如何验证房间是否已满？

时间：2014-07-07 09:02:31

标签： php mysql sql database

我目前正在开发预订系统。我现在遇到的问题是找出公寓是否已经预订完毕。在我的数据库中，我有一张桌子，上面有所有公寓及其详细信息。我试图获得所有公寓的日期，例如预订了4间卧室。我正在运行以下sql以返回所有4卧室公寓的预订日期。

SELECT * 
FROM  `apartment_booking` AS ab
JOIN apartment AS a ON ( a.id = apartmentId ) 
JOIN booking AS b ON ( b.id = bookingId ) 
WHERE bedrooms =  '4'
ORDER BY checkIn

返回sql是

id  CheckIn     checkOut     userId
74  2014-04-15  2014-04-22    1
75  2014-04-15  2014-04-22    1
102 2014-06-03  2014-07-07    1
71  2014-06-16  2014-06-23    1
114 2014-07-19  2014-08-02    1
121 2014-07-20  2014-08-02    1
57  2014-07-22  2014-08-05    1
122 2014-07-28  2014-08-02    1
117 2014-08-03  2014-08-10    1

由于我在系统中有4间公寓，有4间卧室，我想知道所有四间卧室都预订的日期。

输出到日期2014-07-28至2014-08-02的示例已完全预订，因为该日期范围共有四个预订。

数据库：

CREATE TABLE `apartment` (
  `id` int(11) NOT NULL auto_increment,
  `code` varchar(4) NOT NULL,
  `bedrooms` int(11) NOT NULL,
  `description` varchar(500) default NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=26 ;

--
-- Dumping data for table `apartment`
--

INSERT INTO `apartment` (`id`, `code`, `bedrooms`, `description`) VALUES
(1, '1c', 3, ''),
(4, '4d', 4, NULL),
(5, '5b', 2, NULL),
(10, '10c', 3, NULL),
(11, '11b', 2, NULL),
(12, '12d', 4, NULL),
(13, '13c', 3, NULL),
(14, '14a', 1, 'Yo'),
(15, '15b', 2, NULL),
(16, '16b', 2, NULL),
(17, '17d', 4, NULL),
(22, '22d', 4, NULL),


CREATE TABLE `apartment_booking` (
  `id` int(11) NOT NULL auto_increment,
  `apartmentId` int(11) NOT NULL,
  `bookingId` int(11) NOT NULL,
  `ref` varchar(50) NOT NULL,
  `pax` int(11) NOT NULL default '1',
  `remarks` varchar(500) default NULL,
  `guestFullName` varchar(30) default NULL,
  `guestCountry` varchar(2) default NULL,
  `guestFlightDetails` varchar(200) default NULL,
  PRIMARY KEY  (`id`),
  KEY `apartmentId` (`apartmentId`),
  KEY `bookingId` (`bookingId`),
  KEY `ref` (`ref`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=179 ;

--
-- Dumping data for table `apartment_booking`
--

INSERT INTO `apartment_booking` (`id`, `apartmentId`, `bookingId`, `ref`, `pax`, `remarks`, `guestFullName`, `guestCountry`, `guestFlightDetails`) VALUES
(164, 1, 140, 'Hotelbeds', 5, '', 'Andrew Robertson', 'MT', '')
(165, 21, 141, 'Hotelbeds', 6, '', 'Pipitone', 'MT', ''),
(166, 5, 142, 'maltaholidaylets', 2, '', 'holly turpin', 'MT', ''),
(167, 12, 143, 'direct003', 4, '', 'Bernard Walch', 'MT', ''),
(168, 17, 144, 'meetingpoint', 4, '', 'Edvin Modigh', 'MT', ''),
(169, 23, 145, 'direct', 3, '', 'Andrea bacchetti', 'MT', ''),
(172, 25, 148, 'direct', 5, '', 'Wimold Peters', 'MT', ''),
(173, 20, 149, '7228110687', 4, '', 'Ms. Benedetta Tombari', 'MT', ''),
(174, 23, 149, '7228110687 meetingpoint', 2, '', 'Ms. Milena Moretti', 'MT', ''),
(175, 25, 150, 'meetingpoint', 6, '', 'N Burdett', 'MT', ''),
(176, 8, 151, 'Hotelbeds', 2, '', 'tito titti', 'MT', ''),
(177, 1, 152, 'meetingpoint', 3, '', 'Stephen Mckenna', 'MT', ''),
(178, 16, 153, 'mhcs', 4, '', 'Wojclech Blaszak', 'MT', '');

-- --------------------------------------------------------

--
-- Table structure for table `booking`
--

CREATE TABLE `booking` (
  `id` int(11) NOT NULL auto_increment,
  `reference` varchar(20) NOT NULL,
  `dateTime` datetime NOT NULL,
  `checkIn` date NOT NULL,
  `checkOut` date NOT NULL,
  `userId` int(11) default NULL,
  PRIMARY KEY  (`id`),
  KEY `agent` (`userId`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=154 ;

--
-- Dumping data for table `booking`
--

INSERT INTO `booking` (`id`, `reference`, `dateTime`, `checkIn`, `checkOut`, `userId`) VALUES
(136, 'euroresort booking.b', '2014-07-02 09:30:08', '2014-08-04', '2014-08-11', 1),
(137, '7014505534', '2014-07-02 09:32:05', '2014-07-19', '2014-07-24', 1),
(138, 'BR4277518', '2014-07-02 09:45:02', '2014-08-09', '2014-08-16', 1),
(139, '100206154', '2014-07-02 10:11:45', '2014-07-27', '2014-08-03', 1),
(140, '120-135249-95', '2014-07-02 10:13:14', '2014-07-02', '2014-07-03', 1),
(141, '120-135181-94', '2014-07-02 10:14:31', '2014-08-10', '2014-08-17', 1),
(142, '000548MHL', '2014-07-02 12:38:54', '2014-08-25', '2014-09-01', 1),
(143, 'direct003', '2014-07-02 15:48:04', '2014-08-11', '2014-08-22', 1),
(144, 'SH3049361', '2014-07-02 15:52:18', '2014-08-05', '2014-08-14', 1),
(145, 'direct009', '2014-07-03 08:27:56', '2014-07-19', '2014-07-26', 1),
(148, 'direct010', '2014-07-04 08:12:13', '2014-07-08', '2014-07-22', 1),
(149, '7228110687', '2014-07-04 13:28:16', '2014-08-10', '2014-08-16', 1),
(150, '7308310623', '2014-07-07 08:39:04', '2014-08-11', '2014-08-20', 1),
(151, '120-135677-92', '2014-07-07 08:43:06', '2014-08-22', '2014-08-29', 1),
(152, '100209964', '2014-07-07 10:59:16', '2014-08-05', '2014-08-12', 1),
(153, 'mhcs', '2014-07-07 13:07:22', '2014-08-08', '2014-08-16', 1);

2 个答案:

答案 0 :(得分：1)

有点复杂。

以下查询生成0到999之间的数字范围，并将每个数字添加为每个预订的checkIn日期的天数，其中结果日期小于或等于公寓预订的checkOut日期有4个房间。这应该每天预订每间公寓一排。

然后计算每个日期的预订ID数量，并与4间卧室的公寓数量（来自子查询）进行比较。然后HAVING子句丢弃所有预订的公寓数量与4个房间的公寓数量不同的日期。

SELECT aBookedDate, sub2.apartment_cnt, COUNT(id) AS all_booking_cnt
FROM
(
    SELECT booking.id, DATE_ADD(booking.checkIn, INTERVAL iCnt DAY) AS aBookedDate
    FROM
    (
        SELECT units.i + tens.i * 10 + hundreds.i * 100 AS iCnt
        FROM (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9)units
        CROSS JOIN (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9)tens
        CROSS JOIN (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9)hundreds
    ) sub0
    CROSS JOIN booking
    INNER JOIN apartment_booking ON booking.id = apartment_booking.bookingId
    INNER JOIN apartment ON apartment.id = apartment_booking.apartmentId
    WHERE DATE_ADD(booking.checkIn, INTERVAL iCnt DAY) <= booking.checkOut
    AND apartment.bedrooms = 4
) sub1
CROSS JOIN 
(
    SELECT COUNT(*) AS apartment_cnt
    FROM apartment 
    WHERE bedrooms = 4
) sub2
GROUP BY aBookedDate
HAVING all_booking_cnt = sub2.apartment_cnt

SQL小提琴： -

http://www.sqlfiddle.com/#!2/6edbe/5

答案 1 :(得分：0)

您需要左外连接，因此您还会显示未预订的公寓。

SELECT * 
FROM  `apartment_booking` AS ab
JOIN apartment AS a ON ( a.id = apartmentId ) 
JOIN booking AS b ON ( b.id = bookingId ) 
WHERE bedrooms =  '4' and userId is null
ORDER BY checkIn

具有空用户ID的用户将为空（即未预订），因为该公寓没有连接预订。你没有对结构说得太多，所以我认为你删除预订而不是保留历史。如果您保留所有历史记录，则需要使用今天的日期查看日期。

SELECT * 
FROM  `apartment_booking` AS ab
JOIN apartment AS a ON ( a.id = apartmentId ) 
LEFT OUTER JOIN booking AS b ON ( b.id = bookingId ) 
WHERE bedrooms =  '4' and checkOut > NOW()
ORDER BY checkIn

编辑：

看起来应该是这样的，我稍后会尝试准备一个小提琴：

SELECT 
    (COUNT( 
        SELECT * 
        FROM  `apartment_booking` AS ab
        JOIN apartment AS a ON ( a.id = apartmentId ) 
        LEFT OUTER JOIN booking AS b ON ( b.id = bookingId ) 
        WHERE bedrooms =  '4' and checkIn <= <<<SOMEDATEHERE>>> and checkOut >= <<<<SOMEOTHERDATEHERE>>>>>
     ) >= 4);
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