postgresql下单时的累计金额

Question

我有一个订单表，其中包含下订单和完成时间的日期时间：

orderid	userid	price	status	createdat	doneat
1	128	100	completed	2/16/21 18:40:45	2/21/21 07:59:46
2	128	150	completed	2/21/21 05:27:29	2/23/21 11:58:23
3	128	100	completed	9/3/21 08:38:14	9/10/21 14:24:35
4	5	100	completed	5/28/22 23:28:07	6/26/22 06:10:35
5	5	100	canceled	7/8/22 22:28:57	8/10/22 06:55:17
6	5	100	completed	7/25/22 13:46:38	8/10/22 06:57:20
7	5	5	completed	8/7/22 18:07:07	8/12/22 06:56:23

我想要一个新列，它是创建订单时每个用户的累计总额（总价）：

orderid	userid	price	status	createdat	doneat	cumulative total when placed (per user)
1	128	100	completed	2/16/21 18:40:45	2/21/21 07:59:46	0
2	128	150	completed	2/21/21 05:27:29	2/23/21 11:58:23	0
3	128	100	completed	9/3/21 08:38:14	9/10/21 14:24:35	250
4	5	100	completed	5/28/22 23:28:07	6/26/22 06:10:35	0
5	5	100	canceled	7/8/22 22:28:57	8/10/22 06:55:17	100
6	5	100	completed	7/25/22 13:46:38	8/10/22 06:57:20	100
7	5	5	completed	8/7/22 18:07:07	8/12/22 06:56:23	100

逻辑是将在当前行创建日期之前完成的所有订单的每个用户的价格相加。对于orderid=2，虽然是用户的第二笔订单，但是在其createdat datetime 2/21/21 05:27:29之前没有完成的订单，所以下单时的累计总数为0。

[5,6,7] 中的 orderid 也是如此。对于这些订单和该用户 ID，在其创建日期之前完成的唯一订单是订单 4，因此下达时的累计总数为 100。

PowerBI中的逻辑是这样的：

SUMX (
filter(
orders, 
earlier orders.userid = orders.userid && orders.doneat < orders.createdat && order.status = 'completed'),
orders.price)

有人对如何在 postgresql 中实现这一点有任何提示吗？

我尝试过这样的事情，但没有奏效。

select (case when o.doneat < o.createdat over (partition by o.userid, o.status order by o.createdat)
    then sum(o.price) over (partition by o.userid, o.status ORDER BY o.doneat asc ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)
        end) as cumulativetotal_whenplaced
    from order o

谢谢

Answer 1

您可以将每一行复制到：

1. 一个“原始”（我们将用标志keep = true 装饰），具有会计价值val = 0（到目前为止）和时间t = createdat；
2. 一个“重复”（keep = false），它的价格（如果状态为'completed'）为val 和一个时间t = doneat。

   然后只需考虑正确的位即可：

   select orderid, userid, price, status, createdat, doneat, cumtot
   from (
     select *, sum(val) over (partition by userid order by t, keep desc) as cumtot
     from (
       select *, createdat as t, 0 as val, true as keep from foo
       union all
       select *, doneat as t,
         case when status = 'completed' then price else 0 end as val,
         false as keep
       from foo
     ) as a
   ) as a
   where keep
   order by orderid;
   

   示例：DB Fiddle。

   RedShift 的注意事项：上面的窗口表达式需要替换为：

      ...
      select *, sum(val) over (
        partition by userid order by t, keep desc
        rows unbounded preceding) as cumtot
      ...
   

   您的数据的结果：

   orderid	userid	price	status	createdat	doneat	cumtot
   1	128	100	completed	2021-02-16T18:40:45.000Z	2021-02-21T07:59:46.000Z	0
   2	128	150	completed	2021-02-21T05:27:29.000Z	2021-02-23T11:58:23.000Z	0
   3	128	100	completed	2021-09-03T08:38:14.000Z	2021-09-10T14:24:35.000Z	250
   4	5	100	completed	2022-05-28T23:28:07.000Z	2022-06-26T06:10:35.000Z	0
   5	5	100	canceled	2022-07-08T22:28:57.000Z	2022-08-10T06:55:17.000Z	100
   6	5	100	completed	2022-07-25T13:46:38.000Z	2022-08-10T06:57:20.000Z	100
   7	5	5	completed	2022-08-07T18:07:07.000Z	2022-08-12T06:56:23.000Z	100

   笔记：这种跨时间的会计类型实际上对许多极端情况（各种订单重叠，一些开始和结束，而另一些仍在进行中等）是稳健的。它是我应该在某天描述的快速间隔压缩算法的基础所以。

   奖励：尝试找出分区窗口按t 排序的原因（相当明显）并且还通过keep desc（不太明显）。