如何使用日期间隔创建SQL组
我有以下疑问:如何使用日期间隔创建SQL组,sql,oracle,group-by,aggregate,aggregate-functions,Sql,Oracle,Group By,Aggregate,Aggregate Functions,我有以下疑问: 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”类就诊的所有患者的列表 “患者就诊表”还有一个“就诊日期”列,用于存储就诊日期。我的问题:是否可以在不删除group by语句的情况下修改上述查询,以查询至少两次就诊的所有患者,以及其中任何一次就诊间隔为60
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”类就诊的所有患者的列表
“患者就诊表”还有一个“就诊日期”列,用于存储就诊日期。我的问题:是否可以在不删除group by语句的情况下修改上述查询,以查询至少两次就诊的所有患者,以及其中任何一次就诊间隔为60
天数
谢谢
注:我正在使用Oracle,如果有内置功能,我也可以使用 我没有oracle要测试,但我认为这会起作用
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'
我没有甲骨文测试,但我认为这将工作
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'
任何两个日期,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
如果你是说连续,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,有点像3。Oracle LAG版本可能更快。任何两个日期,那么第一次和最后一次访问都符合条件吗
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
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
如果你是说连续,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,有点像3。Oracle LAG版本可能更快。您可能指的是这两次连续访问中的任何一次,而不仅仅是这两次访问中的任何一次,即您不是在寻找一个使用MAX和Min的答案。为什么限制保留组?@Martin最好是任意两次dates@Shannon我有这个限制,因为这个语句是通过XML到SQL编译器。保留Group by语句将使其成为代码生成器的一个简单更改。您可能是指这两次连续访问中的任何一次,而不仅仅是这两次访问中的任何一次,即,您不是在寻找一个使用MAX和MIN的答案。为什么保留Group by的限制?@Martin最好是任意两次dates@Shannon我有这个限制是因为此语句是通过XML到SQL编译器生成的代码。保留Group by语句将使其成为对代码生成器的简单更改。首先,如果未通过子查询返回,则无法选择on DATEFILD。其次,如果未通过子查询返回,则无法选择on DATEFILD。
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