SQL - get counts based on rolling window per unique id - sql

Question

I'm working with a table that has an id and date column. For each id, there's a 90-day window where multiple transactions can be made. The 90-day window starts when the first transaction is made and the clock resets once the 90 days are over. When the new 90-day window begins triggered by a new transaction I want to start the count from the beginning at one. I would like to generate something like this with the two additional columns (window and count) in SQL:
id date window count
name1 7/7/2019 first 1
name1 12/31/2019 second 1
name1 1/23/2020 second 2
name1 1/23/2020 second 3
name1 2/12/2020 second 4
name1 4/1/2020 third 1
name2 6/30/2019 first 1
name2 8/14/2019 first 2
I think getting the rank of the window can be done with a CASE statement and MIN(date) OVER (PARTITION BY id). This is what I have in mind for that:
CASE WHEN MIN(date) OVER (PARTITION BY id) THEN 'first'
WHEN DATEDIFF(day, date, MIN(date) OVER (PARTITION BY id)) <= 90 THEN 'first'
WHEN DATEDIFF(day, date, MIN(date) OVER (PARTITION BY id)) > 90 AND DATEDIFF(day, date, MIN(date) OVER (PARTITION BY id)) <= 180 THEN 'third'
WHEN DATEDIFF(day, date, MIN(date) OVER (PARTITION BY id)) > 180 AND DATEDIFF(day, date, MIN(date) OVER (PARTITION BY id)) <= 270 THEN 'fourth'
ELSE NULL END
And incrementing the counts within the windows would be ROW_NUMBER() OVER (PARTITION BY id, window)?

Answer 1

You cannot solve this problem with window functions only. You need to iterate through the dataset, which can be done with a recursive query:
with
tab as (
select t.*, row_number() over(partition by id order by date) rn
from mytable t
)
cte as (
select id, date, rn, date date0 from tab where rn = 1
union all
select t.id, t.date, t.rn, greatest(t.date, c.date + interval '90' day)
from cte c
inner join tab t on t.id = c.id and t.rn = c.rn + 1
)
select
id,
date,
dense_rank() over(partition by id order by date0) grp,
count(*) over(partition by id order by date0, date) cnt
from cte
The first query in the with clause ranks records having the same id by increasing date; then, the recursive query traverses the data set and computes the starting date of each group. The last step is numbering the groups and computing the window count.

Answer 2

GMB is totally correct that a recursive CTE is needed. I offer this as an alternative form for two reasons. First, because it uses SQL Server syntax, which appears to be the database being used in the question. Second, because it directly calculates window and count without window functions:
with t as (
select t.*, row_number() over (partition by id order by date) as seqnum
from tbl t
),
cte as (
select t.id, t.date, dateadd(day, 90, t.date) as window_end, 1 as window, 1 as count, seqnum
from t
where seqnum = 1
union all
select t.id, t.date,
(case when t.date > cte.window_end then dateadd(day, 90, t.date)
else cte.window_end
end) as window_end,
(case when t.date > cte.window_end then window + 1 else window end) as window,
(case when t.date > cte.window_end then 1 else cte.count + 1 end) as count,
t.seqnum
from cte join
t
on t.id = cte.id and
t.seqnum = cte.seqnum + 1
)
select id, date, window, count
from cte
order by 1, 2;
Here is a db<>fiddle.