An alternative solution that uses window functions rather than recursive CTEs
SELECT
employmentid,
MIN(startdate) as startdate,
NULLIF(MAX(COALESCE(enddate,'9999-01-01')), '9999-01-01') as enddate
FROM (
SELECT
employmentid,
startdate,
enddate,
DATEADD(
DAY,
-COALESCE(
SUM(DATEDIFF(DAY, startdate, enddate)+1) OVER (PARTITION BY employmentid ORDER BY startdate ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING),
0
),
startdate
) as grp
FROM @t
) withGroup
GROUP BY employmentid, grp
ORDER BY employmentid, startdate
This works by calculating a grp value that will be the same for all consecutive rows. This is achieved by:
- Determine totals days the span occupies (+1 as the dates are inclusive)
SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM @t
- Cumulative sum the days spanned for each employment, ordered by startdate. This gives us the total days spanned by all the previous employment spans
- We coalesce with 0 to ensure we dont have NULLs in our cumulative sum of days spanned
- We do not include current row in our cumulative sum, this is because we will use the value against
startdate rather than enddate (we cant use it against enddate because of the NULLs)
SELECT *, COALESCE(
SUM(daysSpanned) OVER (
PARTITION BY employmentid
ORDER BY startdate
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
)
,0
) as cumulativeDaysSpanned
FROM (
SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM @t
) inner1
- Subtract the cumulative days from the
startdate to get our grp. This is the crux of the solution.
- If the start date increases at the same rate as the days spanned then the days are consecutive, and subtracting the two will give us the same value.
- If the startdate increases faster than the days spanned then there is a gap and we will get a new
grp value greater than the previous one.
- Although
grp is a date, the date itself is meaningless we are using just as a grouping value
SELECT *, DATEADD(DAY, -cumulativeDaysSpanned, startdate) as grp
FROM (
SELECT *, COALESCE(
SUM(daysSpanned) OVER (
PARTITION BY employmentid
ORDER BY startdate
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
)
,0
) as cumulativeDaysSpanned
FROM (
SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM @t
) inner1
) inner2
With the results
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| employmentid | startdate | enddate | daysSpanned | cumulativeDaysSpanned | grp |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 5 | 2007-12-03 00:00:00.000 | 2011-08-26 00:00:00.000 | 1363 | 0 | 2007-12-03 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 5 | 2013-05-02 00:00:00.000 | NULL | NULL | 1363 | 2009-08-08 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 30 | 2006-10-02 00:00:00.000 | 2011-01-16 00:00:00.000 | 1568 | 0 | 2006-10-02 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 30 | 2011-01-17 00:00:00.000 | 2012-08-12 00:00:00.000 | 574 | 1568 | 2006-10-02 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 30 | 2012-08-13 00:00:00.000 | NULL | NULL | 2142 | 2006-10-02 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 66 | 2007-09-24 00:00:00.000 | NULL | NULL | 0 | 2007-09-24 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
- Finally we can
GROUP BY grp to get the get rid of the consecutive days.
- Use
MIN and MAX to get the new startdate and endate
- To handle the NULL
enddate we give them a large value to get picked up by MAX then convert them back to NULL again
SELECT
employmentid,
MIN(startdate) as startdate,
NULLIF(MAX(COALESCE(enddate,'9999-01-01')), '9999-01-01') as enddate
FROM (
SELECT *, DATEADD(DAY, -cumulativeDaysSpanned, startdate) as grp
FROM (
SELECT *, COALESCE(
SUM(daysSpanned) OVER (
PARTITION BY employmentid
ORDER BY startdate
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
)
,0
) as cumulativeDaysSpanned
FROM (
SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM @t
) inner1
) inner2
) inner3
GROUP BY employmentid, grp
ORDER BY employmentid, startdate
To get the desired result
+--------------+-------------------------+-------------------------+
| employmentid | startdate | enddate |
+--------------+-------------------------+-------------------------+
| 5 | 2007-12-03 00:00:00.000 | 2011-08-26 00:00:00.000 |
+--------------+-------------------------+-------------------------+
| 5 | 2013-05-02 00:00:00.000 | NULL |
+--------------+-------------------------+-------------------------+
| 30 | 2006-10-02 00:00:00.000 | NULL |
+--------------+-------------------------+-------------------------+
| 66 | 2007-09-24 00:00:00.000 | NULL |
+--------------+-------------------------+-------------------------+
- We can combine the inner queries to get the query at the start of this answer. Which is shorter, but less explainable
Limitations of all this required that
- there are no overlaps of startdate and enddate for an employment. This could produce collisions in our
grp.
- startdate is not NULL. However this could be overcome by replacing NULL start dates with small date values
- Future developers can decipher the window black magic you performed