Find the true start end dates for customers that have multiple accounts in SQL Server 2014 - sql

Question

I have a checking account table that contains columns Cust_id (customer id), Open_Date (start date), and Closed_Date (end date). There is one row for each account. A customer can open multiple accounts at any given point. I would like to know how long the person has been a customer.
eg 1:
CREATE TABLE [Cust]
(
[Cust_id] [varchar](10) NULL,
[Open_Date] [date] NULL,
[Closed_Date] [date] NULL
)
insert into [Cust] values ('a123', '10/01/2019', '10/15/2019')
insert into [Cust] values ('a123', '10/12/2019', '11/01/2019')
Ideally I would like to insert this into a table with just one row, that says this person has been a customer from 10/01/2019 to 11/01/2019. (as he opened his second account before he closed his previous one.
Similarly eg 2:
insert into [Cust] values ('b245', '07/01/2019', '09/15/2019')
insert into [Cust] values ('b245', '10/12/2019', '12/01/2019')
I would like to see 2 rows in this case- one that shows he was a customer from 07/01 to 09/15 and then again from 10/12 to 12/01.
Can you point me to the best way to get this?

Answer 1

I would approach this as a gaps and islands problem. You want to group together groups of adjacents rows whose periods overlap.
Here is one way to solve it using lag() and a cumulative sum(). Everytime the open date is greater than the closed date of the previous record, a new group starts.
select
cust_id,
min(open_date) open_date,
max(closed_date) closed_date
from (
select
t.*,
sum(case when not open_date <= lag_closed_date then 1 else 0 end)
over(partition by cust_id order by open_date) grp
from (
select
t.*,
lag(closed_date) over (partition by cust_id order by open_date) lag_closed_date
from cust t
) t
) t
group by cust_id, grp
In this db fiddle with your sample data, the query produces:
cust_id | open_date | closed_date
:------ | :--------- | :----------
a123 | 2019-10-01 | 2019-11-01
b245 | 2019-07-01 | 2019-09-15
b245 | 2019-10-12 | 2019-12-01

Answer 2

I would solve this with recursion. While this is certainly very heavy, it should accommodate even the most complex account timings (assuming your data has such). However, if the sample data provided is as complex as you need to solve for, I highly recommend sticking with the solution provided above. It is much more concise and clear.
WITH x (cust_id, open_date, closed_date, lvl, grp) AS (
SELECT cust_id, open_date, closed_date, 1, 1
FROM (
SELECT cust_id
, open_date
, closed_date
, row_number()
OVER (PARTITION BY cust_id ORDER BY closed_date DESC, open_date) AS rn
FROM cust
) AS t
WHERE rn = 1
UNION ALL
SELECT cust_id, open_date, closed_date, lvl, grp
FROM (
SELECT c.cust_id
, c.open_date
, c.closed_date
, x.lvl + 1 AS lvl
, x.grp + CASE WHEN c.closed_date < x.open_date THEN 1 ELSE 0 END AS grp
, row_number() OVER (PARTITION BY c.cust_id ORDER BY c.closed_date DESC) AS rn
FROM cust c
JOIN x
ON x.cust_id = c.cust_id
AND c.open_date < x.open_date
) AS t
WHERE t.rn = 1
)
SELECT cust_id, min(open_date) AS first_open_date, max(closed_date) AS last_closed_date
FROM x
GROUP BY cust_id, grp
ORDER BY cust_id, grp
I would also add the caveat that I don't run on SQL Server, so there could be syntax differences that I didn't account for. Hopefully they are minor, if present.

Answer 3

you can try something like that:
select distinct
cust_id,
(select min(Open_Date)
from Cust as b
where b.cust_id = a.cust_id and
a.Open_Date <= b.Closed_Date and
a.Closed_Date >= b.Open_Date
),
(select max(Closed_Date)
from Cust as b
where b.cust_id = a.cust_id and
a.Open_Date <= b.Closed_Date and
a.Closed_Date >= b.Open_Date
)
from Cust as a
so, for every row - you're selecting minimal and maximal dates from all overlapping ranges, later distinct filters out duplicates