从一列到第二列中选择最大值答案

【问题标题】：Select maximum value from one column by second column从一列到第二列中选择最大值
【发布时间】：2015-02-13 19:54:16
【问题描述】：

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

我必须编写查询返回每个位置的最高薪员工列表，包括他们的姓名、薪水和薪水+奖金。排名应该基于工资加奖金。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

但是我得到了错误的名字并且查询不返回没有奖金的员工（empid = 5 在员工表中），根据位置（25000 + 0 奖金）具有最高工资。

【问题讨论】：

标签： mysql sql

【解决方案1】：

你可以这样做

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 group by 行为，并且在大多数其他数据库中会失败（如果启用了设置 ONLY_FULL_GROUP_BY，在 MySQL 的更高版本中也会失败）。

我建议使用如下查询：

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;

您可以在此处确定每个地点的最高薪水+奖金，并将该结果用作联接中的派生表，以过滤出每个地点的总数最高的员工。

看到这个SQL Fiddle

【讨论】：

【解决方案2】：

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

【讨论】：

是的，这是有效的，但对于没有奖金的员工，我得到的总数（空）不仅仅是薪水

【解决方案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查看它

【讨论】：

【解决方案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

【讨论】：

ISNULL 接受一个参数。应该是 IFNULL 或 COALESCE 而不是