T SQL - Count People with Visits in 3 consecutive months - sql-server

Question

With the following data:
Declare #t Table
(
Name Varchar(1),
VisitDate Date
)
Insert Into #t select 'A','2017-01-05'
Insert Into #t select 'A','2017-03-05'
Insert Into #t select 'A','2017-04-05'
Insert Into #t select 'A','2017-05-05'
Insert Into #t select 'A','2017-08-05'
Insert Into #t select 'B','2017-03-05'
Insert Into #t select 'C','2017-01-05'
Insert Into #t select 'C','2017-02-05'
Insert Into #t select 'C','2017-04-05'
Insert Into #t select 'D','2017-01-05'
Insert Into #t select 'D','2017-02-05'
Insert Into #t select 'D','2017-03-05'
Insert Into #t select 'D','2017-06-05'
Insert Into #t select 'B','2018-01-05'
Insert Into #t select 'B','2018-02-05'
Insert Into #t select 'B','2018-03-05'
Insert Into #t select 'E','2018-01-05'
Insert Into #t select 'E','2018-02-05'
Insert Into #t select 'E','2018-03-05'
Insert Into #t select 'E','2018-06-05'
I need to write a query that will return the Year & Names that have VisitDates in any three consecutive months in any year.
Based on the data, I expect to see:
2017 A
2017 D
2018 B
2018 E
To be honest, I don't know where to start with this using SQL.
I would appreciate any help I can get.
Thanks!!

Answer 1

You can avoid the joins, or parsing the whole data set multiple times, by using the same method as used in gaps-and-islands.
http://rextester.com/SYHJ40676
WITH
sequenced AS
(
SELECT
Name,
YEAR(VisitDate) AS VisitYear,
MONTH(VisitDate) AS VisitMonth,
ROW_NUMBER()
OVER (PARTITION BY Name, YEAR(VisitDate)
ORDER BY MONTH(VisitDate)
)
AS MonthSequenceID
FROM
#t
GROUP BY
Name,
YEAR(VisitDate),
MONTH(VisitDate)
)
SELECT DISTINCT
Name,
VisitYear
FROM
sequenced
GROUP BY
Name,
VisitYear,
VisitMonth - MonthSequenceID
HAVING
COUNT(*) >= 3

Answer 2

just join the two following months to the data and see where it goes:
SELECT DATEPART(year, m1.VisitDate), m1.Name
FROM #t m1
JOIN #t m2 on m2.Name = m1.Name AND DATEPART(month, m2.VisitDate) = DATEPART(month, m1.VisitDate) + 1
JOIN #t m3 on m3.Name = m1.Name AND DATEPART(month, m3.VisitDate) = DATEPART(month, m1.VisitDate) + 2
since it was asked in the comment, how to solve this problem with a year overlap, this should work:
SELECT DATEPART(year, m1.VisitDate), m1.Name
FROM #t m1
JOIN #t m2 on m2.Name = m1.Name AND EOMONTH(m1.VisitDate,1) = EOMONTH(m2.VisitDate)
JOIN #t m3 on m3.Name = m1.Name AND EOMONTH(m1.VisitDate,2) = EOMONTH(m3.VisitDate)
doc on EOMONTH: https://learn.microsoft.com/en-us/sql/t-sql/functions/eomonth-transact-sql?view=sql-server-2017
edit: my answer is just a quick hack and highly inperformant and has errors when there are multiple instances per month.
I suggest using this answer: https://stackoverflow.com/a/52669713/4903754

Answer 3

wrote my code as per the syntax of postgres SQL 9.5.0
first I have created the flag for consecutive months and by using that flag retrieved the required data.lag(),lead()
We need to compare the dates weather they are in consecutive or not for that I'm using lag(),lead() functions.
with temp as (
select name,visitdate,
coalesce(lag(visitdate) over (partition by name order by visitdate),lead(visitdate) over (partition by name order by visitdate))check1,
coalesce(lead(visitdate) over (partition by name order by visitdate),lag(visitdate) over (partition by name order by visitdate)) check2
from TT
order by 1
),
t2 as (
select name,
case
when
(DATE_PART('year', visitdate::date) - DATE_PART('year', check1::date)) * 12 +
(DATE_PART('month', visitdate::date) - DATE_PART('month', check1::date))=1
or
(DATE_PART('year', check2::date) - DATE_PART('year', visitdate::date)) * 12 +
(DATE_PART('month', check2::date) - DATE_PART('month', visitdate::date))=1
then 1 else 0
end as flag
from temp)
select name ,count(1) from t2 where flag=1 group by name having count(1)>=3