Selecting the number of last consecutive days from timestamp (excepting today) - mysql

Question

I have a table A_DailyLogins with the columns ID (auto increment), Key (userid) and Date (timestamp). I want a query which would return the number of last consecutive days from those timestamp based on the Key, for example if he has a row for yesterday, one for two days ago and another one for three days ago, but the last one isn't from four days ago, it would return 3, because this is the number of last days the user was logged in.
My attempt was to create a query selecting the last 7 rows of the players ordered by Date DESC (this is what I wanted in the first place, but then I thought that it would be great to have all the last consecutive days), and then I retrieved the query result and compared the dates (converted to year/month/day with functions from that language [Pawn]) and increased the number of consecutive days when a date is before the other one with one day. (but this is extremely slow compared to what I think that can be done directly only with MySQL)
The closest thing I found is this: Check for x consecutive days - given timestamps in database . But it still isn't how I want it to be, it's still pretty different. I tried to modify it, but it is way too hard for me, I don't have that much experience in MySQL.

Answer 1

context
let consecutive login period be a period where the user is logged in on all days ( has an entry in A_DailyLogins on every day in period ) where there is no entry in A_DailyLogins immediately before or after the consecutive login period with the same user
and number of consecutive days be the difference between the maximum and minumum dates in a consecutive login period
the maximum date of a consecutive login period has no login entry immediately after ( sequentially ) to it..
the minimum date of a consecutive login period has no login entry immediately previous ( sequentially ) to it..
plan
left join A_DailyLogins to itself using same user and sequential dates where right is null to find maximums
analogous logic to find minimums
calculate row ordering over minimums and maximums with appropriate order by
join maximums and minimums on row number
filter where maximum login is yesterday/today
calculate date_diff between maximum and minimum in range
left join users to above resultset and coalesce over the case where user does not have a consecutive login period ending yesterday/today
input
+----+------+------------+
| ID | Key | Date |
+----+------+------------+
| 25 | eric | 2015-12-23 |
| 26 | eric | 2015-12-25 |
| 27 | eric | 2015-12-26 |
| 28 | eric | 2015-12-27 |
| 29 | eric | 2016-01-01 |
| 30 | eric | 2016-01-02 |
| 31 | eric | 2016-01-03 |
| 32 | nusa | 2015-12-27 |
| 33 | nusa | 2015-12-29 |
+----+------+------------+
query
select all_users.`Key`,
coalesce(nconsecutive, 0) as nconsecutive
from
(
select distinct `Key`
from A_DailyLogins
) all_users
left join
(
select
lower_login_bounds.`Key`,
lower_login_bounds.`Date` as from_login,
upper_login_bounds.`Date` as to_login,
1 + datediff(least(upper_login_bounds.`Date`, date_sub(current_date, interval 1 day))
, lower_login_bounds.`Date`) as nconsecutive
from
(
select curr_login.`Key`, curr_login.`Date`, #rn1 := #rn1 + 1 as row_number
from A_DailyLogins curr_login
left join A_DailyLogins prev_login
on curr_login.`Key` = prev_login.`Key`
and prev_login.`Date` = date_add(curr_login.`Date`, interval -1 day)
cross join ( select #rn1 := 0 ) params
where prev_login.`Date` is null
order by curr_login.`Key`, curr_login.`Date`
) lower_login_bounds
inner join
(
select curr_login.`Key`, curr_login.`Date`, #rn2 := #rn2 + 1 as row_number
from A_DailyLogins curr_login
left join A_DailyLogins next_login
on curr_login.`Key` = next_login.`Key`
and next_login.`Date` = date_add(curr_login.`Date`, interval 1 day)
cross join ( select #rn2 := 0 ) params
where next_login.`Date` is null
order by curr_login.`Key`, curr_login.`Date`
) upper_login_bounds
on lower_login_bounds.row_number = upper_login_bounds.row_number
where upper_login_bounds.`Date` >= date_sub(current_date, interval 1 day)
and lower_login_bounds.`Date` < current_date
) last_consecutive
on all_users.`Key` = last_consecutive.`Key`
;
output
+------+------------------+
| Key | last_consecutive |
+------+------------------+
| eric | 2 |
| nusa | 0 |
+------+------------------+
valid as run on 2016-01-03
sqlfiddle