我有酒店预订系统,当用户输入新的预订日期时,我需要检索所有未预订的房间,
预订表包含room_id,startdate,period
我写了这个查询,只需要简化它
select *
from rooms
where id in (
select room_id
from reservations
where '2018-6-11' not BETWEEN date_at and date_add(date_at, INTERVAL period day)
and '2018-6-30' not between date_at and date_add(date_at, INTERVAL period day)
and date_at not between '2018-6-11' and '2018-6-30'
and date_add(date_at ,INTERVAL period day ) not between '2018-6-11' and '2018-6-30' )
or id not in (select room_id from reservations)
答案 0 :(得分:1)
如何将IN
子句转换为等效连接?如果它适用于您的情况,请参阅以下查询:
SELECT *
FROM rooms rm
JOIN
(SELECT room_id rmid
FROM reservations
WHERE '2018-6-11' NOT BETWEEN date_at AND date_add(date_at, INTERVAL period DAY)
AND '2018-6-30' NOT BETWEEN date_at AND date_add(date_at, INTERVAL period DAY)
AND date_at NOT BETWEEN '2018-6-11' AND '2018-6-30'
AND date_add(date_at ,INTERVAL period DAY ) NOT BETWEEN '2018-6-11' AND '2018-6-30'
) tb1
ON rm.ID = tb.rmid
JOIN reservations rv
ON rm.ID <> rv.room_id;
答案 1 :(得分:0)
让我们将其分解为2个子问题。
在查询中,我假设$start_date
和$end_date
是PHP变量,其中存储了用户选择的值。
(select *
from rooms where room_id NOT IN (select room_id from reservations))
UNION
(select *
from rooms where room_id IN (
select room_id
from reservations
where $start_date not BETWEEN date_at and date_add(date_at, INTERVAL period day)
and $end_date not between date_at and date_add(date_at, INTERVAL period day)
and ($start_date > date_add(date_at, INTERVAL period day) OR $end_date < date_at)
))
答案 2 :(得分:0)
您不需要进行2次IN
次查询。
由于在您的第一个IN
查询中,您只有NOT
个子句,因此可以将这些子句更改为NOT IN
并删除子查询中的NOT
。
这也将删除第二个OR IN
。
select *
from rooms
where id not in (
select room_id
from reservations
where '2018-6-11' BETWEEN date_at and date_add(date_at, INTERVAL period day)
or '2018-6-30' between date_at and date_add(date_at, INTERVAL period day)
or date_at between '2018-6-11' and '2018-6-30'
or date_add(date_at ,INTERVAL period day ) between '2018-6-11' and '2018-6-30' ))
此查询基本上转换为: 给我所有没有预订的房间,我的开始日期或结束日期在预订时间之间。
修改:
感谢@ vivek_23我意识到查询不能简化内部条件 - 尽管如此,这个版本将对数据库的查询量减少到2个。