7.7、多表查询案例

— 准备数据
create table dept(
id int auto_increment comment ‘ID’ primary key,
name varchar(50) not null comment ‘部门名称’
)comment ‘部门表’;

create table emp(
id int auto_increment comment ‘ID’ primary key,
name varchar(50) not null comment ‘姓名’,
age int comment ‘年龄’,
job varchar(20) comment ‘职位’,
salary int comment ‘薪资’,
entrydate date comment ‘入职时间’,
managerid int comment ‘直属领导ID’,
dept_id int comment ‘部门ID’
)comment ‘员工表’;

— 添加外键
alter table emp add constraint fk_emp_dept_id foreign key (dept_id) references dept(id);
INSERT INTO dept (id, name) VALUES (1, ‘研发部’), (2, ‘市场部’),(3, ‘财务部’), (4, ‘销售部’), (5, ‘总经办’), (6, ‘人事部’);
INSERT INTO emp (id, name, age, job,salary, entrydate, managerid, dept_id) VALUES
(1, ‘金庸’, 66, ‘总裁’,20000, ‘2000-01-01’, null,5),
(2, ‘张无忌’, 20, ‘项目经理’,12500, ‘2005-12-05’, 1,1),
(3, ‘杨逍’, 33, ‘开发’, 8400,’2000-11-03’, 2,1),
(4, ‘韦一笑’, 48, ‘开发’,11000, ‘2002-02-05’, 2,1),
(5, ‘常遇春’, 43, ‘开发’,10500, ‘2004-09-07’, 3,1),
(6, ‘小昭’, 19, ‘程序员鼓励师’,6600, ‘2004-10-12’, 2,1),
(7, ‘灭绝’, 60, ‘财务总监’,8500, ‘2002-09-12’, 1,3),
(8, ‘周芷若’, 19, ‘会计’,48000, ‘2006-06-02’, 7,3),
(9, ‘丁敏君’, 23, ‘出纳’,5250, ‘2009-05-13’, 7,3),
(10, ‘赵敏’, 20, ‘市场部总监’,12500, ‘2004-10-12’, 1,2),
(11, ‘鹿杖客’, 56, ‘职员’,3750, ‘2006-10-03’, 10,2),
(12, ‘鹤笔翁’, 19, ‘职员’,3750, ‘2007-05-09’, 10,2),
(13, ‘方东白’, 19, ‘职员’,5500, ‘2009-02-12’, 10,2),
(14, ‘张三丰’, 88, ‘销售总监’,14000, ‘2004-10-12’, 1,4),
(15, ‘俞莲舟’, 38, ‘销售’,4600, ‘2004-10-12’, 14,4),
(16, ‘宋远桥’, 40, ‘销售’,4600, ‘2004-10-12’, 14,4),
(17, ‘陈友谅’, 42, null,2000, ‘2011-10-12’, 1,null);

create table salgrade(
grade int,
losal int,
hisal int
) comment ‘薪资等级表’;

insert into salgrade values (1,0,3000);
insert into salgrade values (2,3001,5000);
insert into salgrade values (3,5001,8000);
insert into salgrade values (4,8001,10000);
insert into salgrade values (5,10001,15000);
insert into salgrade values (6,15001,20000);
insert into salgrade values (7,20001,25000);
insert into salgrade values (8,25001,30000);

create table student(
id int auto_increment primary key comment ‘主键ID’,
name varchar(10) comment ‘姓名’,
no varchar(10) comment ‘学号’
) comment ‘学生表’;
insert into student values (null, ‘黛绮丝’, ‘2000100101’),(null, ‘谢逊’, ‘2000100102’),(null, ‘殷天正’, ‘2000100103’),(null, ‘韦一笑’, ‘2000100104’);

create table course(
id int auto_increment primary key comment ‘主键ID’,
name varchar(10) comment ‘课程名称’
) comment ‘课程表’;
insert into course values (null, ‘Java’), (null, ‘PHP’), (null , ‘MySQL’) , (null, ‘Hadoop’);

create table student_course(
id int auto_increment comment ‘主键’ primary key,
studentid int not null comment ‘学生ID’,
courseid int not null comment ‘课程ID’,
constraint fk_courseid foreign key (courseid) references course (id),
constraint fk_studentid foreign key (studentid) references student (id)
)comment ‘学生课程中间表’;
insert into student_course values (null,1,1),(null,1,2),(null,1,3),(null,2,2),(null,2,3),(null,3,4);

练习 根据需求，完成SQL语句的编写
1. 查询员工的姓名、年龄、职位、部门信息。
2. 查询年龄小于30岁的员工姓名、年龄、职位、部门信息。
3. 查询拥有员工的部门ID、部门名称。
4. 查询所有年龄大于40岁的员工, 及其归属的部门名称; 如果员工没有分配部门, 也需要展示出来。
5. 查询所有员工的工资等级。
6. 查询 “研发部” 所有员工的信息及工资等级。
7. 查询 “研发部” 员工的平均工资。
8. 查询工资比 “灭绝” 高的员工信息。
9. 查询比平均薪资高的员工信息。
10. 查询低于本部门平均工资的员工信息。
11. 查询所有的部门信息, 并统计部门的员工人数。
12. 查询所有学生的选课情况, 展示出学生名称, 学号, 课程名称

参考答案：
— 1. 查询员工的姓名、年龄、职位、部门信息 （隐式内连接）
— 表: emp , dept
— 连接条件: emp.dept_id = dept.id
select e.name , e.age , e.job , d.name from emp e , dept d where e.dept_id = d.id;

— 2. 查询年龄小于30岁的员工的姓名、年龄、职位、部门信息（显式内连接）
— 表: emp , dept
— 连接条件: emp.dept_id = dept.id
select e.name , e.age , e.job , d.name from emp e inner join dept d on e.dept_id = d.id where e.age < 30;

— 3. 查询拥有员工的部门ID、部门名称
— 表: emp , dept
— 连接条件: emp.dept_id = dept.id
select distinct d.id , d.name from emp e , dept d where e.dept_id = d.id;

— 4. 查询所有年龄大于40岁的员工, 及其归属的部门名称; 如果员工没有分配部门, 也需要展示出来
— 表: emp , dept
— 连接条件: emp.dept_id = dept.id
— 外连接
select e.*, d.name from emp e left join dept d on e.dept_id = d.id where e.age > 40 ;

— 5. 查询所有员工的工资等级
— 表: emp , salgrade
— 连接条件 : emp.salary >= salgrade.losal and emp.salary <= salgrade.hisal
select e. , s.grade , s.losal, s.hisal from emp e , salgrade s where e.salary >= s.losal and e.salary <= s.hisal;
select e. , s.grade , s.losal, s.hisal from emp e , salgrade s where e.salary between s.losal and s.hisal;

— 6. 查询 “研发部” 所有员工的信息及 工资等级
— 表: emp , salgrade , dept
— 连接条件 : emp.salary between salgrade.losal and salgrade.hisal , emp.dept_id = dept.id
— 查询条件 : dept.name = ‘研发部’
select e.* , s.grade from emp e , dept d , salgrade s where e.dept_id = d.id and ( e.salary between s.losal and s.hisal ) and d.name = ‘研发部’;

— 7. 查询 “研发部” 员工的平均工资
— 表: emp , dept
— 连接条件 : emp.dept_id = dept.id
select avg(e.salary) from emp e, dept d where e.dept_id = d.id and d.name = ‘研发部’;

— 8. 查询工资比 “灭绝” 高的员工信息。
— a. 查询 “灭绝” 的薪资
select salary from emp where name = ‘灭绝’;
— b. 查询比她工资高的员工数据
select * from emp where salary > ( select salary from emp where name = ‘灭绝’ );

— 9. 查询比平均薪资高的员工信息
— a. 查询员工的平均薪资
select avg(salary) from emp;
— b. 查询比平均薪资高的员工信息
select * from emp where salary > ( select avg(salary) from emp );

— 10. 查询低于本部门平均工资的员工信息
— a. 查询指定部门平均薪资 1
select avg(e1.salary) from emp e1 where e1.dept_id = 1;
select avg(e1.salary) from emp e1 where e1.dept_id = 2;

— b. 查询低于本部门平均工资的员工信息
select * from emp e2 where e2.salary < ( select avg(e1.salary) from emp e1 where e1.dept_id = e2.dept_id );

— 11. 查询所有的部门信息, 并统计部门的员工人数
select d.id, d.name , ( select count() from emp e where e.dept_id = d.id ) ‘人数’ from dept d;
select count() from emp where dept_id = 1;

— 12. 查询所有学生的选课情况, 展示出学生名称, 学号, 课程名称
— 表: student , course , student_course
— 连接条件: student.id = student_course.studentid , course.id = student_course.courseid
select s.name , s.no , c.name from student s , student_course sc , course c where s.id = sc.studentid and sc.courseid = c.id ;