Finden Sie das erste Vorkommen und die folgenden Vorkommen

Question

Hallo Alle Ich habe die untenstehende Tabelle, und ich möchte Menschen, die zum ersten Mal im Jahr 2014 kamen und die Besuche nach 2 Monaten ihres ersten Besuchs und in weniger als 13 Monaten hatte nach ihrem ersten Besuch.

ID firstname surname dob event_key event_date status 
1 Mark Zeus 12-Dec-67 10 02-Feb-14 Appointment 
1 Mark Zeus 12-Dec-67 10 02-Feb-14 Attend 
1 Mark Zeus 12-Dec-67 11 06-Jul-14 Appointment 
1 Mark Zeus 12-Dec-67 11 06-Jul-14 Attend 
1 Mark Zeus 12-Dec-67 12 23-Nov-14 Appointment 
1 Mark Zeus 12-Dec-67 12 23-Nov-14 Attend 
1 Mark Zeus 12-Dec-67 13 23-Nov-14 Appointment 
1 Mark Zeus 12-Dec-67 13 23-Nov-14 Attend 
2 Jane Adams 24-Jul-57 14 15-Mar-15 Appointment 
2 Jane Adams 24-Jul-57 14 15-Mar-15 Attend 
2 Jane Adams 24-Jul-57 15 20-Apr-14 Appointment 
2 Jane Adams 24-Jul-57 15 20-Apr-14 Attend 
2 Jane Adams 24-Jul-57 16 04-Jan-15 Appointment 
2 Jane Adams 24-Jul-57 16 04-Jan-15 Attend 
3 Rohan Clark 05-Jan-78 17 22-Apr-09 Appointment 
3 Rohan Clark 05-Jan-78 17 22-Apr-09 Attend 
3 Rohan Clark 05-Jan-78 19 21-Mar-14 Appointment 
3 Rohan Clark 05-Jan-78 19 21-Mar-14 Attend 
3 Rohan Clark 05-Jan-78 20 30-Jun-15 Appointment 
3 Rohan Clark 05-Jan-78 20 30-Jun-15 Attend 
3 Rohan Clark 05-Jan-78 21 01-Jan-16 Appointment 
3 Rohan Clark 05-Jan-78 21 01-Jan-16 Attend 
4 James Gohan 03-Jun-78 22 17-Aug-14 Appointment 
4 James Gohan 03-Jun-78 22 17-Aug-14 Attend 
4 James Gohan 03-Jun-78 23 30-Dec-14 Appointment 
4 James Gohan 03-Jun-78 23 30-Dec-14 Attend 
4 James Gohan 03-Jun-78 24 03-Mar-15 Appointment 
4 James Gohan 03-Jun-78 24 03-Mar-15 Attend 
4 James Gohan 03-Jun-78 25 12-Apr-16 Appointment 
4 James Gohan 03-Jun-78 25 12-Apr-16 Attend 
5 Isabel Robyn 07-Jul-56 26 07-Jul-15 Appointment 
5 Isabel Robyn 07-Jul-56 26 07-Jul-15 Attend 
5 Isabel Robyn 07-Jul-56 27 12-May-16 Appointment 
5 Isabel Robyn 07-Jul-56 27 12-May-16 Attend 
5 Isabel Robyn 07-Jul-56 28 18-Feb-17 Appointment 
5 Isabel Robyn 07-Jul-56 28 18-Feb-17 Attend 

Unten ist meine Abfrage, die eine verschachtelte Tabelle, die alle Menschen mit ihrem ersten Besuch zurückkehren können

SELECT    * 

FROM    #temp t1 

WHERE    t1.[event_date] BETWEEN '01-Jan-2014' AND '31-Dec-2016' 

        AND t1.[ID]     
        IN 
         (
          SELECT DISTINCT t2.[ID] 
          FROM #temp t2 
          WHERE t2.[event_date] <= DATEADD(month, 13, t2.[event_date]) AND t2.[event_date] >= DATEADD(month, 2, [event_date]) 
         ) 

ORDER BY   t1.[ID], t1.[event_key] 

-- DROP TABLE   #temp 

Answer 1

Sie nicht funktioniert hat haben, und die anderen Besucher mit dem vergleichen.

select distinct t1.id from Table1 t1 inner join 
    (select id, min(event_date) firstVisit from Table1 
    group by id 
    having min(event_date) between '2014-01-01' and '2014-12-31') t2 
    on t2.id=t1.id 
where DATEDIFF(MONTH,t2.firstVisit,t1.event_date) > 2 
    and DATEDIFF(MONTH,t2.firstVisit,t1.event_date) < 13 

das gibt zurück:

1 
2 
4 

Answer 2

Ich würde dies mit bedingter Aggregation tun:

select id 
from (select t.*, 
      row_number() over (partition by id order by event_date) as seqnum 
     from #temp t 
    ) t 
group by id 
having year(min(event_date)) = 2014 and 
     max(case when seqnum = 2 then event_date end) >= dateadd(month, 2, event_date) and 
     max(case when seqnum = 2 then event_date end) < dateadd(month, 13, event_date);