如何统计当前记录日期之前的记录？

Question

我有一张这样的桌子：

-----------------
id | date
-----------------
1  | 2023-10-07
2  | 2023-10-08
3  | 2023-10-14

并且我想获取截至当前记录的记录计数，其中日期小于当前记录的星期日。所以我希望输出看起来像这样：

---------------------
week_number | count
---------------------
2           | 2
3           | 3

我尝试了不同的方法，但结果是一周内的条目计数器。就好像他不查看当前记录之前的所有记录，而只查看同一周内的记录。

WITH weekly AS (
    SELECT
        get_week_number(date::DATE) AS week_number,
        COUNT(*) AS count
    FROM
        inventory
    WHERE
        date <= date_trunc('week', date)::date + INTERVAL '6 days'
    GROUP BY
        week_number
)

SELECT
    week_number, count
FROM
    inventory
        JOIN weekly ON get_week_number(date::DATE) = weekly.week_number
WHERE
    date <= date_trunc('week', date)::date + INTERVAL '6 days'
GROUP BY
    week_number, count
ORDER BY
    week_number;

Answer 1

您可能正在寻找窗口函数。 db<>fiddle 的演示

select id, 
       date, 
       extract(week from date) AS week_number,
       count(*)over(order by date) AS count
from inventory order by id;

id	日期	周数	数
1	2023-10-07	40	1
2	2023-10-08	40	2
3	2023-10-14	41	3

select week_number, 
       sum(count)over(order by week_number) 
from( select extract(week from date::DATE) AS week_number,
             count(*) AS count
       from inventory
       group by week_number) AS subquery;

周数	总和
40	2
41	3

默认窗口frame子句是

RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW

，这意味着

count()over()

将按照您想要的方式获取当前记录的记录计数。

Answer 2

这个计算

日期小于当前记录周日的记录计数到当前记录

SELECT *
     , extract(week FROM date + 1) AS week_nr  -- fix off-by-1
     , count(*) OVER (ORDER BY date_bin('7 days', date, timestamp '2023-1-1')
                      RANGE BETWEEN UNBOUNDED PRECEDING AND '1 day' PRECEDING) AS count_til_last_week
FROM   inventory;

我假设“当前记录的星期日”您的意思是从星期日开始的几周。 ISO 周从周一开始。 date_trunc()

 相应地截断周。使用 

date_bin()

 “截断”到星期日。参见：

• 使 date_trunc() 在星期日而不是星期一开始

在提取周数之前向日期添加一天，以修复相同的相差 1 的错误。 （年底前后可能会出现极端情况问题。）

要获取具有

“日期小于当前记录的星期日” 的行数，您必须在“上周”停止计数。 RANGE BETWEEN UNBOUNDED PRECEDING AND '1 day' PRECEDING

 在将每个日期减少到一周中的星期日后实现了这一目标。

详情请参阅
有关窗口函数的手册。 每周

总计

：

SELECT extract(year FROM sun) AS year
     , extract(week FROM sun + interval '1 day') AS week_number
     , sum(row_count) OVER (ORDER BY sun
                            ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING) AS count_til_last_week
FROM  (
   SELECT date_bin('7 days', date, timestamp '2023-1-1') AS sun
        , count(*) AS row_count
   FROM   inventory
   GROUP  BY 1
   ) sub;

请注意，没有年份的周数会很快导致混乱或错误的聚合。因此，我根据完整日期进行计数并另外显示年份。 （那里没有减 1。）

小提琴