Question

我正在使用Oracle db 11g

我有一张桌子＆＃39; EMPLOYEES＆＃39;像这样...

ID  JOB_ID      SALARY
100 AD_PRES     24000
101 AD_VP       17000
102 AD_VP       17000
103 IT_PROG     9000
104 IT_PROG     6000
107 IT_PROG     4200
124 ST_MAN      5800
141 ST_CLERK    3500
142 ST_CLERK    3100
143 ST_CLERK    2600
144 ST_CLERK    2500
149 SA_MAN      10500
174 SA_REP      11000
176 SA_REP      8600
178 SA_REP      7000
200 AD_ASST     4400
201 MK_MAN      13000
202 MK_REP      6000
205 AC_MGR      12000
206 AC_ACCOUNT  8300

我希望得到每份工作（job_id）的最高平均工资及其工作。

我首先尝试了这个并导致错误

SELECT MAX(AVG(salary)) AS max_avg_salary, job_id
FROM employees
GROUP BY job_id;

ORA-00937: not a single-group group function
00937. 00000 -  "not a single-group group function"
*Cause:    
*Action:

我终于使用此代码

了

WITH emp AS (SELECT AVG(salary) AS avg_salary, job_id FROM employees GROUP BY job_id)
SELECT e1.avg_salary AS max_avg_salary, e1.job_id
FROM emp e1 JOIN (SELECT MAX(avg_salary) AS max_avg_salary FROM emp) e2
ON e1.avg_salary = e2.max_avg_salary;

MAX_AVG_SALARY  JOB_ID
24000           AD_PRES

我想知道的是......

为什么我的第一个代码会出错？
有没有比我的代码更好（更简单或更简单）的方式？

Answer 1

这个怎么样？

select job_id, salary
  from (  select job_id,
                 avg (salary) salary,
                 rank () over (order by avg (salary) desc) rnk
            from employees
        group by job_id)
 where rnk = 1;

Answer 2

[已编辑] ...除非您只想获得一个答案，否则其他受访者的解决方案应该有效，我会提供一个变体：

<强>非Oracle：

SELECT TOP 1 *
    FROM
        (SELECT AVG(salary) AS avg_salary, job_id
            FROM employees
            GROUP BY job_id
        ) a
    ORDER BY a.avg_salary DESC

Littlefoot非常正确地指出，由于TOP 1，这对Oracle 无效。你应该选择他的解决方案。对于任何非甲骨文的人，我都会留在这里：

然而，提问者本人，C Park，使用ROWNUM建议使用与Oracle兼容的变体。他已经对它进行了测试，它对他有用，所以这将有效。

<强>甲骨文：

SELECT *
    FROM
        (SELECT AVG(salary) AS avg_salary, job_id
            FROM employees
            GROUP BY job_id
        ) a
    ORDER BY a.avg_salary DESC
    WHERE ROWNUM = 1

我希望这会有所帮助。

Answer 3

您可以像这样修改您的查询，

SELECT MAX(avg_salary) 
  FROM (SELECT AVG(salary) AS avg_salary, job_id
          FROM employees
         GROUP BY job_id);

或者如果您还要显示作业ID，

SELECT avg_salary, job_id
  FROM (SELECT AVG(salary) AS avg_salary, job_id
          FROM employees
         GROUP BY job_id
         ORDER BY avg_salary DESC)
 WHERE ROWNUM = 1;

具有嵌套组功能的SQL

3 个答案: