对于每一行，根据日期 SQL BigQuery 计算之前有多少行落后

Question

我在 BigQuery 中有一个带有还款订阅计划的表，如下所示：

ID	子月号	to_be_paid_date	实际付款日期	迟到了
156	1	2020-03-01	2020-03-01	不
156	2	2020-04-01	2021-06-02	是的
156	3	2020-05-01	2020-06-07	是的
156	4	2020-06-01	2021-06-07	是的

对于每个客户id ，都有订阅月份编号和我们希望他们支付订阅费用的日期。 我知道哪些付款比预期迟到，但我想知道在下一次付款到期时之前的许多个月都没有付款。

例如，到第 4 个月的订阅到期时 (2020-06-01)，第 2 个月和第 3 个月仍未付款。 所以我试图计算像这样的东西num_past_overdue ：

ID	子月号	to_be_paid_date	实际付款日期	num_past_overdue
156	1	2020-03-01	2020-03-01	-
156	2	2020-04-01	2021-06-02	0
156	3	2020-05-01	2020-06-07	1
156	4	2020-06-01	2021-06-07	2

我尝试使用 LEAD function 和 CASE WHEN，但它只给我上个月是否已支付的信息，而不是在下个月到期时未支付前几个月的信息。

WITH payments as (
SELECT *
,LEAD(to_be_paid_date) OVER (PARTITION BY id ORDER BY id,  to_be_paid_date) AS next_due_date  
FROM table)

SELECT *
, CASE WHEN DATE_DIFF(actual_payment_date, next_due_date, DAY)> 0 THEN True
          ELSE False END AS overdue_when_next_was_due
FROM payments

Answer 1

试试下面的方法：

with t1 as (
  select 156 as id, 1 as sub_month_number, date('2020-03-01')   as to_be_paid_date, date('2020-03-01')  as actual_payment_date, 'no' as was_late,
  union all select 156 as id, 2 as sub_month_number, date('2020-04-01') as to_be_paid_date, date('2021-06-02') as actual_payment_date, 'yes' as was_late,
  union all select 156 as id, 3 as sub_month_number, date('2020-05-01') as to_be_paid_date, date('2020-06-07') as actual_payment_date, 'yes' as was_late,
  union all select 156 as id, 4 as sub_month_number, date('2020-06-01') as to_be_paid_date, date('2021-06-07') as actual_payment_date, 'yes' as was_late,
)

select 
  *,
  sum(
    case 
      when was_late='yes' then 1 
      else 0 
    end
    ) over(order by actual_payment_date) as num_past_overdue
from t1

Output：