Bạn có thể sử dụng logic như vậy với sum tổng hợp và lag cửa sổ hoạt động như bên dưới:
select clientid, min(startdate) as startdate, max(enddate) as enddate
from
(
select tt.*, sum(grp) over (order by clientid, startdate) sm
from
(
with t(clientid, startdate, enddate) as
(
select 1, date'2017-10-10', date'2017-10-12' from dual union all
select 1, date'2017-10-12', date'2017-10-13' from dual union all
select 1, date'2017-10-13', date'2017-10-17' from dual union all
select 1, date'2017-11-10', date'2017-11-17' from dual union all
select 1, date'2017-11-17', date'2017-11-23' from dual union all
select 1, date'2017-12-12', date'2017-12-14' from dual union all
select 2, date'2017-11-10', date'2017-11-15' from dual union all
select 2, date'2017-12-01', date'2017-12-02' from dual union all
select 2, date'2017-12-02', date'2017-12-05' from dual
)
select clientid,
decode(nvl(lag(enddate) over
(order by enddate),startdate),startdate,0,1)
as grp, --> means prev. value equals or not
row_number() over (order by clientid, enddate) as rn, startdate, enddate
from t
) tt
order by rn
)
group by clientid, sm
order by clientid, enddate;
CLIENTID STARTDATE ENDDATE
---------- ---------- ----------
1 10.10.2017 17.10.2017
1 10.11.2017 23.11.2017
1 12.12.2017 14.12.2017
2 10.11.2017 15.11.2017
2 01.12.2017 05.12.2017
Thực thi truy vấn từng bước để hiểu rõ hơn
