Sql query to Count Total Consecutive Years from latest year - sql

Question

I have a table Temp:
CREATE TABLE Temp
(
[ID] [int],
[Year] [INT],
)
**ID Year**
1 2016
1 2016
1 2015
1 2012
1 2011
1 2010
2 2016
2 2015
2 2014
2 2012
2 2011
2 2010
2 2009
3 2016
3 2015
3 2004
3 1999
4 2016
4 2015
4 2014
4 2010
5 2016
5 2014
5 2013
I want to calculate the total consecutive years starting from the most recent Year.
Result should look like this:
ID Total Consecutive Yrs
1 2
2 3
3 2
4 3
5 1

Answer 1

select ID,
-- returns a sequence without gaps for consecutive years
first_value(year) over (partition by ID order by year desc) - year +1 as x,
-- returns a sequence without gaps
row_number() over (partition by ID order by year desc) as rn
from Temp
e.g. for ID=1:
1 2016 1 1
1 2015 2 2
1 2012 5 3
1 2011 6 4
1 2010 7 5
As long as there's no gap, both sequences increase the same.
Now check for equal sequences and count the rows:
with cte as
(
select ID,
-- returns a sequence without gaps for consecutive years
first_value(year) over (partition by ID order by year desc) - year + 1 as x,
-- returns a sequence without gaps
row_number() over (partition by ID order by year desc) as rn
from Temp
)
select ID, count(*)
from cte
where x = rn -- no gap
group by ID
Edit:
Based on your year zero comment:
with cte as
(
select ID, year,
-- returns a sequence without gaps for consecutive years
first_value(year) over (partition by ID order by year desc) - year + 1 as x,
-- returns a sequence without gaps
row_number() over (partition by ID order by year desc) as rn
from Temp
)
select ID,
-- remove the year zero from counting
sum(case when year <> 0 then 1 else 0 end)
from cte
where x = rn
group by ID

Answer 2

You can use lead and get this counts as below:
Select top (1) with ties Id, RowN as [Total Consecutive Years] from (
Select *, Num = case when ([year]- lead(year) over(partition by Id order by [Year] desc) > 1) then 0 else 1 end
, RowN = Row_Number() over (partition by Id order by [Year] desc)
from temp
) a
where a.Num = 0
order by row_number() over(partition by Id order by RowN)
Output as below:
+----+-------------------------+
| Id | Total Consecutive Years |
+----+-------------------------+
| 1 | 2 |
| 2 | 3 |
| 3 | 2 |
| 4 | 3 |
| 5 | 1 |
+----+-------------------------+

Answer 3

You can do this using window functions:
select id, count(distinct year)
from (select t.*,
dense_rank() over (partition by id order by year + seqnum desc) as grp
from (select t.*,
dense_rank() over (partition by id order by year desc) as seqnum
from temp t
) t
) t
where grp = 1
group by id;
This assumes that "most recent year" is per id.

Answer 4

Gordon Linoff,
Your code is awesome!
Your code pulls consecutive years from the most recent year.
I modified it to pull overall max consecutive years.
Posted here in case anyone else needs it:
--overall max consecutive years
select id,max(yr_cnt) max_consecutive_years
from (
select id, grp,count(seqnum) yr_cnt
from (select t.*,
dense_rank() over (partition by id order by year + seqnum desc) as grp
from (select t.*,
dense_rank() over (partition by id order by year desc) as seqnum
from temp t
) t
) t
group by id,grp) t2
group by id;