DATEDIFF excluding summer months - sql

Question

We are running reports for a seasonal business, with expected lulls during the summer months. For some metrics, we'd essentially like to pretend that those months don't even exist.
Thus consider the default behavior of:
SELECT DATEDIFF(MONTH, '2015-05-01', '2015-06-01') -- answer = 1
SELECT DATEDIFF(MONTH, '2015-05-01', '2015-07-01') -- 2
SELECT DATEDIFF(MONTH, '2015-05-01', '2015-08-01') -- 3
SELECT DATEDIFF(MONTH, '2015-05-01', '2015-09-01') -- 4
We want to ignore June and July, so we would like those answers to look like this:
SELECT DATEDIFF(MONTH, '2015-05-01', '2015-06-01') -- answer = 1
SELECT DATEDIFF(MONTH, '2015-05-01', '2015-07-01') -- 1
SELECT DATEDIFF(MONTH, '2015-05-01', '2015-08-01') -- 1
SELECT DATEDIFF(MONTH, '2015-05-01', '2015-09-01') -- 2
What is the easiest way to accomplish this? I'd like a pure SQL solution, rather than something using TSQL, but writing a custom function such as NOSUMMER_DATEDIFF could also work.
Also, keep in mind the reports will span multiple years, so the solution should be able to handle that.

Answer 1

If you are only interested month differences, then I would suggest a trick here. Count the number of months since some date 0, but ignore the summer months. For example:
'2015-05-01' --> 2015 * 10 + 5 = 20155
'2015-06-01' --> 2015 * 10 + 6 = 20156
'2015-07-01' --> 2015 * 10 + 6 = 20156
'2015-08-01' --> 2015 * 10 + 6 = 20156
'2015-09-01' --> 2015 * 10 + 7 = 20157
This is a fairly easy calculation:
select (case when month(date2) <= 6 then year(date2) * 10 + month(date2)
when month(date2) in (7, 8) then year(date2) * 10 + 6
else year(date2) * 10 + (month(date2) - 2)
end)
For the difference:
select ((case when month(date2) <= 6 then year(date2) * 10 + month(date2)
when month(date2) in (7, 8) then year(date2) * 10 + 6
else year(date2) * 10 + (month(date2) - 2)
end) -
(case when month(date1) <= 6 then year(date1) * 10 + month(date1)
when month(date1) in (7, 8) then year(date1) * 10 + 6
else year(date1) * 10 + (month(date1) - 2)
end)
)

Answer 2

To able to achieve that, you have to "split" dates ranges to an "array" of dates for every single range of dates. CTE might be helpful in this case.
See:
--your table which holds dates ranges
DECLARE #dates TABLE(id INT IDENTITY(1,1), dFrom DATE, dTo DATE)
INSERT INTO #dates (dFrom, dTo)
VALUES('2015-05-01', '2015-06-01'),
('2015-05-01', '2015-07-01'),
('2015-05-01', '2015-08-01'),
('2015-05-01', '2015-09-01')
--summer month table
DECLARE #summermonths TABLE(summMonth INT)
INSERT INTO #summermonths(summMonth)
VALUES(6), (7)
--here Common Table Expressions is in action to "split" dates ranges to an array of dates for every single date range
;WITH CTE AS
(
SELECT id, DATEADD(MM, 0, dFrom) AS ndFrom, dTo, CASE WHEN MONTH(DATEADD(MM, 0, dFrom)) = 6 OR MONTH(DATEADD(MM, 0, dFrom)) = 7 THEN 0 ELSE 1 END AS COfMonth
FROM #dates
WHERE DATEADD(MM, 1, dFrom)<=dTo
UNION ALL
SELECT id, DATEADD(MM, 1, ndFrom) AS ndFrom, dTo, CASE WHEN MONTH(DATEADD(MM, 1, ndFrom)) = 6 OR MONTH(DATEADD(MM, 1, ndFrom)) = 7 THEN 0 ELSE 1 END AS COfMonth
FROM CTE
WHERE DATEADD(MM, 1, ndFrom)<=dTo
)
SELECT t1.id, t2.dFrom, t2.dTo, SUM(t1.COfMonth) AS MyDateDiff
FROM CTE AS t1 INNER JOIN #dates AS t2 ON t1.id = t2.id
GROUP BY t1.id, t2.dFrom , t2.dTo
Result:
id dFrom dTo MyDateDiff
1 2015-05-01 2015-06-01 1
2 2015-05-01 2015-07-01 1
3 2015-05-01 2015-08-01 2
4 2015-05-01 2015-09-01 3 --not 2, because of 5, 8, 9
Got it?
Note: a solution might be differ in case of dFrom and dTo is not the first date of month.