如何確認房間是否已滿？

Question

我目前正在研究訂票系統。 我目前在查找公寓是否已滿時遇到問題。 在我的數據庫中，我有一個表格，其中包含所有公寓及其詳細信息。 我正在嘗試獲取所有帶有4間卧室的公寓的預定日期。 我正在運行以下sql以返回所有4間卧室公寓的預訂日期。

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

因為我在系統中有4間公寓，其中有4間卧室，所以我想獲得所有4間卧室都被預訂的日期。

輸出示例顯示日期為2014-07-28到2014-08-02的書已被預訂滿，因為在該日期范圍內，總共有四筆預訂。

數據庫：

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

它變得有點復雜。

以下查詢生成一個介於0到999之間的數字范圍，並將每個數字作為天數添加到每個預訂的入住日期中，其中得出的日期小於或等於四個房間的公寓的預訂的退房日期。 這應該使每間公寓每天預訂一排。

然后計算每個日期的預訂ID數量，並將其與4間卧室的公寓數量進行比較（來自子查詢）。 然后，HAVING子句會丟棄預訂的公寓數量與4個房間的公寓數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擺弄它：-

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

Answer 2

您需要左外部聯接，因此您還要顯示未預訂的公寓。

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

用戶ID為空的用戶將為空（也就是未預訂），因為沒有預訂與該公寓相關。 您沒有對結構進行足夠的說明，所以我認為您是刪除預訂而不是保留歷史記錄。 如果保留所有歷史記錄條目，則需要用今天的日期來檢查日期。

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

編輯：

它看起來應該像這樣，稍后我將嘗試准備一個小提琴：

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);