从给定的不同用户的时间间隔找到最合适的时间.
Rows: 5
fid userid FromDateTime ToDateTime flag
62 1 2012-07-18 01:48:20 2012-07-18 02:55:20 1
63 1 2012-07-18 10:30:46 2012-07-18 12:54:46 1
64 1 2012-07-18 18:50:24 2012-07-18 20:35:24 1
67 1 2012-07-18 15:03:36 2012-07-18 16:03:36 1
68 2 2012-07-18 21:10:47 2012-07-18 23:10:47 1
user1是免费的
2012-07-18 01:48:20 to 2012-07-18 02:55:20,2012-07-18 10:30:46 to 2012-07-18 12:54:46
......
用户2仅在此时间段之间免费：
2012-07-18 21:10:47 to 2012-07-18 23:10:47
现在我想找出一个用户可以安排会议的最佳时间间隔.
最佳答案
要查找user1和user2都是免费的,请尝试以下操作：
原文链接：https://www.f2er.com/mysql/434174.htmlselect
a.datetime_start as user1start,a.datetime_end as user1end,b.datetime_start as user2start,b.datetime_end as user2end,case when a.datetime_start > b.datetime_start then a.datetime_start
else b.datetime_start end as avail_start,case when a.datetime_end>b.datetime_end then b.datetime_end
else a.datetime_end end as avail_end
from users a inner join users b on
a.datetime_start<=b.datetime_end and a.datetime_end>=b.datetime_start
and a.userid={user1} and b.userid={user2}
select max(datetime_start) as avail_start,min(datetime_end) as avail_end
from(
select *,@rn := CASE WHEN @prev_start <=datetime_end and @prev_end >=datetime_start THEN @rn ELSE @rn+1 END AS rn,@prev_start := datetime_start,@prev_end := datetime_end
from(
select * from users2 m
where exists ( select null
from users2 o
where o.datetime_start <= m.datetime_end and o.datetime_end >= m.datetime_start
and o.id <> m.id
)
and m.userid in (2,4,3,5)
order by m.datetime_start) t,(SELECT @prev_start := -1,@rn := 1,@prev_end=-1) AS vars
) c
group by rn
having count(rn)=4 ;
需要根据用户数更改(2,5)中的m.userid并使count(rn)= 4.