数据:
1、取得每个部门最高薪水的人员名称
第一步:先找出每个部门的最高薪水(按部门编号分组),把它当做临时表:
select deptno,max(sal) as maxsal from emp group by deptno;
第二步:
select
e.ename,e.sal,d.dname
from
emp e
join
(select deptno,max(sal) as maxsal from emp group by deptno) t
on
t.maxsal=e.sal and t.deptno=e.deptno
join
dept d
on
d.deptno=e.deptno;
2、哪些人的薪水在部门的平均薪水之上
第一步:找出部门的平均薪水
select deptno,avg(sal) as avgsal from emp group by deptno;
第二步:
select
e.ename,e.sal
from
emp e
join
(select deptno,avg(sal) as avgsal from emp group by deptno) t
on
e.sal>t.avgsal and e.deptno=t.deptno;
3、取得部门中(所有人的)平均的薪水等级
第一步:找出员工的薪水等级
select e.ename,e.deptno,s.grade from emp e join salgrade s on e.sal between s.losal and s.hisal;
第二步:求平均
select
deptno,avg(grade) as avg_grade
from
emp e
join
salgrade s
on
e.sal between s.losal and s.hisal
group by
deptno;
4、不准用组函数(Max),取得最高薪水
第一种方法:
薪水降序后取第一个数据
select ename,sal from emp order by sal desc limit 1;
第二种方法:
第一步:表的自连接,找出表中的薪水,这个薪水要小于另一表的薪水,这时候薪水最大值就不在里面了。
select
distinct e1.sal
from
emp e1
join
emp e2
on
e1.sal < e2.sal;
第二步:
select ename,sal from emp where sal not in (select distinct e1.sal from emp e1 join emp e2 on e1.sal < e2.sal);
5、取得平均薪水最高的部门的部门编号
方法一:
select deptno,avg(sal) as avgsal from emp group by deptno;
select deptno,max(avgsal) from (select deptno,avg(sal) as avgsal from emp group by deptno)t;
方法二:
找出平均薪水降序取第一个
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,e.deptno,avg(e.sal) as avgsal
from
emp e
left join
dept d
on
e.deptno=d.deptno
group by
deptno
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
t.dname,t.avgsal,s.grade
from
(select d.dname,d.deptno,avg(e.sal) as avgsal from dept d join emp e on d.deptno=e.deptno group by deptno) t
join
salgrade s
on
t.avgsal between s.losal and s.hisal
where
(select grade from salgrade s where (select avg(sal) as avgsal from emp group by deptno order by avgsal limit 1) between s.losal and s.hisal) = s.grade;
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
日期可以升序或降序排
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
s.grade,count(*) as number
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
e1.ename as '员工',ifnull(e2.ename,'Big boss') as '领导'
from
emp e1
left join
emp e2
on
e1.mgr=e2.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
e1.empno,e1.ename,d.dname
from
emp e1
join
emp e2
on
e1.hiredate < e2.hiredate and e1.mgr = e2.empno
join
dept d
on
e1.deptno = d.deptno;
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.
外连接
select
d.dname,e.*
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 e.* from emp e where e.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.
第一步:找出部门人数
select deptno,count(*) as number from emp group by deptno;
第二步:
select
e.ename,d.dname,t.number
from
emp e
join
dept d
on
e.deptno=d.deptno
join
(select deptno,count(*) as number from emp group by deptno) t
on
t.deptno=e.deptno
where
e.job='clerk';
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.
第一步:找出工作的最低薪水
select job,min(sal) from emp group by job;
select
t1.job,t1.minsal,t2.job_number
from
(select job,min(sal) as minsal from emp group by job) t1
join
(select job,count(*) as job_number from emp group by job) t2
on
t1.job=t2.job
where
t1.minsal>1500;
简单方法:
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.
脑筋急转弯
select 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
e1.ename as '员工',d.dname as '部门',s.grade as '薪水等级',e2.ename as '领导'
from
emp e1
left join
emp e2
on
e1.mgr=e2.empno
join
salgrade s
on
e1.sal between s.losal and s.hisal
join
dept d
on
d.deptno=e1.deptno
where
e1.sal>(select avg(sal) as avgsal from emp);
23、列出与"SCOTT"从事相同工作的所有员工及部门名称.
select
e.ename,d.dname
from
emp e
join
dept d
on
e.deptno=d.deptno
where
e.job=(select job from emp where ename='scott') and 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 ename,sal from emp where sal in (select sal from emp where deptno=30) and deptno <> 30;
结果为空
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
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.dname,count(e.ename) as employee_number,ifnull(avg(e.sal),0) as avgsal,avg(timestampdiff(year,hiredate,now()))
from
emp e
right join
dept d
on
e.deptno=d.deptno
group by
e.deptno;
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.* ,count(e.ename) from emp e right join dept d on e.deptno=d.deptno group by deptno;
29、列出各种工作的最低工资及从事此工作的雇员姓名
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、列出各个部门的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;
32、求出员工领导的薪水超过3000的员工名称与领导名称
select
e1.ename as '员工',e2.ename as '领导',e2.sal
from
emp e1
join
emp e2
on
e1.mgr=e2.empno
where
e2.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.dname,count(e.ename),ifnull(sum(e.sal),0)
from
emp e
right join
dept d
on
e.deptno=d.deptno
where
d.dname like '%S%'
group by
d.deptno;
34、给任职日期超过30年的员工加薪10%.
SET SQL_SAFE_UPDATES = 0;
update emp set sal=sal*1.1 where timestampdiff(year,hiredate,now())>30;
select * from emp;