Find Customers With 4 Consecutive Years of Giving (Including Gaps) - sql

Question

I have a table similar to below:
+------------+-----------+
| CustomerID | OrderYear |
+------------+-----------+
| 1 | 2012 |
| 1 | 2013 |
| 1 | 2014 |
| 1 | 2017 |
| 1 | 2018 |
| 2 | 2012 |
| 2 | 2013 |
| 2 | 2014 |
| 2 | 2015 |
| 2 | 2017 |
+------------+-----------+
How would I identify which CustomerIDs have 4 consecutive years of giving? (In the above, only customer 2.) As you can see, some records will have gaps in order years.
I started down the row of trying to utilize some combination of ROW_NUMBER/LAG/LEAD with no luck to this point.
Very paired down/modified attempt...
WITH CTE
AS
(
SELECT T.ConstituentLookupID,
T.FISCALYEAR,
COUNT(T.FISCALYEAR) OVER (PARTITION BY T.ConstituentLookupID) AS
YearCount,
FIRST_VALUE(T.FISCALYEAR) OVER(PARTITION BY T.ConstituentLookupID ORDER
BY T.FISCALYEAR DESC) - T.FISCALYEAR + 1 as X,
ROW_NUMBER() OVER(PARTITION BY T.ConstituentLookupID ORDER BY
T.FISCALYEAR DESC) AS RN
FROM #Temp AS T)
SELECT CTE.ConstituentLookupID,
CTE.FISCALYEAR,
CTE.YearCount,
CTE.X,
CTE.RN,
FROM CTE
WHERE CTE.YearCount >= 4 --Have at least 4 years of giving
AND CTE.X - CTE.RN = 1 --Some kind of way to calculate consecutive years. Doesnt account current year and gaps...;

Answer 1

Assuming no duplicates, you can use lag():
select distinct customerid
from (
select t.*,
lag(orderyear, 3) over(partition by customerid order by orderyear) oderyear3
from mytable t
) t
where orderyear = orderyear3 + 3
A more conventional approach is to use some gaps-and-islands technique. This is convenient if you want the start and end of each series. Here, an island is a series of rows with "adjacent" order years, and you want islands that are at least 4 years long. We can identify the islands by comparing the order year against an incrementing sequence, then use aggregation:
select customerid, min(orderyear) firstorderyear, max(orderyear) lastorderyear
from (
select t.*,
row_number() over(partition by customerid order by orderyear) rn
from mytable t
) t
group by customerid, orderyear - rn
having count(*) >= 4

Answer 2

Assuming you have no more than one row per customer and year, the simplest method is lag():
select customerid, year
from (select t.*,
lag(orderyear, 3) over (partition by customerid order by orderyear) as prev3_year
from t
) t
where prev3_year = year - 3;
The idea is to look 3 years back. If that year is year - 3, then there are four years in a row. If your data can have duplicates, there are tweaks to the logic (they make the query more slightly more complicated).
This could return duplicates, so you might just want:
select distinct customerid
from (select t.*,
lag(orderyear, 3) over (partition by customerid order by orderyear) as prev3_year
from t
) t
where prev3_year = year - 3;

Answer 3

I have a simple solution using row number and group by
SELECT Max(z.customerid),
Count(z.grp)
FROM (SELECT customerid,
orderyear,
orderyear - Row_number()
OVER (
ORDER BY customerid) AS Grp
FROM mytable)z
GROUP BY z.grp
HAVING Count(z.grp) = 4