MySQL 第十二章 练习题

题目及答案

1.取得每个部门最高薪水的人员名称

select depno,max(sal) as maxsal from emp group by deptno;

select e.ename,t.*
from emp e
join (select depno,max(sal) as maxsal from emp group by deptno) t
on t.deptno = e.deptno and t.maxsal = e.sal;

2.哪些人的薪水在部门的平均薪水之上

select deptno,avg(sal) as avgsal from emp group by deptno;

select e.ename,t.*
from emp e
join (select depno,avg(sal) as avgsal from emp group by deptno) t
on t.deptno = e.deptno and t.avgsal < e.sal;

3.取出部门中所有人的平均的薪水等级

select e.ename,e.asl,e.deptno,s.grade
from emp e
join salgrade s
on e.sal between s.losal and s.hisal;

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;

4.取得最高薪水

select enmae,sal from emp order by sal desc limit 1;
select max(sal) from emp;
select sal from emp where sal not in(select distinct a.sal from emp a join emp b on a.sal<b.sal)

5.取得平均薪水最高的部门的部门编号

select deptno,avg(sal) as avgsal from emp group by deptno;
select deptno,avg(sal) as avgsal from emp group by deptno order by avgsal desc limit 1;

select max(t.avgsal) from (select avg(sal) as avgsal from emp group by deptno) t;
select deptno,avg(sal) as avgsal from emp group by deptno having avgsal = (
select deptno,avg(sal) as avgsal from emp group by deptno order by avgsal desc limit 1
);

6.取得平均薪水最高的部门的部门名称

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;

7.求平均薪水的等级最低的部门的部门名称

select deptno,avg(sal) as avgsal from emp group by deptno;

select t.*,s.grade
from (select d.dname,avg(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;

select grade 
from salgrade 
where (select avg(sal) as avgsal from emp group by deptno order by avgsal asc limit 1) between losal and hisal;

select  t.*,s.grade
from (select d.dname,avg(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
where s.grade = (select grade from salgrade where (select avg(sal) as avgsal from emp group by deptno order by avgsal asc limit 1) between losal and hisal);

8.取得比普通员工(员工代码没有在 mgr 字段上出现的) 的最高薪水还要高的领导人姓名

select max(sal) from emp where empno not in(select distinct mgr from emp where mgr is not null);
select ename,sal 
from emp 
where sal > (select max(sal) from emp where empno not in(select distinct mgr from emp where mgr is not null));

9.取得薪水最高的前五名员工

select ename,sal from emp order by sal desc limit 5;

10.取得薪水最高的第六到第十名员工

select ename,sal from emp order by sal desc limit 5, 5;

11.取得最后入职的 5 名员工(日期也可以降序，升序)

select ename,hiredate from emp order by hiredate desc limit 5;

12.取得每个薪水等级有多少员工

select e.ename,e.sal,s.grade 
from emp e 
join salgrade s 
on e.sal between s.losal and s.hisal;

select s.grade ,count(*)
from emp e 
join salgrade s 
on e.sal between s.losal and s.hisal
group by s.grade;

13.面试题：有 3 个表 S(学生表)，C（课程表），SC（学生选课表）
S（SNO，SNAME）代表（学号，姓名）
C（CNO，CNAME，CTEACHER）代表（课号，课名，教师）
SC（SNO，CNO，SCGRADE）代表（学号，课号，成绩）
问题：
1，找出没选过“黎明”老师的所有学生姓名。
2，列出 2 门以上（含2 门）不及格学生姓名及平均成绩。
3，即学过 1 号课程又学过 2 号课所有学生的姓名。

14.列出所有员工及领导的姓名

select a.ename '员工', b.ename '领导'
from emp a
left join emp b
on a.mgr = b.empno;

15.列出受雇日期早于其直接上级的所有员工的编号,姓名,部门名称

select a.ename '员工', a.hiredate, b.ename '领导', b.hiredate, d.dname
from emp a
join emp b
on a.mgr = b.empno
join dept d
on a.deptno = d.deptno
where a.hiredate < b.hiredate;

16. 列出部门名称和这些部门的员工信息, 同时列出那些没有员工的部门

select e.*,d.dname
from emp e
right join dept d
on e.deptno = d.deptno;

17.列出至少有 5 个员工的所有部门

select deptno
from emp
group by deptno
having count(*) >= 5;

18.列出薪金比"SMITH" 多的所有员工信息

select ename,sal from emp where sal > (select sal from emp where ename = 'SMITH');

19.列出所有"CLERK"( 办事员) 的姓名及其部门名称, 部门的人数

select ename,job from emp where job = 'CLERK';

select e.ename,e.job,d.dname
from emp e
join dept d
on e.deptno = d.deptno
where e.job = 'CLERK';

select deptno, count(*) as deptcount from emp group by deptno;

select t1.*,t2.deptcount
from (select e.ename,e.job,d.dname,d.deptno
	from emp e
	join dept d
	on e.deptno = d.deptno
	where e.job = 'CLERK') t1
join (select deptno, count(*) as deptcount from emp group by deptno) t2
on t1.deptno = t2.deptno;

20.列出最低薪金大于 1500 的各种工作及从事此工作的全部雇员人数
按照工作岗位分组求最小值。

select job,count(*) from emp group by job having min(sal) > 1500;

21.列出在部门"SALES"< 销售部> 工作的员工的姓名, 假定不知道销售部的部门编号.

elect ename from emp where deptno = (select deptno from dept where dname = 'SALES');

22.列出薪金高于公司平均薪金的所有员工, 所在部门, 上级领导, 雇员的工资等级.

select e.ename '员工',d.dname,l.ename '领导',s.grade
from emp e
join dept d
on e.deptno = d.deptno
left join emp l
on e.mgr = l.empno
join salgrade s
on e.sal between s.losal and s.hisal
where e.sal > (select avg(sal) from emp);

23.列出与"SCOTT" 从事相同工作的所有员工及部门名称

select job from emp where ename = 'SCOTT';

select e.ename,e.job,d.dname
from emp e
join dept d
on e.deptno = d.deptno
where e.job = (select job from emp where ename = 'SCOTT')
and e.ename <> 'SCOTT';

24.列出薪金等于部门 30 中员工的薪金的其他员工的姓名和薪金.

select distinct sal from emp where deptno = 30;

select ename,sal 
from emp 
where sal in(select distinct sal from emp where deptno = 30) 
and deptno <> 30;

25.列出薪金高于在部门 30 工作的所有员工的薪金的员工姓名和薪金. 部门名称

select max(sal) from emp where deptno = 3
select e.ename,e.sal,d.dname
from emp e
join dept d
on e.deptno = d.deptno
where e.sal > (select max(sal) from emp where deptno = 30);

26.列出在每个部门工作的员工数量, 平均工资和平均服务期限

select 
	d.deptno, count(e.ename) ecount,ifnull(avg(e.sal),0) as avgsal, ifnull(avg(timestampdiff(YEAR, hiredate, now())), 0) as avgservicetime
from emp e
right join dept d
on e.deptno = d.deptno
group by d.deptno;

在mysql当中怎么计算两个日期的“年差”，差了多少年？
TimeStampDiff(间隔类型, 前一个日期, 后一个日期)
间隔类型：SECOND秒，MINUTE分钟，HOUR小时，DAY天，WEEK星期, MONTH月，QUARTE季度,YEAR年
27.列出所有员工的姓名、部门名称和工资。

select e.ename,d.dname,e.sal
from emp e
join dept d
on e.deptno = d.deptno;

28.列出所有部门的详细信息和人数

select d.deptno,d.dname,d.loc,count(e.ename)
from emp e
right join dept d
on e.deptno = d.deptno
group by d.deptno,d.dname,d.loc;

29.列出各种工作的最低工资及从事此工作的雇员姓名

select job,min(sal) as minsal
from emp
group by job;

select e.ename,t.*
from emp e
join (select  job,min(sal) as minsal
	from emp
	group by job) t
on e.job = t.job and e.sal = t.minsal;

30.出各个部门的 MANAGER( 领导) 的最低薪金

select deptno, min(sal)
from emp
where job = 'MANAGER'
group by deptno;

31.列出所有员工的 年工资, 按 年薪从低到高排序

select ename,(sal + ifnull(comm,0)) * 12 as yearsal
from emp
order by yearsal asc;

32.求出员工领导的薪水超过3000的员工名称与领导

select a.ename '员工',b.ename '领导'
from emp a
join emp b
on a.mgr = b.empno
where b.sal > 3000;

33.求出部门名称中, 带’S’字符的部门员工的工资合计、部门人数

select d.deptno,d.dname,d.loc,count(e.ename),ifnull(sum(e.sal),0) as sumsal
from emp e
right join dept d
on e.deptno = d.deptno
where d.dname like '%S%'
group by d.deptno,d.dname,d.loc;

34.给任职日期超过 30 年的员工加薪 10%.

update emp set sal = sal * 1.1 where timestampdiff(YEAR, hiredate, now()) > 30;

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