Question

我们会在不同日期向客户收取账单，并且会以不定期的方式收取款项。我们需要计算付款延迟天数，直到收到特定付款到期的全额付款为止。数据是仅一个客户端0123

的示例数据

table Due (id, fil varchar(12), amount numeric(10, 2), date DATE)

table Received (id, fil varchar(12), amount numeric(10, 2), date DATE)

到期表：

id     fil     amount        date
----------------------------------------
1.     0123   1000.         2019-jan-01
2.     0123   1500          2019-jan-15
3.     0123   1200.         2019-jan-25
4.     0123   1800.         2019-feb-10

表Received：

id.    fil.    amount.       date
-----------------------------------------
1.     0123    1000.         2019-jan-10
2.     0123    500.          2019-jan-20
3.     0123    1300.         2019-jan-25
4.     0123     400.         2019-feb-08
5.     0123    1000.         2019-feb-20

联接表应显示：

fil.     due_date        due_amount.    received_amount date    delay
------------------------------------------------------------------------
0123    2019-jan-01     1000.          1000                     9
0123    2019-jan-15.    1500.          500
0123                                  1300.                    10(since payment completed on 25th jan)
0123    2019-jan-25.    1200.          400.
0123                                  1000.                    26
0123    2019-feb-10.    1800.

我已尽力使计算尽可能准确……如果出现广告错误，请原谅。我只是来写一个脚本来做到这一点，但是也许有人可以建议一个合适的联接。

感谢您的尝试。

Answer 1

正如@DavidHempy所说，这是不可能的，除非知道每笔付款是针对哪张发票。您可以计算自帐户为零以来已经过了多少天，这可能会有所帮助：

with all_activity as (
  select due.date, 
       -1 * amount as amount 
  from due
  union all
  select received.date, 
       amount 
  from received), 
totals as (
 select date, 
        amount, 
        sum(amount) over (order by date), 
        case when sum(amount) over (order by date) >=0 
             then true 
             else false 
        end as nothing_owed
  from all_activity)
select date, 
       amount, 
       sum, 
       date - max(date) filter (where nothing_owed = true) OVER (order by date) 
       as days_since_positive
from totals order by 1
;
    date    |  amount  |   sum    | days_since_positive
------------+----------+----------+---------------------
 2019-01-01 | -1000.00 | -1000.00 |
 2019-01-10 |  1000.00 |     0.00 |                   0
 2019-01-15 | -1500.00 | -1500.00 |                   5
 2019-01-20 |   500.00 | -1000.00 |                  10
 2019-01-25 | -1200.00 |  -900.00 |                  15
 2019-01-25 |  1300.00 |  -900.00 |                  15
 2019-02-08 |   400.00 |  -500.00 |                  29
 2019-02-10 | -1800.00 | -2300.00 |                  31
 2019-02-20 |  1000.00 | -1300.00 |                  41
(9 rows)

您可以扩展此逻辑以找出它们从0开始的最后截止日期。

付款已全额收到

1 个答案: