1 

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