calculating the difference between two rows or alternative - sql

Question

I would like to take this query (see below) and add a where win = in the select statement. here I would like to add a column to show the number of races it took to fulfill the where e.g below where win = 2
I've tried calculating the number between rows but it was wildly wrong on my part
select
date, time, raceid, win
from master
where date = #date
order by time
DATE TIME RACEID WIN
2019-01-06 00:40:00 4445 2
2019-01-06 00:50:00 4432 0
2019-01-06 01:00:00 4441 2
2019-01-06 01:10:00 4446 2
2019-01-06 01:20:00 4433 1
2019-01-06 01:30:00 4439 1
2019-01-06 01:40:00 4447 2
2019-01-06 01:50:00 4434 2
2019-01-06 02:00:00 4442 0
2019-01-06 02:10:00 4448 0
2019-01-06 02:20:00 4435 2
2019-01-06 02:30:00 4443 2
2019-01-06 02:40:00 4449 2
2019-01-06 02:50:00 4436 0
2019-01-06 02:50:00 4444 2
I would like to take this query and add a where win = in the select statement. here I would like to add a column to show the number of races it took to fulfill the where e.g below where win = 2
DATE TIME RACEID WIN RacestoWin
2019-01-06 00:40:00 4445 2 1
2019-01-06 01:00:00 4441 2 2
2019-01-06 01:10:00 4446 2 1
2019-01-06 01:40:00 4447 2 3
2019-01-06 01:50:00 4434 2 1
2019-01-06 02:20:00 4435 2 3
2019-01-06 02:30:00 4443 2 1
2019-01-06 02:40:00 4449 2 1
2019-01-06 02:50:00 4444 2 2
Is there a simple way of doing this? Not the best so any guidance would be greatly appreciated!!

Answer 1

I see. You are counting the rows between the wins. Basically, you want to assign a group. This group is the cumulative number of 2s on or after that record. Then, within each group, you can use row_number() or even aggregation in this case (because you know the last row of the group is "2"):
select date, max(time), 2 as win, count(*) as racestowin
from (select m.*,
sum(case when m.win = 2 then 1 else 0 end) over (partition by m.date order by m.time desc) as grouping
from master m
) m
group by date, grouping;