How to do a SQL group with date gap

I have the following query:

 SELECT patient_id FROM patient_visit where visit_type in ('开发者_Python百科A', 'B', 'C') 
 group by patient_id having count(*) >= 2

To get a list of all patients that had at least two visits of type 'A', 'B', or 'C'.

The patient_visit table also has a visit_date column which stores the date of the visit. My question: is it possible to modify the above query WITHOUT removing the group by statement to query "all patients with at least two visits AND where any of those two visit had a gap of 60 number of days"?

Thanks!

P.S.: i'm using Oracle, if there's a built-in function, I can use that too.

---

Any two dates, so the first and last visits would qualify?

SELECT patient_id
FROM patient_visit
where visit_type in ('A', 'B', 'C') 
group by patient_id
having count(*) >= 2 AND MAX(visit_date) - MIN(visit_date) >= 60

If you meant consecutive, then

SELECT patient_id
FROM patient_visit
where visit_type in ('A', 'B', 'C') 
  AND EXISTS (
    select *
    from patient_visit v
    where v.visit_type in ('A', 'B', 'C')
      and v.patient_id = patient_visit.patient_id
      and v.visit_date >= patient_visit.visit_date + 60)
  AND NOT EXISTS (
    select *
    from patient_visit v2
    where v2.visit_type in ('A', 'B', 'C')
      and v2.patient_id = patient_visit.patient_id
      and v2.visit_date > patient_visit.visit_date
      and v2.visit_date < patient_visit.visit_date + 60)
group by patient_id

This is an expensive query, something of the order O(N³). The Oracle LAG version could be faster.

---

SQL> create table patient_visit (patient_id number(38) not null
  2      , visit_type varchar2(1) not null
  3      , visit_date date not null);

Table created.

SQL> insert into patient_visit
  2  select 1, 'A', date '2010-01-01' from dual
  3  union all select 1, 'D', date '2010-01-02' from dual
  4      -- ignore, by type
  5  union all select 1, 'C', date '2010-01-01' + 60 from dual
  6      -- 1 is included
  7  union all select 1, 'B', date '2011-01-01' from dual
  8      -- don't include 1 more than once
  9  union all select 2, 'A', date '2010-01-01' from dual
 10  union all select 2, 'B', date '2010-01-02' from dual
 11      -- breaks up 60 day gap.
 12  union all select 2, 'C', date '2010-01-01' + 60 from dual;

7 rows created.

SQL> commit;

Commit complete.

SQL> select patient_id
  2  from (select patient_id
  3          , visit_date
  4          , lag(visit_date) over (partition by patient_id
  5              order by visit_date) prior_visit_date
  6      from patient_visit
  7      where visit_type in ('A', 'B', 'C'))
  8  where visit_date - prior_visit_date >= 60
  9  group by patient_id;

PATIENT_ID
----------
         1

SQL> spool off

---

I dont have oracle to test but I think this will work

select patient_id from 
   (SELECT patient_id, dateField FROM patient_visit where visit_type in ('A','B', 'C') 
   group by patient_id having count(*) >= 2) as temp 
where temp.dateField > '2011-01-01'