34道查询语句的练习题：
初始表格
emp
+-------+--------+-----------+------+------------+---------+---------+--------+
| EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO |
+-------+--------+-----------+------+------------+---------+---------+--------+
| 7369 | SMITH | CLERK | 7902 | 1980-12-17 | 800.00 | NULL | 20 |
| 7499 | ALLEN | SALESMAN | 7698 | 1981-02-20 | 1600.00 | 300.00 | 30 |
| 7521 | WARD | SALESMAN | 7698 | 1981-02-22 | 1250.00 | 500.00 | 30 |
| 7566 | JONES | MANAGER | 7839 | 1981-04-02 | 2975.00 | NULL | 20 |
| 7654 | MARTIN | SALESMAN | 7698 | 1981-09-28 | 1250.00 | 1400.00 | 30 |
| 7698 | BLAKE | MANAGER | 7839 | 1981-05-01 | 2850.00 | NULL | 30 |
| 7782 | CLARK | MANAGER | 7839 | 1981-06-09 | 2450.00 | NULL | 10 |
| 7788 | SCOTT | ANALYST | 7566 | 1987-04-19 | 3000.00 | NULL | 20 |
| 7839 | KING | PRESIDENT | NULL | 1981-11-17 | 5000.00 | NULL | 10 |
| 7844 | TURNER | SALESMAN | 7698 | 1981-09-08 | 1500.00 | 0.00 | 30 |
| 7876 | ADAMS | CLERK | 7788 | 1987-05-23 | 1100.00 | NULL | 20 |
| 7900 | JAMES | CLERK | 7698 | 1981-12-03 | 950.00 | NULL | 30 |
| 7902 | FORD | ANALYST | 7566 | 1981-12-03 | 3000.00 | NULL | 20 |
| 7934 | MILLER | CLERK | 7782 | 1982-01-23 | 1300.00 | NULL | 10 |
+-------+--------+-----------+------+------------+---------+---------+--------+
dept
+--------+------------+----------+
| DEPTNO | DNAME | LOC |
+--------+------------+----------+
| 10 | ACCOUNTING | NEW YORK |
| 20 | RESEARCH | DALLAS |
| 30 | SALES | CHICAGO |
| 40 | OPERATIONS | BOSTON |
+--------+------------+----------+
salgrade
+-------+-------+-------+
| GRADE | LOSAL | HISAL |
+-------+-------+-------+
| 1 | 700 | 1200 |
| 2 | 1201 | 1400 |
| 3 | 1401 | 2000 |
| 4 | 2001 | 3000 |
| 5 | 3001 | 9999 |
+-------+-------+-------+
mysql中 有专门计算日期差的函数 timestampdiff(间隔类型，起始日期，当前日期)
1.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资：
select max(sal), deptno from emp group by deptno;
+----------+--------+
| max(sal) | deptno |
+----------+--------+
| 5000.00 | 10 |
| 3000.00 | 20 |
| 2850.00 | 30 |
+----------+--------+
第二部将上面的表当成临时表 t 和emp表 进行对比
select
e.ename, t.*
from
( select max(sal) as 'max' , deptno from emp group by deptno) t
join
emp e
on
t.deptno = e.deptno and t.max = e.sal;
+-------+---------+--------+
| ename | max | deptno |
+-------+---------+--------+
| BLAKE | 2850.00 | 30 |
| SCOTT | 3000.00 | 20 |
| KING | 5000.00 | 10 |
| FORD | 3000.00 | 20 |
+-------+---------+--------+
2.那些人的薪水在部门平均水平之上
第一步 找出每个部门的平均薪水：
select deptno, avg(sal) from emp group by deptno;
+--------+-------------+
| deptno | avg(sal) |
+--------+-------------+
| 10 | 2916.666667 |
| 20 | 2175.000000 |
| 30 | 1566.666667 |
+--------+-------------+
第二步 找出高于这个水平的人：
select e.ename, e.sal, t.* from emp e join (select deptno, avg(sal) as avg from emp e group by deptno) t on e.sal>t.avg and e.deptno=t.deptno;
+-------+---------+--------+-------------+
| ename | sal | deptno | avg |
+-------+---------+--------+-------------+
| ALLEN | 1600.00 | 30 | 1566.666667 |
| JONES | 2975.00 | 20 | 2175.000000 |
| BLAKE | 2850.00 | 30 | 1566.666667 |
| SCOTT | 3000.00 | 20 | 2175.000000 |
| KING | 5000.00 | 10 | 2916.666667 |
| FORD | 3000.00 | 20 | 2175.000000 |
+-------+---------+--------+-------------+
3.取得部门中所有人的平均薪水等级：
select
e.deptno, avg(s.grade)
from
emp e
join
salgrade s
on
e.sal between s.losal and s.hisal
group by
deptno;
+--------+--------------+
| deptno | avg(s.grade) |
+--------+--------------+
| 10 | 3.6667 |
| 20 | 2.8000 |
| 30 | 2.5000 |
+--------+--------------+
4.不用组函数（max）取得最高薪水
第一种降序取最高值 limit：
select
ename, sal
from
emp
order by
sal
desc
limit 1;
+-------+---------+
| ename | sal |
+-------+---------+
| KING | 5000.00 |
+-------+---------+
第二种 max（）函数 这里不让用
第三种：自连接
//这个查询条件能找到除了最大值之外的所有值 a b两张表相同 两表里面不会有比最大值更大的数字出现
select
distinct a.sal
from
emp a
join
emp b
on
a.sal5.取得平均薪水最高的部门编号
第一种： 求平均值 然后用排序 limit 函数取值
select
deptno, avg(sal) as avg
from emp
group by deptno
order by avg desc
limit 1;
+--------+-------------+
| deptno | avg |
+--------+-------------+
| 10 | 2916.666667 |
+--------+-------------+
方法二：max函数
把找到的平均值当作临时表 找最大值
select max(t.avg)
from
(
select
deptno, avg(sal) as avg
from emp
group by deptno) t;
+-------------+
| max(t.avg) |
+-------------+
| 2916.666667 |
+-------------+
6.取得平均薪水最高的部门名称：
emp 表的部门名称等于 dept表的部门名称
--------------------------
select
e.deptno, avg(e.sal) as avg , d.dname
from emp e
join
dept d
on
e.deptno=d.deptno
group by deptno
order by avg desc
limit 1;
+--------+-------------+------------+
| deptno | avg | dname |
+--------+-------------+------------+
| 10 | 2916.666667 | ACCOUNTING |
+--------+-------------+------------+
7.取得平均薪水等级最低的部门名称
找出部门的平均薪水 和部门名称
select avg(sal), deptno from emp group by deptno;
+-------------+--------+
| avg(sal) | deptno |
+-------------+--------+
| 2916.666667 | 10 |
| 2175.000000 | 20 |
| 1566.666667 | 30 |
+-------------+--------+
第二部找出平均水平的等级： 上一把找出来的表当作临时表
select
t.*, s.grade
from
(select avg(sal) as avg, deptno from emp group by deptno) t
join salgrade s
on
t.avg between s.losal and s.hisal;
最后：通过临时表找到工资等级 然后通过 工资排序找到最低的工资
select t.*, s.grade
from
(select avg(sal) as avg, e.deptno, dname
from
emp e
join
dept d
on
e.deptno = d.deptno
group by
deptno) t
join
salgrade s
on
t.avg between s.losal and s.hisal
order by
avg
limit 1;
8.取得比普通员工（员工代码没有在mgr字段上出现的）最高薪水还要高的领导人的姓名
//mgr是领导的编号 员工编号不在这个范围
找出所有领导的编号
select distinct mgr from emp;
+------+
| mgr |
+------+
| 7902 |
| 7698 |
| 7839 |
| 7566 |
| NULL |
| 7788 |
| 7782 |
+------+
找出员工的最高薪水 not in 在使用的时候后面的小括号里如果有 null 要手动排除掉
select
ename, max(sal)
from emp
where empno
not in
(select
distinct mgr
from emp
where mgr
is not
null);
+-------+----------+
| ename | max(sal) |
+-------+----------+
| SMITH | 1600.00 |
+-------+----------+
比普通员工最高薪水高的一定是领导
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));
+-------+---------+
| ename | sal |
+-------+---------+
| JONES | 2975.00 |
| BLAKE | 2850.00 |
| CLARK | 2450.00 |
| SCOTT | 3000.00 |
| KING | 5000.00 |
| FORD | 3000.00 |
+-------+---------+