How to implement IF condition in two relational tables? - mysql

Question

I have two relational tables, and I would like to filter data using IF condition. The problem is that using LEFT JOIN I got records that cannot be grouped.
The tables that I have are:
calendar
bookers
The first table consists of lessons that can be booked by more people, and the second table contains data who booked each lesson. The IF condition that I would like to implement is: return '2' if lesson is booked by specific user, return '1' if lesson is booked, but by another user, and return '0' if lesson is not booked.
What I would like to get according to above tables is given in the figure below.
Expected result
But, when I use LEFT JOIN to link those tables, I got record for every user that booked specific lesson.
SELECT calendar.id, calendarId, lessonType, description,
CASE
WHEN bookedBy then IF(bookedBy = 8, '2', '1')
ELSE '0'
END AS bb,
(select count(bookedBy) from bookers where calendar.id = bookers.lessonId) as nOfBookers
FROM calendar
LEFT JOIN bookers ON calendar.id = bookers.lessonId
WHERE `calendarId`= 180
Without the LEFT JOIN (fiddle), counts are shown properly, but I cannot include IF condition, because the table bookers is not defined.
I would appreciate any help. Thank you very much in advance.
Here is the Fiddle.
CREATE TABLE `calendar` (
`id` int(11) NOT NULL,
`calendarId` varchar(50) NOT NULL,
`lessonType` varchar(255) DEFAULT NULL,
`description` varchar(255) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
INSERT INTO `calendar`
(`id`, `calendarId`, `lessonType`, `description`)
VALUES
(1, '180', 'A', ''),
(2, '180', 'A', ''),
(3, '180', 'A', ''),
(4, '180', 'B', ''),
(5, '180', 'B', ''),
(6, '180', 'B', ''),
(7, '180', 'B', ''),
(8, '180', 'B', ''),
(9, '180', 'B', '');
CREATE TABLE `bookers` (
`id` int(11) NOT NULL,
`lessonId` int(11) DEFAULT NULL,
`bookedBy` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
--
-- Dumping data for table `bookers`
--
INSERT INTO `bookers` (`id`, `lessonId`, `bookedBy`) VALUES
(4, 1, 8),
(5, 2, 8),
(6, 2, 28),
(7, 2, 17),
(8, 3, 11);
--
-- Indexes for dumped tables
--
ALTER TABLE `calendar`
ADD PRIMARY KEY (`id`),
ADD UNIQUE KEY `id` (`id`);
--
-- Indexes for table `bookers`
--
ALTER TABLE `bookers`
ADD PRIMARY KEY (`id`);
--
-- AUTO_INCREMENT for dumped tables
--
--
-- AUTO_INCREMENT for table `bookers`
--
ALTER TABLE `bookers`
MODIFY `id` int(11) NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=9;
COMMIT;
select version();

Answer 1

Try this:
SELECT id, calendarid, lessontype, description,
CASE WHEN FIND_IN_SET(8,vbb)>0 THEN 2
WHEN vbb IS NOT NULL THEN 1
ELSE 0 END AS bb,
nOfBookers
FROM
(SELECT c.id, calendarId, lessonType, GROUP_CONCAT(bookedby) AS vbb, description,
(SELECT COUNT(bookedby) FROM bookers WHERE c.id = bookers.lessonId) AS nOfBookers
FROM calendar c
LEFT JOIN bookers b ON c.id = b.lessonId
WHERE `calendarId`= 180
GROUP BY c.id, calendarId, lessonType, description) A;
In addition to your original LEFT JOIN attempt, I've added GROUP_CONCAT(bookedby) AS vbb which will return a comma separated bookedby value; which is 17,28,8. After that, I make the query as a sub-query and do CASE expression with FIND_IN_SET function on vbb to look for specific bookedby.
Here's an update fiddle: https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=0933e9fc3cb7445311c34c6705d11637