Question

我有以下查询：

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

获取至少两次“A”，“B”或“C”类型访问的所有患者的清单。

patient_visit表还有一个visit_date列，用于存储访问日期。我的问题：是否有可能修改上述查询而无需删除group by语句来查询“至少两次访问的所有患者以及这两次访问中的任何一个有60的差距天数“？

谢谢！

P.S。：我正在使用Oracle，如果有内置函数，我也可以使用它。

Answer 1

任何两个日期，所以第一次和最后一次访问都符合条件？

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

如果你的意思是连续的，那么

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

这是一个昂贵的查询，有点O（N ³）。 Oracle LAG版本可能更快。

Answer 2

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

Answer 3

我没有oracle进行测试，但我认为这将有效

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'

如何使用日期间隔执行SQL组

3 个答案: