Question

我有两张桌子：

EMPLOYEES
=====================================================
ID    NAME    SUPERVISOR    LOCATION    SALARY
-----------------------------------------------------
34       John                  AL          100000
17       Mike    34            NY          75000
5        Alan    34            LE          25000
10       Dave    5             NY          20000

BONUS
========================================
ID        Bonus
----------------------------------------
17        5000
34        5000
10        2000

我必须编写查询，其中返回每个位置中收入最高的员工的名单，包括他们的姓名，工资和工资+奖金。排名应基于工资加奖金。所以我写了这个查询：

select em.name as name, em.salary as salary, bo.bonus as bonus, max(em.salary+bo.bonus) as total
from employees as em
join bonus as bo on em.empid = bo.empid 
group by em.location

但是我得到错误的名字和查询不会返回没有奖金的员工（员工表中的empid = 5），这些员工的薪水最高（25000 + 0奖金）。

Answer 1

也许尝试使用left join：

select em.name as name, em.salary as salary, bo.bonus as bonus, max(em.salary+bo.bonus) as total
from employees as em
left join bonus as bo on em.empid = bo.empid 
group by em.location

Answer 2

你可以做

select 
  em.location, 
  em.name as name, 
  em.salary as salary, 
  IFNULL(bo.bonus,0)) as bonus, 
  max(em.salary+IFNULL(bo.bonus,0)) as total
from employees as em
left join bonus as bo on em.empid = bo.empid 
group by em.location;

但是，此查询依赖于特定于MySQL的行为，并且在大多数其他数据库中也会失败（如果启用了设置ONLY_FULL_GROUP_BY，也会在MySQL的更高版本中失败）。

我建议使用下面的查询：

select 
  em.location, 
  em.name as name, 
  em.salary as salary, 
  IFNULL(bo.bonus,0)) as bonus, 
  highest.total
from employees as em 
left join bonus as bo on em.empid = bo.empid 
join (
    select 
      em.location, 
      max(em.salary+IFNULL(bo.bonus,0)) as total
    from employees as em 
    left join bonus as bo on em.empid = bo.empid 
    group by em.location
) highest on em.LOCATION = highest.LOCATION and em.salary+IFNULL(bo.bonus,0) = highest.total;

在此确定每个位置的最高工资+奖金，并将该结果用作联接中的派生表，以过滤掉每个位置总数最高的员工。

请参阅此SQL Fiddle

Answer 3

应该是：

select 
   em.name as name, 
   em.salary as salary, 
   COALESCE(bo.bonus,0) as bonus, 
   max(em.salary + COALESCE(bo.bonus,0) ) as total
from employees as em
left join bonus as bo on em.empid = bo.empid 
group by em.location

您可以在SQLFiddle

中查看

Answer 4

请试试这个：

 select em.name as name, em.salary as salary,ISNULL(bo.bonus,0) as bonus,
     max(em.salary+ISNULL(bo.bonus,0)) as total
    from employees as em
    left join bonus as bo on em.ID = bo.ID 
    group by  em.name,em.salary, bo.bonus order by MAX(em.salary+ISNULL(bo.bonus,0)) Desc

从一列到第二列中选择最大值

4 个答案: