第1题
- 取得每个部门最高薪水的人员名称
第一步：取得每个部门最高薪水
select max(sal) topsal, deptno from emp group by deptno;
第二步：将上面第一步的查询结果当做一张临时表t，进行表连接，条件是：t.deptno=e.deptno and t.maxsal=e.sal
select e.ename, t.* from emp e join (select max(sal) topsal, deptno from emp group by deptno) t on e.deptno=t.deptno and e.sal=t.topsal;
第2题
哪些人的薪水在部门的平均薪水之上
select t.*, e.ename, e.sal from emp e join (select avg(sal) avgsal, deptno from emp group by deptno) t on e.sal>t.avgsal and t.deptno=e.deptno;
第3题
取得每个部门平均薪水的等级
select t.*, s.grade from salgrade s join (select deptno, avg(sal) avgsal from emp group by deptno) t on t.avgsal between s.losal and s.hisal;
第4题
取得部门中（所有人的）平均的薪水等级
select avg(t.grade), t.deptno from (select e.ename, e.sal, e.deptno, s.grade from emp e join salgrade s on e.sal between s.losal and s.hisal) t group by t.deptno;
select e.deptno, avg(s.grade) from emp e join salgrade s on e.sal between s.losal and s.hisal group by e.deptno;
第5题
不准用组函数（Max），取得最高薪水（给出两种解决方案）
自己想的：
select max(sal) from emp;
select a.sal from emp a where not exists (select b.sal from emp b where b.sal > a.sal);
第一种方案：按照薪资降序排列，取第一个。
select ename, sal from emp order by sal desc limit 1;
第二种方案：采用表的自连接方式。
select ename, sal from emp where sal not in (select distinct a.sal from emp a join emp b on a.sal<b.sal);
第6题
取得平均薪水最高的部门的部门编号（至少给出两种解决方案）
第一种方案：降序排列取第一个（其实不准确， 比如有多个部门平均工资都是最高，这种情况可以用 having）
select deptno, avg(sal) avgsal from emp group by deptno order by avgsal desc limit 1;
第二种方案：max函数。
第一步：获取不同部门平均工资
select deptno, avg(sal) avgsal from emp group by deptno;
第二步：获取最大工资
这个重命名 t 是必需的
select max(avgsal) maxsal from (select deptno, avg(sal) avgsal from emp group by deptno) t;
第三步：
select deptno, avg(sal) avgsal from emp group by deptno having avgsal=(select max(avgsal) maxsal from (select deptno, avg(sal) avgsal from emp group by deptno) t);
select deptno, avg(sal) avgsal from emp group by deptno having avgsal=(select avg(sal) avgsal from emp group by deptno order by avgsal desc limit 1);
第7题
取得平均薪水最高的部门的部门名称
select deptno, avg(sal) avgsal from emp group by deptno order by avgsal desc limit 1;
select deptno, avg(sal) avgsal from emp group by deptno having avgsal=(select avg(sal) avgsal from emp group by deptno order by avgsal desc limit 1);
select t.avgsal, d.dname from dept d join (select deptno, avg(sal) avgsal from emp group by deptno having avgsal=(select avg(sal) avgsal from emp group by deptno order by avgsal desc limit 1)) t on t.deptno=d.deptno;
select d.dname,avg(e.sal) as avgsal from emp e join dept d on e.deptno=d.deptno group by d.dname order by avgsal desc limit 1;
第8题
求平均薪水的等级最低的部门的部门名称
select t.*,s.grade from (select d.dname,avg(e.sal) as avgsal from emp e join dept d on e.deptno = d.deptno group by d.dname) t join salgrade s on t.avgsal between s.losal and s.hisal order by s.grade asc limit 1;
