Selecting the most recent date - sql

Question

I have data structured like this:
ID | Enrolment_Date | Appointment1_Date | Appointment2_Date | .... | Appointment150_Date |
112 01/01/2015 01/02/2015 01/03/2018 01/08/2018
113 01/06/2018 01/07/2018 NULL NULL
114 01/04/2018 01/05/2018 01/06/2018 NULL
I need a new variable which counts the number of months between the enrolment_date and the most recent appointment. The challenge is is that all individuals have a different number of appointments.
Update: I agree with the comments that this is poor table design and it needs to be reformatted. Could proposed solutions please include suggested code on how to transform the table?

Answer 1

Since the OP is currently stuck with this bad design, I will point out a temporary solution. As others have suggested, you really must change the structure here. For now, this will suffice:
SELECT '['+ NAME + '],' FROM sys.columns WHERE OBJECT_ID = OBJECT_ID ('TableA') -- find all columns, last one probably max appointment date
SELECT ID,
Enrolment_Date,
CASE WHEN Appointment150_Date IS NOT NULL THEN DATEDIFF (MONTH, Enrolment_Date, Appointment150_Date)
WHEN Appointment149_Date IS NOT NULL THEN DATEDIFF (MONTH, Enrolment_Date, Appointment149_Date)
WHEN Appointment148_Date IS NOT NULL THEN DATEDIFF (MONTH, Enrolment_Date, Appointment148_Date)
WHEN Appointment147_Date IS NOT NULL THEN DATEDIFF (MONTH, Enrolment_Date, Appointment147_Date)
WHEN Appointment146_Date IS NOT NULL THEN DATEDIFF (MONTH, Enrolment_Date, Appointment146_Date)
WHEN Appointment145_Date IS NOT NULL THEN DATEDIFF (MONTH, Enrolment_Date, Appointment145_Date)
WHEN Appointment144_Date IS NOT NULL THEN DATEDIFF (MONTH, Enrolment_Date, Appointment144_Date) -- and so on
END AS NumberOfMonths
FROM TableA
This is a very ugly temporary solution and should be considered as such.

Answer 2

You will need to restructure your data, the given structure is poor database design. Create two separate tables - one called users and one called appointments. The users table contains the user id, enrollment date and any other specific user information. Each row in the appointments table contains the user's unique id and a specific appointment date. Structuring your tables like this will make it easier to write a query to get days/months since last appointment.
For example:
Users Table:
ID, Enrollment_Date
1, 2018-01-01
2, 2018-03-02
3, 2018-05-02
Appointments Table:
ID, Appointment_Date
1, 2018-01-02
1, 2018-02-02
1, 2018-02-10
2, 2018-05-01
You would then be able to write a query to join the two tables together and calculate the difference between the enrollment date and min value of the appointment date.

Answer 3

It is better if you can create two tables.
Enrolment Table (dbo.Enrolments)
ID | EnrolmentDate
1 | 2018-08-30
2 | 2018-08-31
Appointments Table (dbo.Appointments)
ID | EnrolmentID | AppointmentDate
1 | 1 | 2018-09-02
2 | 1 | 2018-09-03
3 | 2 | 2018-09-01
4 | 2 | 2018-09-03
Then you can try something like this.
If you want the count of months from Enrolment Date to the final appointment date then use below query.
SELECT E.ID, E.EnrolmentDate, A.NoOfMonths
FROM dbo.Enrolments E
OUTER APPLY
(
SELECT DATEDIFF(mm, E.EnrolmentDate, MAX(A.AppointmentDate)) AS NoOfMonths
FROM dbo.Appointments A
WHERE A.EnrolmentId = E.ID
) A
And, If you want the count of months from Enrolment Date to the nearest appointment date then use below query.
SELECT E.ID, E.EnrolmentDate, A.NoOfMonths
FROM dbo.Enrolments E
OUTER APPLY
(
SELECT DATEDIFF(mm, E.EnrolmentDate, MIN(A.AppointmentDate)) AS NoOfMonths
FROM dbo.Appointments A
WHERE A.EnrolmentId = E.ID
) A
Try this on sqlfiddle

Answer 4

You have a lousy data structure, as others have noted. You really one a table with one row per appointment. After all, what happens after the 150th appointment?
select t.id, t.Enrolment_Date,
datediff(month, t.Enrolment_Date, m.max_Appointment_Date) as months_diff
from t cross apply
(select max(Appointment_Date) as max_Appointment_Date
from (values (Appointment1_Date),
(Appointment2_Date),
. . .
(Appointment150_Date)
) v(Appointment_Date)
) m;