我正在尝试编写一个脚本，该脚本在客户7天内每次花费超过1200欧元时都非常重要。一旦客户超过1200欧元的门槛，累计金额应重置。例如，如果客户在第3天超过了1200欧元，则该数字为1，并且应在第4天重置累计金额。
我也看到过类似的问题，它们涉及重置的累计金额。这些解决方案都无法在7天滚动条件下使用。
示例数据集
create table test2
(
yyyymmdd DATE not null,
account_id NUMBER,
vol_eur NUMBER
);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('21-01-2018 11:16:19', 'dd-mm-yyyy hh24:mi:ss'), 57642, 1500);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('06-01-2018 09:51:23', 'dd-mm-yyyy hh24:mi:ss'), 57645, 190);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('21-01-2018 07:09:35', 'dd-mm-yyyy hh24:mi:ss'), 57645, 300);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('03-01-2018 14:58:14', 'dd-mm-yyyy hh24:mi:ss'), 57646, 1000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('17-01-2018 13:30:44', 'dd-mm-yyyy hh24:mi:ss'), 57646, 130);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('03-01-2018 18:33:33', 'dd-mm-yyyy hh24:mi:ss'), 57647, 1000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('04-01-2018 08:44:33', 'dd-mm-yyyy hh24:mi:ss'), 57647, 270);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('05-01-2018 19:28:08', 'dd-mm-yyyy hh24:mi:ss'), 57647, 800);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('13-01-2018 12:24:21', 'dd-mm-yyyy hh24:mi:ss'), 57647, 700);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('15-01-2018 10:52:50', 'dd-mm-yyyy hh24:mi:ss'), 57647, 1000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('27-01-2018 12:07:20', 'dd-mm-yyyy hh24:mi:ss'), 57647, 500);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('10-01-2018 21:14:46', 'dd-mm-yyyy hh24:mi:ss'), 57647, 690);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('30-01-2018 15:39:17', 'dd-mm-yyyy hh24:mi:ss'), 57647, 5500);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('05-01-2018 19:43:38', 'dd-mm-yyyy hh24:mi:ss'), 57649, 300);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('06-01-2018 17:54:30', 'dd-mm-yyyy hh24:mi:ss'), 57649, 150);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('15-01-2018 19:38:36', 'dd-mm-yyyy hh24:mi:ss'), 57649, 1000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('20-01-2018 13:26:34', 'dd-mm-yyyy hh24:mi:ss'), 57649, 1150);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('06-01-2018 17:09:54', 'dd-mm-yyyy hh24:mi:ss'), 57651, 300);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('28-01-2018 17:31:14', 'dd-mm-yyyy hh24:mi:ss'), 57651, 250);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('04-01-2018 13:39:06', 'dd-mm-yyyy hh24:mi:ss'), 57654, 150);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('07-01-2018 13:18:26', 'dd-mm-yyyy hh24:mi:ss'), 57654, 200);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('13-01-2018 19:44:08', 'dd-mm-yyyy hh24:mi:ss'), 57654, 150);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('21-01-2018 16:18:05', 'dd-mm-yyyy hh24:mi:ss'), 57654, 150);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('28-01-2018 10:53:03', 'dd-mm-yyyy hh24:mi:ss'), 57654, 60);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('01-01-2018 12:09:00', 'dd-mm-yyyy hh24:mi:ss'), 57655, 1000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('01-01-2018 17:01:27', 'dd-mm-yyyy hh24:mi:ss'), 57655, 1000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('02-01-2018 19:30:31', 'dd-mm-yyyy hh24:mi:ss'), 57655, 200);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('21-01-2018 15:52:29', 'dd-mm-yyyy hh24:mi:ss'), 57655, 1000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('21-01-2018 16:58:52', 'dd-mm-yyyy hh24:mi:ss'), 57655, 500);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('11-01-2018 14:26:30', 'dd-mm-yyyy hh24:mi:ss'), 57661, 2000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('12-01-2018 21:54:25', 'dd-mm-yyyy hh24:mi:ss'), 57661, 500);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('06-01-2018 16:46:25', 'dd-mm-yyyy hh24:mi:ss'), 57666, 5000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('10-01-2018 18:27:51', 'dd-mm-yyyy hh24:mi:ss'), 57666, 5000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('14-01-2018 18:52:14', 'dd-mm-yyyy hh24:mi:ss'), 57666, 5000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('20-01-2018 12:19:07', 'dd-mm-yyyy hh24:mi:ss'), 57666, 5000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('24-01-2018 18:38:40', 'dd-mm-yyyy hh24:mi:ss'), 57666, 2990);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('30-01-2018 18:36:01', 'dd-mm-yyyy hh24:mi:ss'), 57666, 1980);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('19-01-2018 18:48:44', 'dd-mm-yyyy hh24:mi:ss'), 57671, 2000);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('19-01-2018 23:41:56', 'dd-mm-yyyy hh24:mi:ss'), 57671, 100);
insert into test2 (yyyymmdd, account_id, vol_eur)
values (to_date('21-01-2018 19:22:51', 'dd-mm-yyyy hh24:mi:ss'), 57671, 5000);
commit;
答案 0 :(得分：1)
具有递归cte的一个选项。
with rownums as (select t.*,row_number() over(partition by id order by dt) as rnum
from tbl t
)
,rsum as (select id,dt,val,rnum,val as cumsum,0 as dt_diff
from rownums
where rnum = 1
union all
select r.id
,r.dt
,r.val
,r.rnum
,case when dt_diff + rs.dt - r.dt > 7 then r.val
when dt_diff + rs.dt - r.dt <= 7 and r.val + rs.cumsum < 1200 then r.val+rs.cumsum
else 0 end
,case when dt_diff + rs.dt - r.dt > 7 then 0
else dt_diff + rs.dt - r.dt end
from rsum rs
join rownums r on r.id = rs.id and r.rnum = rs.rnum+1
)
select id,dt,val,case when cumsum = 0 and lag(cumsum,1) over(partition by id order by dt) <= 1200 then val+lag(cumsum,1) over(partition by id order by dt)
when cumsum = 0 and lag(cumsum,1) over(partition by id order by dt) > 1200 then val
else cumsum end as res
from rsum
order by 1,2
rownums中每个用户ID的行号。rownums cte中选择第一行作为锚行，然后遍历其余行，与锚行连接，一次向前看一行。 case表达式在这里检查条件。0将累积总和设置为rsum，这表示基于7天之内的总和超过1200或基于新的7天开始的新组开始。使用lag最终计算这些行上的值。