这是一种“第三好”的解决方案,因为一旦你有了实际时间,它会跟踪你比计划提前或延迟多少,并将其应用于最接近的计划时间而没有实际时间:
with q1 as (
select
t.stop_id, sched_time, act_time,
nvl2(act_time, t.sched_time - t.act_time, null) ahead,
sum (nvl2(act_time, 1, 0)) over
(partition by 1 order by stop_id) as actual_count
from schedule t
)
select
stop_id, sched_time,
act_time,
nvl (act_time, sched_time - min (ahead) over
(partition by actual_count)) as act_time2
from q1
搜索结果可能不完全符合您的要求,但是这可能是您可以基于此构建的内容:
STOP_ID SCHED_TIME ACT_TIME ACT_TIME2
001 13:47 13:45 13:45
002 13:50 13:48
003 13:52 13:53 13:53
004 13:59 14:00
005 14:01 14:02
006 14:04 14:04 14:04
-- 2014年7月24日编辑--
假设您已将时间转换为整数,如您所建议的(30秒=1),我稍微尝试了一下。这是一个可怕的解决方案,但我认为它实现了您的建议。我不确定它是否比您的过程循环更快。我很好奇它是否如此。Oracle的分析函数非常好用,但您可以看到我确实在使用它们来实现您所描述的功能:
with q1 as (
select
t.stop_id, t.sched_time, t.act_time,
sum (nvl2(act_time, 1, 0)) over
(partition by 1 order by stop_id) as group_id,
lead (sched_time) over (order by stop_id) as next_sched
from schedule2 t
), q2 as (
select
stop_id, sched_time, act_time, group_id, next_sched,
next_sched - sched_time as elapsed,
row_number() over (partition by group_id order by stop_id) as stops,
min (act_time) over (partition by group_id) as min_time,
min (sched_time) over (partition by group_id) as min_sched
from q1
), q3 as (
select
stop_id, sched_time, act_time, group_id, stops, min_time,
min_sched, next_sched,
sum (elapsed) over (partition by group_id order by stop_id) as elapsed,
max (stops) over (partition by group_id) as grp_stops,
lead (min_time, 1) over (order by stop_id) as next_grp_actual,
lead (min_sched, 1) over (order by stop_id) as next_grp_sched
from q2
), q4 as (
select
stop_id, sched_time, act_time, stops, grp_stops,
min_time, lag (elapsed, 1, 0) over
(partition by group_id order by stop_id) as elapsed,
max (next_grp_sched) over (partition by group_id) - min_sched
as time_btw_sched,
max (next_grp_actual) over (partition by group_id) - min_time
as time_btw_actuals
from q3
)
select
stop_id, sched_time, act_time,
nvl (act_time, min_time + (elapsed / time_btw_sched) *
time_btw_actuals) as act_time2
from q4
这是你的示例所得到的结果:
id sched actual actual (calc)
001 1654 1650 1650
002 1660 1659.6
003 1664 1666 1666
004 1678 1678.83333333333
005 1682 1682.5
006 1688 1688 1688
我认为这个任务可以用编程语言进行封装,这样会更加简洁高效。虽然我只熟练掌握C#和Perl两种语言,但是它们都能很好地完成这个任务。