这部分主要是为了帮助大家回忆回忆MySQL的基本语法，数据库来自于MySQL的官方简化版，题目也是网上非常流行的35题。这些基础习题基本可以涵盖面试中需要现场写SQL的问题。上期帮助大家完成了热身运动，接下来让我们继续练习。
给出最高薪水（给出三种解决方案）
Solution1 ：
mysql> select ename,max(sal) from emp;
+--------+----------+
| ename | max(sal) |
+--------+----------+
| SIMITH | 5000.00 |
+--------+----------+
Solution2：
按照薪水降序排列，取第一个
mysql> select ename,sal from emp order by sal desc limit 1;
+-------+---------+
| ename | sal |
+-------+---------+
| KING | 5000.00 |
+-------+---------+
Solution3：
最大值不小于表中任何一个工资，所以不会出现在此临时表中
mysql> select ename,sal from emp where sal not in
(select a.sal from emp a join emp b on a.sal<b.sal);
+-------+---------+
| ename | sal |
+-------+---------+
| KING | 5000.00 |
+-------+---------+
取得平均薪水最高的部门和部门编号
解法一
1,先取得每个部门的平均薪水
mysql> select deptno, avg(sal) as avgsal from emp group by deptno;
+--------+-------------+
| deptno | avgsal |
+--------+-------------+
| 20 | 2175.000000 |
| 30 | 1566.666667 |
| 10 | 2916.666667 |
+--------+-------------+
2,再取得最高的平均薪水
mysql> select avg(sal) as avgsal from emp group by deptno order by avgsal desc limit 1;
+--------+-------------+
| deptno | avgsal |
+--------+-------------+
| 20 | 2175.000000 |
| 30 | 1566.666667 |
| 10 | 2916.666667 |
+--------+-------------+
3,然后取得和最高平均薪水相同的部门（因为可能有多个部门并列最高）
mysql> select deptno,avg(sal) as avgsal from emp group by deptno having avgsal=
-> (select avg(sal) as avgsal from emp group by deptno order by avgsal desc limit 1);
+--------+-------------+
| deptno | avgsal |
+--------+-------------+
| 10 | 2916.666667 |
+--------+-------------+
解法二 ： 使用max（）函数
mysql> select deptno,avg(sal) as avgsal
from emp group by deptno having avgsal=(select max(t.avgsal) from (select avg(sal) as avgsal from emp group by deptno) t);
+--------+-------------+
| deptno | avgsal |
+--------+-------------+
| 10 | 2916.666667 |
+--------+-------------+
取得平均薪水最高的部门名称
解法一
第一步获取每个部门的平均薪资，
(root@localhost) [employees]>select a.deptno, avg(a.sal) from emp a group by a.deptno;
+--------+-------------+
| deptno | avg(a.sal) |
+--------+-------------+
| 20 | 2175.000000 |
| 30 | 1566.666667 |
| 10 | 2916.666667 |
+--------+-------------+
3 rows in set (0.16 sec)
(root@localhost) [employees]>
再双表联查（emp和dept）,获得每个部门的平均薪资，部门编号和部门名称
(root@localhost) [employees]>select b.dname,a.deptno, avg(a.sal) from emp a join dept b on a.deptno = b.deptno group by a.deptno;
+-------------+--------+-------------+
| dname | deptno | avg(a.sal) |
+-------------+--------+-------------+
| RESEARCHING | 20 | 2175.000000 |
| SALES | 30 | 1566.666667 |
| ACCOUNTING | 10 | 2916.666667 |
+-------------+--------+-------------+
3 rows in set (0.03 sec)
(root@localhost) [employees]>
最后降序排列，只取第一个
mysql> select b.dname,avg(a.sal) as avgsal from emp a join dept b on a.deptno=b.deptno group by b.dname having avgsal=(select max(t.avgsal) from (select avg(sal) as avgsal from emp group by deptno) t);
+------------+-------------+
| dname | avgsal |
+------------+-------------+
| ACCOUNTING | 2916.666667 |
+------------+-------------+
解法二 ： 使用max（）
原理和上一题第二问大同小异。
mysql> select b.dname,avg(a.sal) as avgsal from emp a join dept b on a.deptno=b.deptno group by b.dname having avgsal=(select max(t.avgsal) from (select avg(sal) as avgsal from emp group by deptno) t);
+------------+-------------+
| dname | avgsal |
+------------+-------------+
| ACCOUNTING | 2916.666667 |
+------------+-------------+
往期文章
MySQL基础练习题：习题456
MySQL基础练习题：习题2-3
MySQL基础练习题：创建数据库
