[英]SQL - find all combinations
从这样的数据集中,我需要在同一个房间中获得所有可能的案例组合,以便没有案例与另一个重叠。
room case start end
a 1 2019-11-27 09:00 2019-11-27 10:15
a 2 2019-11-27 10:30 2019-11-27 12:00
a 3 2019-11-27 12:00 2019-11-27 12:30
b 4 2019-11-27 08:30 2019-11-27 10:30
b 5 2019-11-27 10:00 2019-11-27 12:00
b 6 2019-11-27 11:00 2019-11-27 12:20
预期的结果是
room combination cases
a 1 1
a 2 1,2
a 3 1,2,3
a 4 2
a 5 2,3
a 6 3
b 1 4
b 2 4,6
b 3 5
b 4 6
我能够获得哪种情况可以与其他情况相结合的单行结果,例如:
room case combinewithcase
a 1 2
a 1 3
a 2 1
a 2 3
a 3 1
a 3 2
b 4 6
b 6 4
我已经尝试了一些递归,但我无法获得我需要的结果类型,我很感激任何人都可以分享的任何指导。
这是使用字符串聚合的一种方法。 要按时间获得重叠:
select rd.room, rd.dte,
(select string_agg(t2.case, ',') within group (order by t2.case)
from t t2
where t2.room = t.room and
t2.start <= t.dte and
t2.end > t.dte
) as cases
from ((select room, start as dte from t
) union -- on purpose to remove duplicates
(select room, end from t
)
) rd;
然后,您可以将其用作子查询/CTE 来枚举组合:
select room, cases, count(*),
row_number() over (partition by room order by combination) as combination
from (select rd.room, rd.dte,
(select string_agg(t2.case, ',') within group (order by t2.case)
from t t2
where t2.room = t.room and
t2.start <= t.dte and
t2.end > t.dte
) as cases
from ((select room, start as dte from t
) union -- on purpose to remove duplicates
(select room, end from t
)
) rd
) rd
group by room, cases
order by room, cases;
编辑:
在早期版本的 SQL Server 中,您可以使用 XML 方法进行字符串聚合:
select room, cases, count(*),
row_number() over (partition by room order by combination) as combination
from (select rd.room, rd.dte,
stuff( (select concat(',', t2.case)
from t t2
where t2.room = t.room and
t2.start <= t.dte and
t2.end > t.dte
order by t2.case
for xml path
), 1, 1, '')
) as cases
from ((select room, start as dte from t
) union -- on purpose to remove duplicates
(select room, end from t
)
) rd
) rd
group by room, cases
order by room, cases;
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.