我问了一个类似的问题。假设我有下表结构。p1,p2,p3这3个属性表示为一个键。我想每天比较每把钥匙。例如,从第1天到第2天,删除,添加abe和aby。
P1 P2 P3 DAY KEY
a b c 1 abc
a b e 2 abe
a b y 2 aby
a b x 5 abx
a b c 5 abc
我正在考虑生成一个结果集,该结果集具有跟踪历史的开始/结束日期。预期结果集:
P1 P2 P3 KEY STARTTIME ENDTIME
a b c abc 1 2
a b e abe 2 5
a b y aby 2 5
a b x abx 5 NULL
a b c abc 5 NULL
谢谢你为我之前的帖子提供帮助。我修改了下面的一个答案,以获得添加/删除结果集,但仍然无法转换为上面的开始/结束时间模型。
with base as (
select
'a' as p1,
'b' as p2,
'c' as p3,
1 as day
from dual
union
select
'a' as p1,
'b' as p2,
'y' as p3,
2 as day
from dual
union
select
'a' as p1,
'b' as p2,
'e' as p3,
2 as day
from dual
union
select
'a' as p1,
'b' as p2,
'x' as p3,
5 as day
from dual
union
select
'a' as p1,
'b' as p2,
'c' as p3,
5 as day
from dual
),
calendar as (
select
day,
lead(day) over (order by day asc) as nextday,
lag(day) over (order by day asc) as prevday
from
(select distinct day from base)
),
data as (
select
p1,
p2,
p3,
base.day,
lead(base.day) over (partition by p1, p2, p3 order by base.day asc) as nextrow,
lag(base.day) over (partition by p1, p2, p3 order by base.day asc) as prevrow,
calendar.nextday,
calendar.prevday
from
base
left join
calendar
on calendar.day = base.day
)
select * from data
/
select
d1.p1,
d1.p2,
d1.p3,
d1.day,
d1.nextrow,
d1.prevrow,
'ADD' as op,
d1.day as d
from
data d1
where
prevrow is null or prevrow <> prevday
union
select
d2.p1,
d2.p2,
d2.p3,
d2.day,
d2.nextrow,
d2.prevrow,
'REMOVE' as op,
nextday
from
data d2
where
nextrow is null or nextrow <> nextday
order by
d, op asc
1条答案
按热度按时间nwsw7zdq1#
我认为这是一个缺口和孤岛问题。您可以使用行号之间的差异来标识相邻记录的组。计算
end_day
,需要更多的逻辑,因为日子不是连续的:我用了一个窗口sum()
随着时间的推移first_value()
要获得实际的“下一个”日期:db fiddlde上的演示: