如何获得薪水低于平均工资的所有员工？

Question

我可以得到员工的数量和平均工资，但是当我尝试获得额外的选择时，选择列出支付的员工数量低于平均水平，则失败。

select count(employee_id),avg(salary) 
from employees 
Where salary < avg(salary); 

Answer 1

select count(*), (select avg(salary) from employees) 
from employees 
where salary < (select avg(salary) from employees); 

Answer 2

select TotalNumberOfEmployees, 
      AverageSalary, 
      count(e.employee_id) NumberOfEmployeesBelowAverageSalary 
    from (
      select count(employee_id) TotalNumberOfEmployees, 
        avg(salary) AverageSalary 
       from employees 
     ) preagg 
left join employees e on e.salary < preagg.AverageSalary 
group by TotalNumberOfEmployees, 
      AverageSalary 

注：我使用了LEFT JOIN，所以如果你有3名相等的员工，它会显示0，而不是任何结果（以下低于平均一个人）。

Answer 3

问题是AVG是一个聚合函数。 SQL不够聪明，无法弄清楚如何在行内混合聚合结果。传统的方式是使用连接：

select count(*), avg(e.salary), 
     sum(case when e.salary < const.AvgSalary then 1 else 0 end) as NumBelowAverage 
from employees e cross join 
    (select avg(salary) as AvgSalary from employees) as const 

Answer 4

您不清楚在结果集中想要哪些列，这就很难回答您的问题。明确问题可以提高答案的质量。

你似乎想3个事实：

1. 员工人数。
2. 平均工资。
3. 收入低于平均工资的员工人数。

你展现一个查询其确实为前两个事实工作：

SELECT COUNT(*) AS NumberOfEmployees, 
     AVG(Salary) AS AverageSalary 
    FROM Employees 

什么COUNT(*)和COUNT(Employee_ID)之间的区别？不同之处在于后者只对Employee_ID列中存在非NULL值的行进行计数。一个好的优化器会认识到Employee_ID是一个主键并且不包含NULL值，查询也是一样的。但COUNT(*)更传统，对优化器的依赖性更低。

其他统计可以在选择列表中一个简单的值通过一个子查询生成：

SELECT COUNT(*) AS NumberOfEmployees, 
     AVG(Salary) AS AverageSalary, 
     (SELECT COUNT(*) 
      FROM Employees 
     WHERE Salary < (SELECT AVG(Salary) FROM Employees) 
     ) AS NumberOfEmployeesPaidSubAverageWages 
    FROM Employees 

在许多情况下，这是不恰当的写子查询那样，但是对于指定查询的解释，这很好。

Answer 5

select * from <table name> where salary < (select avg(<salary column name) from <table name>); 

例子：

select * from EMPLOYEE where sal < (select avg(emp_sal) from EMPLOYEE); 

Answer 6

SELECT e.ename,e.deptno,e.sal,d.avg 
FROM emp e,(SELECT deptno, avg(sal) avg 
      FROM emp 
      GROUP BY deptno) d 
WHERE e.deptno=d.deptno 
AND 
e.sal < d.avg