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的所有Leave Type值
在Employ_id = 4的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中找不到任何条目,则它应返回0.
让我举一个例子：我想得到employee_id = 4的输出
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？
解决方法:
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,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
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