问题
I am trying to get to a lapsed_date which is when there are >12 weeks (ie. 84 days) for a given ID between:
1) onboarded_at and current_date (if no applied_at exists) - this means lapsed_now if >84 days
2) onboarded_at and min(applied_at) (if one exists)
3) each consecutive applied_at
4) max(applied_at) and current_date - this means lapsed_now if >84 days
If there are multiple instances where he lapsed, then we only show the latest lapsed date.
The attempt I have works for most but not all cases. Can you assists make it work universally?
Sample set:
CREATE TABLE #t
(
id VARCHAR(10),
rank INTEGER,
onboarded_at DATE,
applied_at DATE
);
INSERT INTO #t VALUES
('A',1,'20180101','20180402'),
('A',2,'20180101','20180403'),
('A',3,'20180101','20180504'),
('B',1,'20180201','20180801'),
('C',1,'20180301','20180401'),
('C',2,'20180301','20180501'),
('C',3,'20180301','20180901'),
('D',1,'20180401',null)
Best attempt:
SELECT onb.id,
onb.rank,
onb.onboarded_at,
onb.applied_at,
onb.lapsed_now,
CASE WHEN lapsed_now = 1 OR lapsed_previous = 1
THEN 1
ELSE 0
END lapsed_ever,
CASE WHEN lapsed_now = 1
THEN DATEADD(DAY, 84, lapsed_now_date)
ELSE min_applied_at_add_84
END lapsed_date
FROM
(SELECT *,
CASE
WHEN DATEDIFF(DAY, onboarded_at, MIN(ISNULL(applied_at, onboarded_at)) over (PARTITION BY id)) >= 84
THEN 1
WHEN DATEDIFF(DAY, MAX(applied_at) OVER (PARTITION BY id), GETDATE()) >= 84
THEN 1
ELSE 0
END lapsed_now,
CASE
WHEN MAX(DATEDIFF(DAY, onboarded_at, ISNULL(applied_at, GETDATE()))) OVER (PARTITION BY id) >= 84
THEN 1
ELSE 0
END lapsed_previous,
MAX(applied_at) OVER (PARTITION BY id) lapsed_now_date,
DATEADD(DAY, 84, MIN(CASE WHEN applied_at IS NULL THEN onboarded_at ELSE applied_at END) OVER (PARTITION BY id)) min_applied_at_add_84
FROM #t
) onb
Current solution:
id rank onboarded_at applied_at lapsed_now lapsed_ever lapsed_date
A 1 2018-01-01 2018-04-02 1 1 2018-07-27
A 2 2018-01-01 2018-04-03 1 1 2018-07-27
A 3 2018-01-01 2018-05-04 1 1 2018-07-27
B 2 2018-02-01 2018-08-01 1 1 2018-10-24
C 1 2018-03-01 2018-04-01 0 1 2018-06-24
C 2 2018-03-01 2018-05-01 0 1 2018-06-24
C 3 2018-03-01 2018-09-01 0 1 2018-06-24
D 1 2018-04-01 null 1 1 2018-06-24
Expected solution:
id rank onboarded_at applied_at lapsed_now lapsed_ever lapsed_date
A 1 2018-01-01 2018-04-02 1 1 2018-07-27 (not max lapsed date)
A 2 2018-01-01 2018-04-03 1 1 2018-07-27
A 3 2018-01-01 2018-05-04 1 1 2018-07-27 (May 4 + 84)
B 1 2018-02-01 2018-08-01 0 1 2018-04-26 (Feb 1 + 84)
C 1 2018-03-01 2018-04-01 0 1 2018-07-24
C 2 2018-03-01 2018-05-01 0 1 2018-07-24 (May 1 + 84)
C 3 2018-03-01 2018-09-01 0 1 2018-07-24
D 1 2018-04-01 null 1 1 2018-06-24
回答1:
Bit of guesswork here, but hopefully this does the trick:
SELECT res.id,
res.rank,
res.onboarded_at,
res.applied_at,
res.lapsed_now,
CASE WHEN lapsed_now = 1 OR lapsed_previous = 1
THEN 1
ELSE 0
END lapsed_ever,
CASE
WHEN lapsed_now = 1
THEN DATEADD(DAY, 84, lapsed_now_date)
WHEN applied_difference_gt84 IS NOT NULL
THEN DATEADD(DAY, 84, applied_difference_gt84)
WHEN DATEDIFF(DAY, min_applied_at_add_84, GETDATE()) < 84
THEN DATEADD(DAY, 84, onboarded_at)
ELSE min_applied_at_add_84
END lapsed_date
FROM (
SELECT *, MAX(applied_difference) OVER (PARTITION BY id ORDER BY rank ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) applied_difference_gt84
FROM
(
SELECT *,
CASE
WHEN DATEDIFF(DAY, onboarded_at, MIN(ISNULL(applied_at, onboarded_at)) over (PARTITION BY id)) >= 84
AND DATEDIFF(DAY, MAX(applied_at) OVER (PARTITION BY id), GETDATE()) >= 84
THEN 1
WHEN DATEDIFF(DAY, ISNULL(MAX(applied_at) OVER (PARTITION BY id), onboarded_at), GETDATE()) >= 84
THEN 1
ELSE 0
END lapsed_now,
CASE
WHEN MAX(DATEDIFF(DAY, onboarded_at, ISNULL(applied_at, GETDATE()))) OVER (PARTITION BY id) >= 84
THEN 1
ELSE 0
END lapsed_previous,
CASE
WHEN DATEDIFF(MONTH, applied_at, LEAD(applied_at, 1) OVER (PARTITION BY id ORDER BY rank)) >= 2
THEN applied_at
ELSE NULL
END applied_difference,
ISNULL(MAX(applied_at) OVER (PARTITION BY id), onboarded_at) lapsed_now_date,
DATEADD(DAY, 84, MIN(CASE WHEN applied_at IS NULL THEN onboarded_at ELSE applied_at END) OVER (PARTITION BY id)) min_applied_at_add_84
FROM #t
) onb
) res
Results:
id rank onboarded_at applied_at lapsed_now lapsed_ever lapsed_date
A 1 2018-01-01 2018-04-02 1 1 2018-07-27
A 2 2018-01-01 2018-04-03 1 1 2018-07-27
A 3 2018-01-01 2018-05-04 1 1 2018-07-27
B 1 2018-02-01 2018-08-01 0 1 2018-04-26
C 1 2018-03-01 2018-04-01 0 1 2018-07-24
C 2 2018-03-01 2018-05-01 0 1 2018-07-24
C 3 2018-03-01 2018-09-01 0 1 2018-07-24
D 1 2018-04-01 (null) 1 1 2018-06-24
It's a bit messy because of the need to calculate the difference between the applied_at dates.
回答2:
@Jim, inspired by your answer, I created the following solution. I think it is easily understandable and intuitive, knowing the lapsed criteria:
SELECT id, onboarded_at, applied_at,
max(case when (zero_applicants is not null and current_date - onboarded_at > 84) or (last_applicant is not null and current_date - last_applicant > 84) then 1 else 0 end) over (partition by id) lapsed_now,
max(case when (zero_applicants is not null and current_date - onboarded_at > 84) or (one_applicant is not null and applied_at - onboarded_at > 84)
or (one_applicant is not null and current_date - applied_at > 84) or (next_applicant is not null and next_applicant- applied_at > 84)
or (last_applicant is not null and current_date - last_applicant > 84) then 1 else 0 end) over(partition by id) lapsed_ever,
max(case when zero_applicants is not null and current_date - onboarded_at > 84 then onboarded_at + 84
when one_applicant is not null and applied_at - onboarded_at > 84 then onboarded_at + 84
when one_applicant is not null and current_date - applied_at > 84 then applied_at + 84
when next_applicant is not null and next_applicant - applied_at > 84 then applied_at + 84
when last_applicant is not null and current_date - last_applicant > 84 then last_applicant + 84
end) over (partition by id) lapsed_date
from (
select *,
case when MAX(applied_at) OVER (PARTITION BY id) is null then onboarded_at end as zero_applicants,
case when count(applied_at) over(partition by id)=1 then onboarded_at end as one_applicant,
case when count(applied_at) over(partition by id)>1 then LEAD(applied_at, 1) OVER (PARTITION BY id ORDER BY applied_at) end as next_applicant,
case when LEAD(applied_at, 1) OVER (PARTITION BY id ORDER BY applied_at) is null then MAX(applied_at) over(partition by id) end as last_applicant
from #t
) res
order by id, applied_at
来源：https://stackoverflow.com/questions/52501443/create-a-lapsed-concept-based-on-logic-across-every-row-per-id