如何确认房间是否已满？

Question

我目前正在研究订票系统。 我目前在查找公寓是否已满时遇到问题。 在我的数据库中，我有一个表格，其中包含所有公寓及其详细信息。 我正在尝试获取所有带有4间卧室的公寓的预定日期。 我正在运行以下sql以返回所有4间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

sql的返回是

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

因为我在系统中有4间公寓，其中有4间卧室，所以我想获得所有4间卧室都被预订的日期。

输出示例显示日期为2014-07-28到2014-08-02的书已被预订满，因为在该日期范围内，总共有四笔预订。

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);

Answer 1

它变得有点复杂。

以下查询生成一个介于0到999之间的数字范围，并将每个数字作为天数添加到每个预订的入住日期中，其中得出的日期小于或等于四个房间的公寓的预订的退房日期。 这应该使每间公寓每天预订一排。

然后计算每个日期的预订ID数量，并将其与4间卧室的公寓数量进行比较（来自子查询）。 然后，HAVING子句会丢弃预订的公寓数量与4个房间的公寓数量不同的日期的所有行。

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摆弄它：-

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

Answer 2

您需要左外部联接，因此您还要显示未预订的公寓。

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

用户ID为空的用户将为空（也就是未预订），因为没有预订与该公寓相关。 您没有对结构进行足够的说明，所以我认为您是删除预订而不是保留历史记录。 如果保留所有历史记录条目，则需要用今天的日期来检查日期。

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：

它看起来应该像这样，稍后我将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);