How to select overlapping date ranges in SQL

Question

I have a table with the following columns : sID, start_date and end_date

Some of the values are as follows:

1   1995-07-28  2003-07-20 
1   2003-07-21  2010-05-04 
1   2010-05-03  2010-05-03 
2   1960-01-01  2011-03-01 
2   2011-03-02  2012-03-13 
2   2012-03-12  2012-10-21 
2   2012-10-22  2012-11-08 
3   2003-07-23  2010-05-02

I only want the 2nd and 3rd rows in my result as they are the overlapping date ranges.

I tried this but it would not get rid of the first row. Not sure where I am going wrong?

select a.sID from table a
inner join table b 
on a.sID = b.sID
and ((b.start_date between a.start_date and a.end_date)
and (b.end_date between a.start_date and b.end_date ))
order by end_date desc

I am trying to do in SQL Server

Answer 1

One way of doing this reasonably efficiently is

WITH T1
     AS (SELECT *,
                MAX(end_date) OVER (PARTITION BY sID ORDER BY start_date) AS max_end_date_so_far
         FROM   YourTable),
     T2
     AS (SELECT *,
                range_start = IIF(start_date <= LAG(max_end_date_so_far) OVER (PARTITION BY sID ORDER BY start_date), 0, 1),
                next_range_start = IIF(LEAD(start_date) OVER (PARTITION BY sID ORDER BY start_date) <= max_end_date_so_far, 0, 1)
         FROM   T1)
SELECT SId,
       start_date,
       end_date
FROM   T2
WHERE  0 IN ( range_start, next_range_start ); 

if you have an index on (sID, start_date) INCLUDE (end_date) this can perform the work with a single ordered scan.