MySQL покинул присоединиться три таблицы

Question

Предполагая, что у меня есть 3 таблицы:

leave_type:

id | name 
1 | maternity leave 
2 | Medical leave 
3 | Casual Leave 
4 | Sick Leave 
5 | honeymoon Leave 

сотрудник

id  |  name 
4  |  Employee 1 
7  |  Employee 2 

employee_leave_blance

id | employee_id | year | leave_type_id | val 
1 | 4    | 2015 | 1   | 9  
2 | 4    | 2015 | 2   | 5 
3 | 7    | 2015 | 1   | 10 
4 | 4    | 2015 | 3   | 4 

Здесь employee_leave_blance.leave_type_id = leave_type.id

Теперь я хочу, чтобы получить все Leave Type значения для всех сотрудников.

Позвольте мне привести пример: я хочу, чтобы получить выход для employee_id = 4, 7

employee_id | name    | val 
4   | maternity leave | 9 
4   | Medical leave  | 5 
4   | Casual Leave  | 4 
4   | Sick Leave  | 0 
4   | honeymoon Leave | 0 
7   | maternity leave | 10 
7   | Medical leave  | 0 
7   | Casual Leave  | 0 
7   | Sick Leave  | 0 
7   | honeymoon Leave | 0 

мой запрос

SELECT 
    lt.name, ifnull(el.val,0) as val , el.employee_id 
FROM leave_type AS lt 
    LEFT JOIN employee_leave_balance AS el ON el.leave_type_id = lt.id 
AND 
    el.year = YEAR(CURDATE()) GROUP BY el.employee_id 

Answer 1

Вы можете сделать это следующим образом:

SELECT employee_id,leave_name,IFNULL(val,0) as val 
FROM 

(SELECT e.id as employee_id,l.id as lid,l.name as leave_name 
FROM employee e CROSS JOIN leave_type l) as T1 LEFT JOIN 

(SELECT val,employee_id as eid,leave_type_id as lid 
FROM employee_leave_blance elb) as T2 ON T1.employee_id=T2.eid AND T1.lid=T2.lid 

ORDER BY employee_id asc,val desc 

Результат:

employee_id leave_name  val 
-------------------------------- 
4   maternity leave 9 
4   Medical leave 5 
4   Casual Leave  4 
4   Sick Leave  0 
4   honeymoon Leave 0 
7   maternity leave 10 
7   Casual Leave  0 
7   honeymoon Leave 0 
7   Sick Leave  0 
7   Medical leave 0 

Образец результата в SQL Fiddle.

Answer 2

я есть обновить мой ответ

Update:

select lt.name,e.id,elb.val from 
     leave_type as lt, employee as e,employee_leave_blance as elb 
     where e.id = elb.employee_id and lb.leave_type_id = lt.id 
     and elb.year= YEAR(CURDATE()) order by (e.id) 

Образец на SQL Fiddle

перспективе это query.i не пробовал, но я надеюсь, что он будет работать нормально

select * from leave_type as lt, employee as e, employee_leave_balance as elb 
where e.id = elb.employee_id and elb.leave_type_id = lt.id and e.id = $id 

Answer 3

Попробуйте этот код

SELECT * FROM leave_type 
INNER JOIN employee_leave_blance on leave_type.id = employee_leave_blance.leave_type_id 
INNER JOIN `employee on employee.id = employee_leave_blance.employee_id