print start and end date in one row for continous or overlaping date ranges in oracle SQL

Question

I would like to print in one row start date and end date for continous or overlaping date ranges.

here is the data

create table orders (
po varchar2(6),
st         


        

Answer 1

Use window functions to see if there is any overlap with previous records. Then a cumulative sum to assign a "grouping" and aggregate.

I like to use a cumulative max for more generic overlaps:

select po, min(startdate), max(enddate)
from (select o.*,
             sum(case when prev_enddate >= startdate then 0 else 1 end) over (partition by po order by startdate) as grouping
      from (select o.*,
                   max(enddate) over (partition by po order by startdate range between unbounded preceding and '1' second preceding) as prev_enddate
            from orders o
           ) o
     ) o
group by po, grouping;

In many cases, you can use lag() instead of max():

select po, min(startdate), max(enddate)
from (select o.*,
             sum(case when prev_enddate >= startdate then 0 else 1 end) over (partition by po order by startdate) as grouping
      from (select o.*,
                   lag(enddate) over (partition by po order by startdate) as prev_enddate
            from orders o
           ) o
     ) o
group by po, grouping;

This works so long as the previous row has the overlap, which is typically the case.

Answer 2

There is an elegant (and efficient) solution using the match_recognize clause (which requires Oracle 12.1 or higher).

select po, startdate, enddate
from   orders
match_recognize (
  partition by po
  order     by startdate
  measures  first(startdate) as startdate, max(enddate) as enddate
  pattern   ( c* n )
  define    c as max(enddate) + 1 >= next(startdate)  
);