How to verify if room is fully booked? - php
I am currently working on a booking system . I'm currently encountering a problem in finding out if an apartment is fully booked. In my database i have a table holding all the apartments and their details. I am trying to get the dates that all apartments for example with 4 bedrooms that are booked. I am running the following sql to return the booked dates of all 4 bedroom apartments.
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
The return of the sql is
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
As i have 4 apartments in the system with four bedrooms i would like to get the dates that all four bedrooms are booked.
Example with the output got the dates 2014-07-28 till 2014-08-02 are fully booked as in that date range there are in total four bookings.
Database:
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);
It gets a bit complicated.
The following query generates a range of numbers from 0 to 999, and adds each number as a number of days to the checkIn date for each booking, where the resulting date is less than or equal to the checkOut date for bookings for apartments with 4 rooms. This should give one row per apartment per day booked.
The number of booking ids for each date is then counted, and compared with the number of apartments with 4 bedrooms (from a sub query). The HAVING clause then discards all rows for dates where the number of aparments booked is not the same as the number of apartments with 4 rooms.
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 fiddle for it:-
http://www.sqlfiddle.com/#!2/6edbe/5
You need left outer join, so you also show the apartments that are not booked.
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
Ones that have null user id will be empty (aka non-booked), since there is no booking connected to that apartment. You didn't say enough about the structure so I take it you delete the bookings rather than keeping history. If you keep all historical entries you need to check the date with today's date instead.
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
EDIT:
It should look something like, I'll try to prepare a fiddle with that later:
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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