mysql - 如何找到超过 2 个用户的匹配时间间隔

Question

从不同用户的给定时间间隔中找到最合适的时间。

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

上表显示了不同用户可用的不同免费时间段，例如:

user1 在

有空

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 仅在此时间段内免费:

2012-07-18 21:10:47   to   2012-07-18 23:10:47 

现在我想找出一个最佳时间间隔，双方用户都可以在这个时间间隔内安排他们的 session 。

Answer 1

要查找 user1 和 user2 何时都空闲，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}

SQL FIDDLE HERE.

编辑:要比较 2 个以上的用户，请尝试以下操作:

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 ;

需要根据用户数改变m.userid in (2,4,3,5)和having count(rn)=4。

SQL FIDDLE HERE