Question

我正在处理一个有订单的订单系统。对于分析部门，我想构建一个视图，该视图可以累积给定日期的所有销售额。

这不是问题，我得到了有效的查询。更复杂的是第二个数字，我想显示当天的累计销售额。

表示如果我在2月1日的销售额为$ 100，则该列应显示为$ 100。如果2月2日的销售额为$ 200，则该列应显示$ 300，依此类推。

这是我到目前为止提出的：

select
    date_trunc('day', o.created_at) :: date,
    sum(o.value) sales_for_day,
    count(o.accepted_at) as num_of_orders_for_day,
    -- sales_for_month_to_date
    -- num_of_orders_for_month_to_date
from
    orders o
where
    status = 'accepted'
group by
    date_trunc('day', o.accepted_at);

Answer 1

只需使用窗口功能：

select date_trunc('day', o.created_at) :: date,
       sum(o.value) as sales_for_day,
       count(o.accepted_at) as num_of_orders_for_day,
       sum(sum(o.value)) over (partition by date_trunc('month', o.accepted_at order by min(o.created_at)) as sales_for_month_to_date
       sum(count(*)) over (partition by date_trunc('month', o.accepted_at order by min(o.created_at)) as num_of_orders_for_month_to_date
from orders o
where status = 'accepted'
group by date_trunc('day', o.accepted_at);

根据您代码中的注释，我推测您需要最新的数字，因此这也会按月划分。

累积值直到最新

1 个答案: