mysql左连接两个表

Question

假设我有2个表：

leave_type

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

employee_leave_blance

id      employee_id      year     leave_type_id     val
1       4                2015      1                9
2       4                2015      2                5
3       8                2015      1                10
4       4                2015      3                4

这里employee_leave_blance.leave_type_id = leave_type.id

现在我想获得employee_id=4所有Leave Type值

在Employ_id employee_id=4 employee_leave_blance表中，只有3个leave_type条目，但在leave_type表中有5个。 我想为一名员工获得5个条目。 如果在employee_leave_blance找不到任何条目，则它应返回0。

让我举一个例子：我想得到employee_id = 4输出

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

为此，我尝试了以下查询：

select
    el.employee_id, lt.name, el.val  from leave_type as lt
    left join employee_leave_blance as el on el.leave_type_id = lt.id
where
    el.year = YEAR(CURDATE()) and el.employee_id = 4

然而它返回以下结果：

employee_id      name                val
4                maternity leave     9
4                Medical leave       5
4                Casual Leave        4

现在我如何获得员工的所有leave_type的值？

Answer 1

不确定这是否是您所需要的，请尝试：

select ifnull(el.employee_id,4) as employee_id,
lt.name,ifnull(el.val,0) as val
from leave_type as lt
left join (select year,employee_id,val,leave_type_id 
                    from employee_leave_balance) as el 
                    on el.leave_type_id = lt.id 
                    and el.year = year(curdate()) 
                    and el.employee_id = 4

实际上，您的查询将适用于一些修改。 只需删除where子句and因为你已经连接了表并使用ifnull函数为空值返回0。 这是您的查询，但有一些改进：

select
    ifnull(el.employee_id,4) as employee_id, lt.name, ifnull(el.val,0) as val  
from leave_type as lt
    left join employee_leave_balance as el on el.leave_type_id = lt.id
and
    el.year = YEAR(CURDATE()) and el.employee_id = 4