Oracle向所有部门的工资高于平均工资大的员工展示

Question

我正在编写一个查询，以找到比其部门中平均工资高的员工。我需要显示该部门的员工ID，薪水，部门ID和平均工资。

我有一个几乎可以工作的查询，但它不断给我“ ORA-00904：“ avg_sal”：无效的标识符“错误”错误。我做得正确吗？为什么我会遇到这个无效的标识符错误？

SELECT employee_id, salary, department_id,
  (SELECT ROUND(AVG(salary),2)
  FROM employees e_inner
  WHERE e_inner.department_id = e.department_id) AS avg_sal
FROM employees e
WHERE salary > avg_sal
ORDER BY avg_sal DESC

Answer 1

我不相信您可以在某个子句中参考列别名（在这种情况下为AVG_SAL）。

您需要重复该内部查询，即：

SELECT employee_id, salary, department_id,
  (SELECT ROUND(AVG(salary),2)
  FROM employees e_inner
  WHERE e_inner.department_id = e.department_id) AS avg_sal
FROM employees e
WHERE salary > 
 (SELECT ROUND(AVG(salary),2)
  FROM employees e_inner
  WHERE e_inner.department_id = e.department_id)
ORDER BY avg_sal DESC

并不是很棒，有两个内部查询，但这是纠正错误的最突出的方法。

更新： 尚未对此进行测试，但请尝试以下操作：

SELECT e.employee_id, e.salary, e.department_id, b.avg_sal
FROM employees e
INNER JOIN
(SELECT department_id, ROUND(AVG(salary),2) AS avg_sal
 FROM employees
 GROUP BY department_id) e_avg ON e.department_id = e_avg.department_id AND e.salary > e_avg.avg_sal
ORDER BY e_avg.avg_sal DESC

Answer 2

使用分析更有效：

select employee_id, salary, department_id, avg_sal
from
(
  SELECT employee_id, salary, department_id, 
    round(avg(salary) over (partition by department_id), 2) avg_sal
  from emp
)
where salary > avg_sal
order by avg_sal desc

Answer 3

您可以将其重写为加入：

SELECT  e1.employee_id
,       e1.salary
,       e1.department_id
,       ROUND(AVG(e2.salary),2) as Avg_Sal
FROM    employees e
JOIN    employees e2
ON      e2.department_id = e.department_id
GROUP BY
        e1.employee_id
,       e1.salary
,       e1.department_id
HAVING  e1.salary > ROUND(AVG(e2.salary),2)

或子查询：

SELECT  *  
FROM    (
        SELECT  employee_id
        ,       salary
        ,       department_id
        ,       (
                SELECT  ROUND(AVG(salary),2)
                FROM    employees e_inner
                WHERE   e_inner.department_id = e.department_id
                ) AS avg_sal
        FROM    employees e
        ) as SubqueryAlias
WHERE   salary > avg_sal

Answer 4

select *
from employees e
join(
      select Round(avg(salary)) AvgSal,department_id,department_name as dept_name
      from employees join departments 
      using (department_id)
      group by department_id,department_name
) dd
using(department_id)
where e.salary > dd.AvgSal;

另一个解决方案

select * 
from employees e, 
(
 select 
    department_id, 
    avg(salary) avg_sal 
 from employees 
 group by department_id
) e1
where e.department_id=e1.department_id 
and e.salary > e1.avg_sal