创建三个表并添加数据：
--部门表
dept部门表(deptno部门编号/dname部门名称/loc地点)
create table dept (
deptno numeric(2),
dname varchar(14),
loc varchar(13)
);
insert into dept values (10, 'ACCOUNTING', 'NEW YORK');
insert into dept values (20, 'RESEARCH', 'DALLAS');
insert into dept values (30, 'SALES', 'CHICAGO');
insert into dept values (40, 'OPERATIONS', 'BOSTON');
--工资等级表
salgrade工资等级表(grade 等级/losal此等级的最低/hisal此等级的最高)
create table salgrade (
grade numeric,
losal numeric,
hisal numeric
);
insert into salgrade values (1, 700, 1200);
insert into salgrade values (2, 1201, 1400);
insert into salgrade values (3, 1401, 2000);
insert into salgrade values (4, 2001, 3000);
insert into salgrade values (5, 3001, 9999);
--员工表
emp员工表(empno员工号/ename员工姓名/job工作/mgr上级编号/hiredate受雇日期/sal薪金/comm佣金/deptno部门编号)
工资 ＝ 薪金 ＋ 佣金
1.表自己跟自己连接
create table emp (
empno numeric(4) not null,
ename varchar(10),
job varchar(9),
mgr numeric(4),
hiredate datetime,
sal numeric(7, 2),
comm numeric(7, 2),
deptno numeric(2)
);
insert into emp values (7369, 'SMITH', 'CLERK', 7902, '1980-12-17', 800, null, 20);
insert into emp values (7499, 'ALLEN', 'SALESMAN', 7698, '1981-02-20', 1600, 300, 30);
insert into emp values (7521, 'WARD', 'SALESMAN', 7698, '1981-02-22', 1250, 500, 30);
insert into emp values (7566, 'JONES', 'MANAGER', 7839, '1981-04-02', 2975, null, 20);
insert into emp values (7654, 'MARTIN', 'SALESMAN', 7698, '1981-09-28', 1250, 1400, 30);
insert into emp values (7698, 'BLAKE', 'MANAGER', 7839, '1981-05-01', 2850, null, 30);
insert into emp values (7782, 'CLARK', 'MANAGER', 7839, '1981-06-09', 2450, null, 10);
insert into emp values (7788, 'SCOTT', 'ANALYST', 7566, '1982-12-09', 3000, null, 20);
insert into emp values (7839, 'KING', 'PRESIDENT', null, '1981-11-17', 5000, null, 10);
insert into emp values (7844, 'TURNER', 'SALESMAN', 7698, '1981-09-08', 1500, 0, 30);
insert into emp values (7876, 'ADAMS', 'CLERK', 7788, '1983-01-12', 1100, null, 20);
insert into emp values (7900, 'JAMES', 'CLERK', 7698, '1981-12-03', 950, null, 30);
insert into emp values (7902, 'FORD', 'ANALYST', 7566, '1981-12-03', 3000, null, 20);
insert into emp values (7934, 'MILLER', 'CLERK', 7782, '1982-01-23', 1300, null, 10);
聚合函数：
sum() max() min() count()
分组函数：
group by .... having ....
列题：
#1.求员工表 所有人的薪水和 as salsum 别名
select
sum(sal) as salsum
from emp;
#2.求每个部门的所有人的薪水和
select
deptno,
sum(sal) as salsum
from emp
group by deptno;
#总结: group by 后面出现多少个字段，那么select 后面也要一模一样
#每/各 是为分组字段
#3.求每个部门的每个岗位的所有人的薪水和
select
deptno,job,
sum(sal) as salsum
from emp
group by deptno,job;
#4.求每个部门的每个岗位的所有人的薪水和，及人数
select
deptno,job,
sum(sal) as salsum,
count(deptno) as pnum
from emp
group by deptno,job;
group by deptno --》3
group by deptno , job
#5.求薪水和大于1500的哪些部门 ==》 每个部门的薪水和 ==》 哪些薪水和大于1500的部门
select
deptno,
sum(sal) as salsum
from emp
group by deptno
having sum(sal)>1500;
#子查询
select
t.*
from
(select
deptno,
sum(sal) as salsum
from emp
group by deptno) as t
where t.salsum > 1500;
关联 left join,right join,inner join
重新创建表：
drop table testa;
create table testa(aid int,aname varchar(100),address varchar(100));
create table testb(bid int,bname varchar(100),age int);
create table testsal(cid int,sal int);
delete from testsal;
insert into testsal values(1,100);
insert into testsal values(2,300);
insert into testsal values(4,700);
insert into testsal values(8,1300);
insert into testsal values(9,3300);
select * from testsal;
insert into testa values(1,'xiao1','SH1');
insert into testa values(2,'xiao2','SH1');
insert into testa values(3,'xiao3',null);
insert into testa values(4,'xiao4','SH2');
insert into testa values(5,'xiao5','SH2');
insert into testb values(1,'xiao1',10);
insert into testb values(2,'xiao2',20);
insert into testb values(3,'xiao3',30);
insert into testb values(4,'xiao4',40);
insert into testb values(4,'xiao44',440);
insert into testb values(7,'xiao7',70);
insert into testb values(8,'xiao8',80);
insert into testb values(9,'xiao9',90);
## left join：
select a.*,b.*from testa a left join testb b on a.aid=b.bid a <-- b
#以左表为主，左表数据最全，右表来匹配的，匹配多少算多少
aid aname bid bname age
1 xiao1 1 xiao1 10
2 xiao2 2 xiao2 20
3 xiao3 3 xiao3 30
4 xiao4 4 xiao4 40
5 xiao5 null null null
## right join：
select a.*,b.*from testa a right join testb b on a.aid=b.bid
#以右表为主，右表数据最全，左表来匹配的，匹配多少算多少 a --> b
1 xiao1 1 xiao1 10
2 xiao2 2 xiao2 20
3 xiao3 3 xiao3 30
4 xiao4 4 xiao4 40
7 xiao7 70
8 xiao8 80
9 xiao9 90
## inner join
select a.*,b.*from testa a inner join testb b on a.aid=b.bid
#等值连接 两边都要存在
1 xiao1 1 xiao1 10
2 xiao2 2 xiao2 20
3 xiao3 3 xiao3 30
4 xiao4 4 xiao4 40
## 笛卡尔集
select a.*,b.*from testa a, testb b;
## 等值连接
select a.*,b.*from testa a, testb b where a.aid = b.bid;
jion相关联的题：
#哪些人 有薪水的有年龄的有地址
select s.*,b.*,a.*
from testsal s
left join testb b on s.cid =b.bid
left join testa a on s.cid =a.aid
where
s.sal is not null
and b.age is not null
and a.address is not null;
select
s.*,b.*,a.*
from testsal s
left join testb b on s.cid =b.bid
left join testa a on b.bid =a.aid
where
s.sal is not null
and b.age is not null
and a.address is not null;
select
s.*,b.*,a.*
from testsal s
left join testb b on s.cid =b.bid
right join testa a on b.bid =a.aid;
#哪些人 有薪水的 有年龄的有地址
select
s.*,t.*
from testsal s
left join
(select
b.*,a.*
from testb b
right join testa a on b.bid =a.aid) t on s.cid = t.aid
where
s.sal is not null
and t.age is not null
and t.address is not null;
select
s.*,t.*
from testsal s
inner join
(select
b.*,a.*
from testb b
inner join testa a on b.bid =a.aid) t on s.cid = t.aid;
面试题：
#哪些部门的薪水最高两位的工种有哪些？（分组取前几位）
#1.每个部门每个工种的薪水和
select * from emp;
create view sal
as
select
deptno,job,sum(sal+ifnull(comm,0)) as sal
from emp group by deptno,job;
select * from sal;
#2.从1结果集 找出哪些部门的工种薪水和最高的前2位工种是什么？
select
a.*
from sal a
where
(
select count(*) from sal b
where a.deptno=b.deptno and a.sal>b.sal
)
#假设取薪水和最高的每个部门的哪个工种?
select * from sal;
insert into emp values (1000, 'BOSS', 'BOSS', NULL, '1981-06-09', 9000, null, 10);
#top 1
select
a.*
from sal a
where (
select count(*) from sal b
where a.deptno=b.deptno and a.sal<b.sal
) = 0
order by a.deptno;
select
a.*
from sal a
where (
select count(*) from sal b where a.deptno=b.deptno and a.sal<b.sal
) <= 1
order by a.deptno asc ,a.sal desc;