MySQL 综合案例2

综合案例2

找出每个岗位的平均薪资，要求显示平均薪资大于1500的，除MANAGER岗位之外，要求按照平均薪资降序排。

mysql> select job, avg(sal)		# 可以给avg(sal)起个别名
    -> from emp
    -> where job <> 'MANAGER'
    -> group by job
    -> having avg(sal) > 1500
    -> order by avg(sal) desc;	# 然后这里直接用别名
+-----------+-------------+
| job       | avg(sal)    |
+-----------+-------------+
| PRESIDENT | 5000.000000 |
| ANALYST   | 3000.000000 |
+-----------+-------------+
2 rows in set (0.00 sec)