嗨，我有一个看起来像这样的数据集：
order id seq act_seq time station
100 2 1 9 2015-09-02 10:18:15.000000 7
100 2 1 10 2015-09-02 10:18:22.000000 7
100 2 1 11 2015-09-02 10:26:48.000000 7
100 2 1 12 2015-09-02 10:35:20.000000 7
100 2 1 13 2015-09-02 17:32:33.000000 7
100 2 1 14 2015-09-02 17:32:39.000000 7
100 2 1 15 2015-09-02 17:32:39.000000 7
100 2 1 16 2015-09-30 04:00:05.000000 7
100 2 1 17 2015-09-30 04:00:05.000000 7
100 2 1 18 2015-09-30 04:00:13.000000 8
100 2 1 19 2015-09-30 04:00:56.000000 8
100 2 1 20 2015-09-30 07:24:00.000000 7
100 2 1 21 2015-09-30 07:24:03.000000 7
100 2 1 22 2015-09-30 07:24:06.000000 7
100 2 1 23 2015-09-30 07:24:03.000000 9
and so on..............
我有一个主键为order，id，seq的订单。对于每个订单，id，seq，每个“站”都有一堆用“ act_seq”标记的活动。
我想找到每个电台的最短时间和最长时间，然后再转换到另一个电台。因此，站点7的输出应为：
order id seq station min_time max_time
100 2 1 7 2015-09-02 10:18:15.000000 2015-09-30 04:00:05.000000
100 2 1 7 2015-09-30 07:24:00.000000 2015-09-30 07:24:06.000000
即站点7的订单已转换为站点8，然后从站点8转换为站点7，因此将有2条记录具有最小和最大时间值。
我该怎么做？
答案 0 :(得分：0)
这可以做到。还要避免使用作为Oracle保留关键字的列名，例如ORDER。
首先在子查询（T2）中创建一个临时列BRK，该临时列将根据lag确定站是否已更改，如1顺序中断。
会根据此新列进行累计和，然后根据此group by列进行CUM_SUM。
SELECT "ORDER",
ID,
SEQ,
STATION,
MIN(TIME) AS MIN_TIME,
MAX(TIME) AS MAX_TIME
FROM
(SELECT T2.*,
SUM(BRK) OVER (ORDER BY ACT_SEQ) AS CUM_SUM
FROM
(SELECT T1.*,
CASE
WHEN STATION <> NXT
THEN 1
ELSE 0
END AS BRK
FROM
( SELECT T.*,LAG(STATION,1) OVER (ORDER BY ACT_SEQ) AS NXT FROM TABLE1 T
) T1
) T2
)
GROUP BY "ORDER",
ID,
SEQ,
STATION,
CUM_SUM
ORDER BY "ORDER",
ID,
SEQ,
STATION,
CUM_SUM;
答案 1 :(得分：0)
这也可以用来获得预期的输出。您需要确定只要前一站是相同的，并使用sum将这些行分组，以找出给定行的最小和最大时间。使用Act_sq作为要排序的主要字段。
with cte as
(
select 100 as Ordern, 2 as ID, 1 as Seq, 9 as Act_sq, '2015-09-02 10:18:15.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 10 as Act_sq, '2015-09-02 10:18:22.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 11 as Act_sq, '2015-09-02 10:26:48.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 12 as Act_sq, '2015-09-02 10:35:20.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 13 as Act_sq, '2015-09-02 17:32:33.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 14 as Act_sq, '2015-09-02 17:32:39.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 15 as Act_sq, '2015-09-02 17:32:39.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 16 as Act_sq, '2015-09-30 04:00:05.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 17 as Act_sq, '2015-09-30 04:00:05.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 18 as Act_sq, '2015-09-30 04:00:13.000000' as timen, 8 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 19 as Act_sq, '2015-09-30 04:00:56.000000' as timen, 8 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 20 as Act_sq, '2015-09-30 07:24:00.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 21 as Act_sq, '2015-09-30 07:24:03.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 22 as Act_sq, '2015-09-30 07:24:06.000000' as timen, 7 as Station from dual union all
select 100 as Ordern, 2 as ID, 1 as Seq, 23 as Act_sq, '2015-09-30 07:24:03.000000' as timen, 9 as Station from dual )
select z.Ordern, z.id, z.Seq, z.Station, z.mintime, z.maxtime
from (
select t.Ordern, t.id, t.Seq, t.Station,t.samevalue , min(t.timen) mintime, max(t.timen) maxtime from (
select Ordern,id,Seq, Station,timen, sum(samestation) over (partition by ordern order by act_sq) samevalue
from (
select Ordern,id,Seq, Station,timen, act_sq,
case when lag(station,1,1) over (partition by ordern order by act_sq) = Station then 0 else 1 end samestation from cte c) tst) t
join ( select Ordern,id,Seq, Station,timen, sum(samestation) over (partition by ordern order by act_sq) samevalue
from (
select Ordern,id,Seq, Station,timen, act_sq,
case when lag(station,1,1) over (partition by ordern order by act_sq) = Station then 0 else 1 end samestation
from cte) new) drv on t.Ordern = drv.Ordern and t.samevalue = drv.samevalue
where t.Station = 7
group by t.Ordern, t.id, t.Seq, t.Station,t.samevalue ) z;
输出：
ORDERN ID SEQ STATION MINTIME MAXTIME
100 2 1 7 2015-09-02 10:18:15.000000 2015-09-30 04:00:05.000000
100 2 1 7 2015-09-30 07:24:00.000000 2015-09-30 07:24:06.000000
答案 2 :(得分：0)
-- Oracle 12c+: Pattern matching
with s (orderid, id, seq, act_seq, dtime, station) as (
select 100, 2, 1, 9 , to_date('2015-09-02 10:18:15', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 10, to_date('2015-09-02 10:18:22', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 11, to_date('2015-09-02 10:26:48', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 12, to_date('2015-09-02 10:35:20', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 13, to_date('2015-09-02 17:32:33', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 14, to_date('2015-09-02 17:32:39', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 15, to_date('2015-09-02 17:32:39', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 16, to_date('2015-09-30 04:00:05', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 17, to_date('2015-09-30 04:00:05', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 18, to_date('2015-09-30 04:00:13', 'yyyy-mm-dd hh24:mi:ss'), 8 from dual union all
select 100, 2, 1, 19, to_date('2015-09-30 04:00:56', 'yyyy-mm-dd hh24:mi:ss'), 8 from dual union all
select 100, 2, 1, 20, to_date('2015-09-30 07:24:00', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 21, to_date('2015-09-30 07:24:03', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 22, to_date('2015-09-30 07:24:06', 'yyyy-mm-dd hh24:mi:ss'), 7 from dual union all
select 100, 2, 1, 23, to_date('2015-09-30 07:24:03', 'yyyy-mm-dd hh24:mi:ss'), 9 from dual)
select r.*
from s
match_recognize (
partition by orderid, id, seq
order by act_seq, dtime
measures
station as station,
first(dtime) as dt_first,
last (dtime) as dt_last
pattern (v+)
define v as station = first(station)
) r
order by orderid, id, seq, station;
ORDERID ID SEQ STATION DT_FIRST DT_LAST
---------- ---------- ---------- ---------- ------------------- -------------------
100 2 1 7 2015-09-02 10:18:15 2015-09-30 04:00:05
100 2 1 7 2015-09-30 07:24:00 2015-09-30 07:24:06
100 2 1 8 2015-09-30 04:00:13 2015-09-30 04:00:56
100 2 1 9 2015-09-30 07:24:03 2015-09-30 07:24:03