如何获得前几周的总金额答案

【问题标题】：How to get total amount per previous weeks如何获得前几周的总金额
【发布时间】：2021-07-30 06:57:27
【问题描述】：

例如，我有这张表：

Date	amount
2021-02-16T21:06:38	10
2021-02-16T21:07:01	5
2021-02-17T01:10:12	-1
2021-02-19T12:00:00	3
2021-02-24T12:00:00	20
2021-02-25T12:00:00	-1

我想要每周所有前几周的总金额。所以这种情况下的结果是：

Date	amount
2021-02-15	0
2021-02-22	17
2021-03-01	36

注意：日期现在是每周的开始时间（星期一）。任何帮助将不胜感激。

【问题讨论】：

标签： sql postgresql window-functions

【解决方案1】：

试试这个：

select week_date, sum(amount) over (order by week_date ) 
from (
SELECT date(date_) + cast(abs(extract(dow FROM date_) -7 ) + 1 as int) "week_date",
sum(amount) "amount"
from example group by 1) t

DEMO

Above Query 将仅涵盖有交易记录的那一周。如果您想涵盖所有缺失的一周，请尝试以下查询：

with cte as (
SELECT date(date_) + cast(abs(extract(dow FROM date_) -7 ) + 1 as int) "week_date",
sum(amount) "amount"
from example group by 1
)
select 
t1."Date",coalesce(sum(cte.amount) over (order by t1."Date"),0)
from cte right join 
(select generate_series(min(week_date)- interval '1 week', max(week_date),interval '1 week') "Date" from cte) t1 on cte.week_date=t1."Date"

DEMO

【讨论】：

【解决方案2】：

使用generate_series() 生成您想要的日期。然后使用left join 将数据带入并以累积和汇总：

select gs.week,
       coalesce(sum(e.amount), 0) as week_amount,
       sum(coalesce(sum(e.amount), 0)) over (order by gs.week) as running_amount
from generate_series('2021-02-15'::date, '2021-03-01'::date, interval '1 week') gs(week) left join
     example e
     on e.date < gs.week and
        e.date >= gs.week - interval '1 week'
group by gs.week
order by gs.week;

Here 是一个 dbfiddle。

【讨论】：