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employee表的全部数据：
SELECT EMPLOYEE_id,salary,avg(salary) FROM test1.employees group by employee_idhaving avg(salary)>8000;
使用这个查询语句想得到平均工资大于8000的平均工资，可是为什么结果不是平均工资呢？
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因为你group by id，每个id下不是只有一个工资么
select EMPLOYEE_id,salary,(select avg(salary) FROM employees) from employees where salary > (select avg(salary) FROM employees);
大概可以，你试试看吧
你是按照employee_id分组的,表记录里面每个employee_id就只有一条记录,你avg不就只有一个值吗.你如果想要全部员工的平均工资就不要groupb by了
理解有问题，平均工资，平均的什么呢，如果是针对employee_id的话，一个人的工资怎么能说平均工资呢。一般来说是某个部门下平均工资大于8000的员工工资吧
select t.salary,t.employee_id,t1.avg_salary from employees t inner join (select department_id,avg(salary) avg_salary from employees group by department_id) t1 on t.department_id = t1.department and t1.avg_salary > 8000
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因为你group by id，每个id下不是只有一个工资么
大概可以，你试试看吧
你是按照employee_id分组的,表记录里面每个employee_id就只有一条记录,你avg不就只有一个值吗.
你如果想要全部员工的平均工资就不要groupb by了
理解有问题，平均工资，平均的什么呢，如果是针对employee_id的话，一个人的工资怎么能说平均工资呢。一般来说是某个部门下平均工资大于8000的员工工资吧
select t.salary,t.employee_id,t1.avg_salary from employees t inner join (select department_id,avg(salary) avg_salary from employees group by department_id) t1 on t.department_id = t1.department and t1.avg_salary > 8000