如何獲得薪水低於平均工資的所有員工？

Question

我可以得到員工的數量和平均工資，但是當我嘗試獲得額外的選擇時，選擇列出支付的員工數量低於平均水平，則失敗。

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名相等的員工，它會顯示0，而不是任何結果（以下低於平均一個人）。

Answer 3

問題是AVG是一個聚合函數。 SQL不夠聰明，無法弄清楚如何在行內混合聚合結果。傳統的方式是使用連接：

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

您不清楚在結果集中想要哪些列，這就很難回答您的問題。明確問題可以提高答案的質量。

你似乎想3個事實：

1. 員工人數。
2. 平均工資。
3. 收入低於平均工資的員工人數。

你展現一個查詢其確實爲前兩個事實工作：

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

什麼COUNT(*)和COUNT(Employee_ID)之間的區別？不同之處在於後者只對Employee_ID列中存在非NULL值的行進行計數。一個好的優化器會認識到Employee_ID是一個主鍵並且不包含NULL值，查詢也是一樣的。但COUNT(*)更傳統，對優化器的依賴性更低。

其他統計可以在選擇列表中一個簡單的值通過一個子查詢生成：

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

在許多情況下，這是不恰當的寫子查詢那樣，但是對於指定查詢的解釋，這很好。

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