Assume I have the following table
id name city salary dept
and I want to select all salaries which are greater than the average salary
尝试这样的事情:
SELECT salary WHERE salary > (SELECT AVG(salary) FROM *)
Assuming it's mysql, only the below two work. (I used a temp table so the names are different from yours)
select * from b where ref > (select avg(ref) from b);
select * from b having ref > (select avg(ref) from b);
This doesn't - select * from b having ref > avg(ref);
Some queries I tried -
mysql> select * from b;
+------+------------+------+
| id | d2 | ref |
+------+------------+------+
| 300 | 2010-12-12 | 3 |
| 300 | 2011-12-12 | 2 |
| 300 | 2012-12-12 | 1 |
| 400 | 2011-12-12 | 1 |
+------+------------+------+
4 rows in set (0.00 sec)
mysql> select * from b having ref > avg(ref);
+------+------------+------+
| id | d2 | ref |
+------+------------+------+
| 300 | 2010-12-12 | 3 |
+------+------------+------+
1 row in set (0.00 sec)
mysql> select * from b having ref > (select avg(ref) from b);
+------+------------+------+
| id | d2 | ref |
+------+------------+------+
| 300 | 2010-12-12 | 3 |
| 300 | 2011-12-12 | 2 |
+------+------------+------+
2 rows in set (0.02 sec)
mysql> select * from b where ref > (select avg(ref) from b);
+------+------------+------+
| id | d2 | ref |
+------+------------+------+
| 300 | 2010-12-12 | 3 |
| 300 | 2011-12-12 | 2 |
+------+------------+------+
2 rows in set (0.00 sec)
mysql> select *,avg(ref) from b having ref > avg(ref);
+------+------------+------+----------+
| id | d2 | ref | avg(ref) |
+------+------------+------+----------+
| 300 | 2010-12-12 | 3 | 1.7500 |
+------+------------+------+----------+
1 row in set (0.00 sec)
If windowed aggregate functions are supported:
SELECT Salary
FROM (
SELECT
Salary,
AVG(Salary) OVER () AS AvgSalary
FROM atable
) s
WHERE Salary > AvgSalary
select empno,e.deptno,sal
from emp e, ( select deptno,avg(sal) avsal
from emp
group by deptno
) a
where e.sal > a.avsal
and e.deptno = a.deptno;
its really easy just use following short command given below
SELECT *FROM table_name WHERE salary > avg(select salary from table_name)
HOPE YOU GET IT :-)
如果表名是Employee(id,name,city,salary)
select salary from Employee where salary > (select ava(salary) from employee)
Assuming emp is the name of the table, which has department id as dept_id
(Group by department)
select e1.* from emp e1 inner join (select avg(sal) avg_sal,dept_id from emp group by
dept_id) as e2 on e1.dept_id=e2.dept_id and e1.sal>e2.avg_sal
Query results shows all employees details whose salary is greater than average salary.
select * from emp where sal > (select avg(sal) from emp)
选择 e.employee_id, e.department_id, e.salary from 雇员 e wheresalary > ( select avg(salary) from雇员 d where e.department_id =d.department_id)
以下将为您工作。
SELECT salary FROM table_name WHERE salary > (SELECT AVG(salary) FROM table_name);
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