php - mysql左连接两个表

Question

假设我有 2 个表:

休假类型

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

此处 employee_leave_blance.leave_type_id = leave_type.id

现在我想获取 employee_id=4 的所有 Leave Type 值

在 employee_leave_blance 表中，employee_id=4 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 中没有找到条目，它应该返回 0。

让我举个例子: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

现在如何获取员工所有 leave_type 的值？

Answer 1

不确定这是否是您需要的，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，您的查询可以进行一些修改。只需删除 and 的 where 子句，因为您已经加入表并使用 ifnull 函数为 null 值返回 0。这是您的查询，但有所改进:

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