TSQL: Continuous period for whole year / each month - sql-server

Question

I try to find all Cust who have membership for at least for one day in each month during 2018.
I came up with solution checking their membership at the beginning / middle / end end of each month like in snippet below, but trying to find more intelligent solution.
I know that I can use tally table for each of 365 days to check this but probably there is more elegant solution ? I'm bit new to SQL, I think I'm missing something in GROUPing area.
In the code snippet shown below, both Cust have at least one day membership.
Desired output:
CustID
------
1
22
Code:
with data as
(
select *
from (values (1, 1, '2017-12-11', '2018-01-16'), (1, 22, '2018-01-28', '2018-03-9' ), (1, 333, '2018-03-1', '2018-12-31') , -- island
(22, 1, '2017-12-31', '2018-01-11'), (22, 2, '2017-2-11', '2019-12-31')) as t (CustID, ContractID, StartDD, EndDD) ---
)
select
isdate(startDD), isdate(EndDD)
from
data
), gaps as
(
select
*,
datediff(day, lag(EndDD, 1, StartDD) over (partition by CustID order by StartDD), StartDD) as BreakDD -- negative is island
from
data
)
select
*,
datepart(month,StartDD) mmS , datepart(month,EndDD) mmE
from
gaps
-- and was active any 1+ day during each of the 12 months in 2018 ????
where
1 = 1
/* and (cast('1/1/2018' as date) between StartDD and EndDD
or cast('1/15/2018' as date) between StartDD and EndDD
or cast('1/31/2018' as date) between StartDD and EndDD)
---- etc.. for each month
and ( cast('12/1/2018' as date) between StartDD and EndDD
or cast('12/15/2018' as date) between StartDD and EndDD
or cast('12/31/2018' as date) between StartDD and EndDD
)
*/
--select CustID, max(BreakDD) Max_Days
--from gaps
--group by CustID

Answer 1

Try this answer.
First create a function to return all the month and year between the given dates.
Function:
--SELECT * FROM dbo.Fn_GetMonthYear('2017-12-11','2018-01-16')
ALTER FUNCTION dbo.Fn_GetMonthYear(#StartDate DATETIME,#EndDate DATETIME)
RETURNS TABLE
AS
RETURN(
SELECT DATEPART(MONTH, DATEADD(MONTH, x.number, #StartDate)) AS [Month]
,DATEPART(YEAR, DATEADD(MONTH, x.number, #StartDate)) AS [Year]
FROM master.dbo.spt_values x
WHERE x.type = 'P'
AND x.number <= DATEDIFF(MONTH, #StartDate, #EndDate)
)
Table Schema:
CREATE TABLE #t(CustID INT, ContractID INT, StartDD date, EndDD date)
INSERT INTO #t values (1, 1, '2017-12-11', '2018-01-16'), (1, 22, '2018-01-28', '2018-03-9' ), (1, 333, '2018-03-1', '2018-12-31') , -- island
(22, 1, '2017-12-31', '2018-01-11'), (22, 2, '2017-2-11', '2019-12-31')
Here is the T-SQL Query for your requirement.
SELECT CustID
,COUNT(DISTINCT [Month]) NoOfMonths
FROM(
SELECT *
FROM #t t
CROSS APPLY dbo.Fn_GetMonthYear(StartDD,EndDD)
)D
WHERE [Year] = 2018
GROUP BY CustID
HAVING COUNT(DISTINCT [Month])=12
Result:
CustID NoOfMonths
1 12
22 12

Answer 2

find all Cust who have membership for at least for one day in each
month during 2018
I think this mean that data must be present between '2018-01-01' and '2018-12-31' for each custid.
CREATE TABLE #t(CustID INT, ContractID INT, StartDD date, EndDD date)
INSERT INTO #t values (1, 1, '2017-12-11', '2018-01-16'), (1, 22, '2018-01-28', '2018-03-9' ), (1, 333, '2018-03-1', '2018-12-31') , -- island
(22, 1, '2017-12-31', '2018-01-11'), (22, 2, '2017-2-11', '2019-12-31')
declare #From Datetime='2018-01-01'
declare #To datetime='2018-12-31'
;with CTE as
(
select CustID,min(StartDD)StartDD
,max(EndDD)EndDD
from #t
group by CustID
)
select CustID,StartDD
,EndDD
from CTE
where StartDD<=#From and EndDD>=#To
This script is not tested across all sample data.
But logic is clear.So it can be corrected accordingly.
So tell for what sample data it is not working.