MySQL Join - Allowing users to attend events - mysql

Question

I am wanting to create a simple system so that a user can attend an event. This will be done in three tables; user, event and attendance. I have created the user and event table.
User table
+---------+----------+
| User_id | Username |
+---------+----------+
| 1 | User1 |
+---------+----------+
| 2 | User2 |
+---------+----------+
_
Event table
+----------+------------+
| Event_id | Event_name |
+----------+------------+
| 1 | Event1 |
+----------+------------+
| 2 | Event2 |
+----------+------------+
Now I am wanting to create a table that shows the following on my webpage.
+-------+--------------+----------+---------+------------+---------------+
| At_id | At_attending | Event_id | User_id | Event_name | User_username |
+-------+--------------+----------+---------+------------+---------------+
| 1 | N | 1 | 1 | Event1 | User1 |
+-------+--------------+----------+---------+------------+---------------+
| 2 | N | 1 | 2 | Event1 | User2 |
+-------+--------------+----------+---------+------------+---------------+
| 3 | N | 2 | 1 | Event2 | User1 |
+-------+--------------+----------+---------+------------+---------------+
| 4 | N | 2 | 2 | Event2 | User2 |
+-------+--------------+----------+---------+------------+---------------+
Each user should have their own "At_attending" for each event where it will either be 'Y' or 'N'. By default it should automatically be 'N'. I'm unsure how to create the Attending table with suitable joins to receive the output I desire on the last table.
Many thanks.

Answer 1

You don't have to create a large table like that one you have put in.
At_id | At_attending | Event_id | User_id |
+------+--------------+----------+---------+
That's enough. The Event_id is then the foreign key to the event table, and User_id is a foreign key to the user table. This will give you an ability to link a row to a specific row on other tables. More on foreign keys.
CREATE TABLE Attenting
(
At_id int NOT NULL,
At_attending BIT(1) DEFAULT 0,
Event_id int,
User_id int,
PRIMARY KEY (At_id),
FOREIGN KEY (Event_id) REFERENCES Event(Event_id)
FOREIGN KEY (User_id) REFERENCES User(User_id)
)
At_attending is of type BIT which means that 0 can be interpret as FALSE and 1 as TRUE. Of course you can use BOOL or BOOLEAN but at MySQL, these types are synonyms for TINYINT(1). A value of zero is considered false. Non-zero values are considered true.
Or you can make yourself easier with CHAR(1) and supporting only T and F (or Y and N). But you have to define a constraint to ensure that the column only accepts both characters.
If you want to query the table as your example, just use JOIN - syntax.
SELECT
`User`.username AS User_username,
`Event`.Event_name,
`Attenting`.*
FROM
`Attenting`
INNER JOIN
`User`
ON
`Attenting`.User_id = `User`.User_id
INNER JOIN
`Event`
ON
`Attenting`.Event_id = `Event`.Event_id
The above will select all data from the three tables and will be shown as your example. The JOIN syntax ensures that other table can be connected to the current table.
Attenting.* will select all columns in the Attenting table.
If you want to select by specific parameters, just add a WHERE on the end.

Answer 2

You're going to need a so-called JOIN table relating users to events. Let's call it user_event. It will have these two columns.
user_id
event_id
That's all it needs. It doesn't need its own ID number or anything else. The two columns together are the primary key for the table. The presence of a row in user_event means the particular user is signed up to attend the particular event.
To generate the kind of display you showed, do this:
SELECT u.user_id,
u.username,
IF(ue.user_id IS NOT NULL, 'Y', 'N') AS attending,
e.event_id,
e.event_name
FROM event AS e
CROSS JOIN user AS u
LEFT JOIN user_event AS ue ON ue.event_id = e.event_id
AND ue.user_id = u.user_id
ORDER BY u.user_id, e.event_id
This will get you a row for every combination of user and event, and a Y or N stating whether the particular user is attending. Here's a SQLfiddle showing it.
http://sqlfiddle.com/#!2/2ac9c/1/0