Getting first and last values from contiguous ranges - sql

Question

I want to get the first enter_date and the last leave_date for contiguous enter_day and leave_day values for each id. Given this example data:
+-----+------------+------------+-----------+-----------+
| id | enter_date | leave_date | enter_day | leave_day |
+-----+------------+------------+-----------+-----------+
| 111 | 2016-07-29 | 2016-12-01 | 1 | 75 |
| 111 | 2016-12-02 | 2017-01-13 | 76 | 95 |
| 111 | 2017-01-17 | 2017-06-02 | 96 | 181 |
| 222 | 2016-07-29 | 2016-12-02 | 1 | 76 |
| 222 | 2017-01-30 | 2017-06-02 | 105 | 181 |
| 333 | 2016-08-01 | 2017-06-02 | 1 | 180 |
+-----+------------+------------+-----------+-----------+
I want the following result:
+-----+------------+------------+
| id | enter_date | leave_date |
+-----+------------+------------+
| 111 | 2016-07-29 | 2017-06-02 |
| 222 | 2016-07-29 | 2016-12-02 |
| 222 | 2017-01-30 | 2017-06-02 |
| 333 | 2016-08-01 | 2017-06-02 |
+-----+------------+------------+
I want one record for ID 111 because there are no gaps between any enter_day and the previous leave_day.
I want both records for ID 222 because there is a gap (days 75 through 104).
EDIT: What I have so far, which isn't giving me the correct leave_date for ID 111:
with cte as (
select a.id, a.enter_date, a.leave_date, b.enter_date next_ed, b.leave_date next_ld
from #tbl a
join #tbl b on b.id = a.id and b.enter_day = a.leave_day + 1
)
select id, min(enter_date) enter_date, max(leave_date) leave_date
from cte
group by id
union
select a.id, a.enter_date, a.leave_date
from #tbl a
left join #tbl b on b.id = a.id and b.enter_day = a.leave_day + 1
left join cte c on c.id = a.id and c.next_ed = a.enter_date and c.next_ld = a.leave_date
where b.id is null and c.id is null
order by 1,3

Answer 1

Below is an example of Gaps-and-Islands for ranges.
I used an ad-hoc tally table, but an actual number/tally would do the trick as well.
If you just run the inner query, you will see very quickly how your sample data of 6 rows explodes into 514 rows. Then is is a small matter of applying the grouped aggregation to get the final results.
Example
Declare #YourTable Table ([id] int,[enter_date] date,[leave_date] date,[enter_day] int,[leave_day] int)
Insert Into #YourTable Values
(111,'2016-07-29','2016-12-01',1,75)
,(111,'2016-12-02','2017-01-13',76,95)
,(111,'2017-01-17','2017-06-02',96,181)
,(222,'2016-07-29','2016-12-02',1,76)
,(222,'2017-01-30','2017-06-02',105,181)
,(333,'2016-08-01','2017-06-02',1,180)
Select ID
,[enter_date] = min([enter_date])
,[leave_date] = max([leave_date])
From (
Select *
,Grp = N - Row_Number() over (Partition By ID Order by N)
From #YourTable A
Join (
Select Top (Select max([leave_day]-[enter_day])+1 From #YourTable)
N=-1+Row_Number() Over (Order By (Select Null))
From master..spt_values n1,master..spt_values n2
) B on B.N between [enter_day] and [leave_day]
) A
Group By [ID],Grp
Order By [ID],min([enter_date])
Returns
ID enter_date leave_date
111 2016-07-29 2017-06-02
222 2016-07-29 2016-12-02
222 2017-01-30 2017-06-02
333 2016-08-01 2017-06-02