每周总金额(Weekly total sums)

我有一个PostgreSQL数据库包含日期表和每天的总数。

mydate       total
2012-05-12      12 
2012-05-14       8 
2012-05-13       4 
2012-05-12      12 
2012-05-15       2 
2012-05-17       1 
2012-05-18       1 
2012-05-21       1 
2012-05-25       1 

现在我需要每周总计给定日期范围。
防爆。 我想从每周总计2012-05-01高达2012-05-31 。

我看这样的输出：

2012-05-01 2012-05-07   0
2012-05-08 2012-05-14  36
2012-05-15 2012-05-22   5
2012-05-23 2012-05-29   1
2012-05-30 2012-05-31   0

Answer 1:

这适用于任意给定的日期范围：

CREATE FUNCTION f_tbl_weekly_sumtotals(_range_start date, _range_end date)
  RETURNS TABLE (week_start date, week_end date, sum_total bigint)
  LANGUAGE sql AS
$func$
SELECT w.week_start, w.week_end, COALESCE(sum(t.total), 0)
FROM  (
   SELECT week_start::date, LEAST(week_start::date + 6, _range_end) AS week_end
   FROM   generate_series(_range_start::timestamp
                        , _range_end::timestamp
                        , interval '1 week') week_start
   ) w
LEFT   JOIN tbl t ON t.mydate BETWEEN w.week_start and w.week_end
GROUP  BY w.week_start, w.week_end
ORDER  BY w.week_start
$func$;

呼叫：

SELECT * FROM f_tbl_weekly_sumtotals('2012-05-01', '2012-05-31');

要点

• 我包裹在为方便起见，功能，所以时间范围必须被一次仅提供。

• 子查询w生产的系列从给定日期范围的第一天开始星期。 上限与封端LEAST留内的上限指定日期范围内的。

• 然后LEFT JOIN的数据表（ tbl在我的例子），以确保所有周的结果，即使没有数据行被发现。

• 其余的应该是显而易见的。 COALESCE输出0 ，而不是NULL的空星期。

• 数据类型必须匹配，我以为mydate date和total int因缺乏信息。 （该sum()的的int是bigint 。）

• 说明我的特定用途的generate_series()

  • 在PostgreSQL的两个日期之间生成的时间序列

Answer 2:

使用此功能

CREATE OR REPLACE FUNCTION last_day(date)
RETURNS date AS
$$
  SELECT (date_trunc('MONTH', $1) + INTERVAL '1 MONTH - 1 day')::date;
$$ LANGUAGE 'sql' IMMUTABLE STRICT;

和generate_series（从8.4起），我们可以创建日期分区。

SELECT wk.wk_start, 
       CAST(
            CASE (extract(month from wk.wk_start) = extract(month from wk.wk_start + interval '6 days'))
            WHEN true THEN wk.wk_start + interval '6 days'
            ELSE last_day(wk.wk_start)
            END 
           AS date) AS wk_end
  FROM
    (SELECT CAST(generate_series('2012-05-01'::date,'2012-05-31'::date,interval '1 week') AS date) AS wk_start) AS wk;

然后用数据放在一起

CREATE TABLE my_tab(mydate date,total integer);
INSERT INTO my_tab 
values    
('2012-05-12'::date,12),
('2012-05-14'::date,8),
('2012-05-13'::date,4),
('2012-05-12'::date,12),
('2012-05-15'::date,2),
('2012-05-17'::date,1),
('2012-05-18'::date,1),
('2012-05-21'::date,1),
('2012-05-25'::date,1); 

WITH month_by_week AS
    (SELECT wk.wk_start, 
       CAST(
            CASE (extract(month from wk.wk_start) = extract(month from wk.wk_start + interval '6 days'))
            WHEN true THEN wk.wk_start + interval '6 days'
            ELSE last_day(wk.wk_start)
            END 
           AS date) AS wk_end
  FROM
    (SELECT CAST(generate_series('2012-05-01'::date,'2012-05-31'::date,interval '1 week') AS date) AS wk_start) AS wk
  )
SELECT month_by_week.wk_start,
       month_by_week.wk_end,
       SUM(COALESCE(mt.total,0))
  FROM month_by_week 
       LEFT JOIN my_tab mt ON  mt.mydate BETWEEN month_by_week.wk_start AND month_by_week.wk_end
 GROUP BY month_by_week.wk_start,
          month_by_week.wk_end
 ORDER BY month_by_week.wk_start;

文章来源: Weekly total sums