Question

我需要的是：我必须列出所有有付款计划的客户（我们每个月都会收取费用）并且有这样的组合：第一笔付款已完成，接下来的2付款仍未支付。财务表有：
ID，PAYMENTPLANID，DATEDUE，AMOUNTEDUE，AMOUNTPAID

我知道在AMOUNTPAID＆gt;时支付行。 0和DATEDUE表示我需要，因为如果客户只支付第一笔费用并在接下来的2个月内停止，我将在第3个月末取消付款计划。

有没有更简单的方法来识别这个不涉及3个子选择的人（我想到的唯一方法）？

示例：

+--------------------------------------------------------+
| ID, PAYMENTPLANID, DATEDUE,     AMOUNTEDUE, AMOUNTPAID |
+--------------------------------------------------------+
| 1,  1,             2017-07-05,  10,         0          |
| 1,  1,             2017-06-05,  10,         0          |
| 1,  1,             2017-05-05,  10,         10         |
| 2,  5,             2017-07-05,  25,         25         |
| 2,  5,             2017-06-05,  25,         0          |
| 2,  5,             2017-05-05,  25,         25         |
+--------------------------------------------------------+

ID 1的付款计划应该在本月取消，因为该人只支付了第一笔付款。付款计划5不应取消。

Answer 1

您可以将lead与可选的第二个参数一起使用，以查看前面的1行和2行，并检查您的条件。

select * --distinct paymentplanid /*if only planid is needed as output*/
from (
select t.*
,lead(amountpaid,1) over(partition by paymentplanid order by datedue) as nxt_1
,lead(amountpaid,2) over(partition by paymentplanid order by datedue) as nxt_2
,lag(amountpaid) over(partition by paymentplanid order by datedue) as prev_1
from tbl t
) t
where amountpaid>0 and prev_1 is null and nxt_1=0 and nxt_2=0

Answer 2

这样的事情应该有用吗？

   select ID , PaymentplandID from
     (select ID , PaymentplandID , max(amount_paid ) over ( order by  ID , 
     PaymentplandID rows between current row and 1 following partion by Datedue  desc  ) 
  as amount  , 
 row_number() over ( order by select ID , PaymentplandID partion by Datedue  asc) as rn 
 from you_tble) t1 
  where t1.amount = 0 and rn = 3

搜索具有3个首次付款的客户，并具有特定组合

2 个答案: