要查找user1和user2都是免费的,请尝试以下操作：
select
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}
编辑：
要比较超过2个用户,请尝试以下内容：
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
需要根据用户数更改(2,5)中的m.userid并使count(rn)= 4.
