要查找user1和user2都是免費(fèi)的,請(qǐng)嘗試以下操作：

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}

編輯：

要比較超過(guò)2個(gè)用戶,請(qǐng)嘗試以下內(nèi)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

需要根據(jù)用戶數(shù)更改(2,5)中的m.userid并使count(rn)= 4.

總結(jié)

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