多表查询

Posted 2020-10-06 孟庆健

一、多表查询

 

#建表
create table department(
id int,
name varchar(20) 
);

create table employee(
id int primary key auto_increment,
name varchar(20),
sex enum(‘male‘,‘female‘) not null default ‘male‘,
age int,
dep_id int
);

#插入数据
insert into department values
(200,‘技术‘),
(201,‘人力资源‘),
(202,‘销售‘),
(203,‘运营‘);

insert into employee(name,sex,age,dep_id) values
(‘egon‘,‘male‘,18,200),
(‘alex‘,‘female‘,48,201),
(‘wupeiqi‘,‘male‘,38,201),
(‘yuanhao‘,‘female‘,28,202),
(‘liwenzhou‘,‘male‘,18,200),
(‘jingliyang‘,‘female‘,18,204)
;


#查看表结构和数据
mysql> desc department;
+-------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+-------------+------+-----+---------+-------+
| id | int(11) | YES | | NULL | |
| name | varchar(20) | YES | | NULL | |
+-------+-------------+------+-----+---------+-------+

mysql> desc employee;
+--------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------+-----------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(20) | YES | | NULL | |
| sex | enum(‘male‘,‘female‘) | NO | | male | |
| age | int(11) | YES | | NULL | |
| dep_id | int(11) | YES | | NULL | |
+--------+-----------------------+------+-----+---------+----------------+

mysql> select * from department;
+------+--------------+
| id | name |
+------+--------------+
| 200 | 技术 |
| 201 | 人力资源 |
| 202 | 销售 |
| 203 | 运营 |
+------+--------------+

mysql> select * from employee;
+----+------------+--------+------+--------+
| id | name | sex | age | dep_id |
+----+------------+--------+------+--------+
| 1 | egon | male | 18 | 200 |
| 2 | alex | female | 48 | 201 |
| 3 | wupeiqi | male | 38 | 201 |
| 4 | yuanhao | female | 28 | 202 |
| 5 | liwenzhou | male | 18 | 200 |
| 6 | jingliyang | female | 18 | 204 |
+----+------------+--------+------+--------+

 

#多表查询
create table department(
id int primary key auto_increment,
name varchar(20)
);

create table employee(
id int primary key auto_increment,
name varchar(20),
sex enum(‘male‘,‘female‘) not null default ‘male‘,
age int,
dep_id int,
foreign key(dep_id) references department(id)
);

#简单查询
select * from department,employee; #笛卡尔积
select * from department,employee where department.id=employee.dep_id;

#内连接：按照on条件只两张表的相同的部分，连接成一张虚拟的表
select * from employee inner join department on department.id=employee.dep_id;
select * from department inner join employee on department.id=employee.dep_id;
#select * from employee,department where department.id=employee.dep_id;



#左链接：在按照on的条件取到两张表共同部分的基础上，保留左表的记录
select * from employee left join department on department.id=employee.dep_id;

#右链接：在按照on的条件取到两张表共同部分的基础上，保留右表的记录
select * from employee right join department on department.id=employee.dep_id;

#full jion：
select * from employee left join department on department.id=employee.dep_id
union
select * from employee right join department on department.id=employee.dep_id;




#子查询：
mysql> select * from employee where dep_id in (select id from department where name in (‘技术‘,‘销售‘));
+----+-----------+--------+------+--------+
| id | name      | sex    | age  | dep_id |
+----+-----------+--------+------+--------+
|  1 | egon      | male   |   18 |    200 |
|  4 | yuanhao   | female |   28 |    202 |
|  5 | liwenzhou | male   |   18 |    200 |
+----+-----------+--------+------+--------+
3 rows in set (0.02 sec)


#查询平均年龄在25岁以上的部门名
select name from department where id in (
select dep_id from employee group by dep_id having avg(age) > 25
);
#查看技术部员工姓名
select name from employee where dep_id = (select id from department where name=‘技术‘);

#查看小于2人的部门名
select name from department where id in (
select dep_id from employee group by dep_id having count(id) < 2
)
union
select name from department where id not in (select distinct dep_id from employee);

#提取空部门                              #有人的部门
select * from department where id not in (select distinct dep_id from employee);


或者：
select name from department where id in
(
select dep_id from employee group by dep_id having count(id) < 2
union
select id from department where id not in (select distinct dep_id from employee)
);

#exists
mysql> select * from employee where exists (select id from department where name=‘hahahahah‘);
Empty set (0.00 sec)

mysql> select * from employee where exists (select id from department where name=‘技术‘);
+----+------------+--------+------+--------+
| id | name       | sex    | age  | dep_id |
+----+------------+--------+------+--------+
|  1 | egon       | male   |   18 |    200 |
|  2 | alex       | female |   48 |    201 |
|  3 | wupeiqi    | male   |   38 |    201 |
|  4 | yuanhao    | female |   28 |    202 |
|  5 | liwenzhou  | male   |   18 |    200 |
|  6 | jingliyang | female |   18 |    204 |
+----+------------+--------+------+--------+
6 rows in set (0.00 sec)

二、练习

 

拷贝表：
create table day43.employee select * from day42.employee


改表名：
alter table employee rename emp;


select 分组字段,count(id) from t1 where 条件 group by 分组字段

max
min
sum
avg

group_concat

#关键字执行优先级
from
where
group by
按照select后的字段取得一张新的虚拟表，有聚合函数则执行聚合函数
having


select max(salary) from t1 where id > 2 group by depart_id having count(id) > 2;
select 333333333 from t1 where id > 2 group by depart_id having 4 > 2;

#练习
select post,count(id),group_concat(name) from emp group by post having count(id) < 2;


select post,avg(salary) as 平均工资 from emp group by post having avg(salary) > 10000;
select post 岗位名,avg(salary) 平均工资 from emp group by post having avg(salary) > 10000;



#order by关键字
select * from emp order by salary;
select * from emp order by salary asc; #升序
select * from emp order by salary desc; #降序

 select * from emp order by age asc,salary desc; #先按照年龄从小到大排，如果年龄分不出胜负（即值相同）再按照salary从大到小排。

#练习：

select * from emp order by age asc,hire_date desc;

select post 岗位名,avg(salary) 平均工资 from emp group by post having avg(salary) > 10000 order by avg(salary) asc;



#limit
mysql> select * from emp limit 10;
#从哪开始，往后取几条
mysql> select * from emp limit 0,3;
mysql> select * from emp limit 3,3;
mysql> select * from emp limit 6,3;


select name,id from emp where id>15 having id > 16;


#distinct
mysql> select * from emp limit 0,3;

 

三、综合练习

init.sql文件内容

 

/*
 数据导入：
 Navicat Premium Data Transfer

 Source Server         : localhost
 Source Server Type    : MySQL
 Source Server Version : 50624
 Source Host           : localhost
 Source Database       : sqlexam

 Target Server Type    : MySQL
 Target Server Version : 50624
 File Encoding         : utf-8

 Date: 10/21/2016 06:46:46 AM
*/

SET NAMES utf8;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------
--  Table structure for `class`
-- ----------------------------
DROP TABLE IF EXISTS `class`;
CREATE TABLE `class` (
  `cid` int(11) NOT NULL AUTO_INCREMENT,
  `caption` varchar(32) NOT NULL,
  PRIMARY KEY (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `class`
-- ----------------------------
BEGIN;
INSERT INTO `class` VALUES (‘1‘, ‘三年二班‘), (‘2‘, ‘三年三班‘), (‘3‘, ‘一年二班‘), (‘4‘, ‘二年九班‘);
COMMIT;

-- ----------------------------
--  Table structure for `course`
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
  `cid` int(11) NOT NULL AUTO_INCREMENT,
  `cname` varchar(32) NOT NULL,
  `teacher_id` int(11) NOT NULL,
  PRIMARY KEY (`cid`),
  KEY `fk_course_teacher` (`teacher_id`),
  CONSTRAINT `fk_course_teacher` FOREIGN KEY (`teacher_id`) REFERENCES `teacher` (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `course`
-- ----------------------------
BEGIN;
INSERT INTO `course` VALUES (‘1‘, ‘生物‘, ‘1‘), (‘2‘, ‘物理‘, ‘2‘), (‘3‘, ‘体育‘, ‘3‘), (‘4‘, ‘美术‘, ‘2‘);
COMMIT;

-- ----------------------------
--  Table structure for `score`
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score` (
  `sid` int(11) NOT NULL AUTO_INCREMENT,
  `student_id` int(11) NOT NULL,
  `course_id` int(11) NOT NULL,
  `num` int(11) NOT NULL,
  PRIMARY KEY (`sid`),
  KEY `fk_score_student` (`student_id`),
  KEY `fk_score_course` (`course_id`),
  CONSTRAINT `fk_score_course` FOREIGN KEY (`course_id`) REFERENCES `course` (`cid`),
  CONSTRAINT `fk_score_student` FOREIGN KEY (`student_id`) REFERENCES `student` (`sid`)
) ENGINE=InnoDB AUTO_INCREMENT=53 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `score`
-- ----------------------------
BEGIN;
INSERT INTO `score` VALUES (‘1‘, ‘1‘, ‘1‘, ‘10‘), (‘2‘, ‘1‘, ‘2‘, ‘9‘), (‘5‘, ‘1‘, ‘4‘, ‘66‘), (‘6‘, ‘2‘, ‘1‘, ‘8‘), (‘8‘, ‘2‘, ‘3‘, ‘68‘), (‘9‘, ‘2‘, ‘4‘, ‘99‘), (‘10‘, ‘3‘, ‘1‘, ‘77‘), (‘11‘, ‘3‘, ‘2‘, ‘66‘), (‘12‘, ‘3‘, ‘3‘, ‘87‘), (‘13‘, ‘3‘, ‘4‘, ‘99‘), (‘14‘, ‘4‘, ‘1‘, ‘79‘), (‘15‘, ‘4‘, ‘2‘, ‘11‘), (‘16‘, ‘4‘, ‘3‘, ‘67‘), (‘17‘, ‘4‘, ‘4‘, ‘100‘), (‘18‘, ‘5‘, ‘1‘, ‘79‘), (‘19‘, ‘5‘, ‘2‘, ‘11‘), (‘20‘, ‘5‘, ‘3‘, ‘67‘), (‘21‘, ‘5‘, ‘4‘, ‘100‘), (‘22‘, ‘6‘, ‘1‘, ‘9‘), (‘23‘, ‘6‘, ‘2‘, ‘100‘), (‘24‘, ‘6‘, ‘3‘, ‘67‘), (‘25‘, ‘6‘, ‘4‘, ‘100‘), (‘26‘, ‘7‘, ‘1‘, ‘9‘), (‘27‘, ‘7‘, ‘2‘, ‘100‘), (‘28‘, ‘7‘, ‘3‘, ‘67‘), (‘29‘, ‘7‘, ‘4‘, ‘88‘), (‘30‘, ‘8‘, ‘1‘, ‘9‘), (‘31‘, ‘8‘, ‘2‘, ‘100‘), (‘32‘, ‘8‘, ‘3‘, ‘67‘), (‘33‘, ‘8‘, ‘4‘, ‘88‘), (‘34‘, ‘9‘, ‘1‘, ‘91‘), (‘35‘, ‘9‘, ‘2‘, ‘88‘), (‘36‘, ‘9‘, ‘3‘, ‘67‘), (‘37‘, ‘9‘, ‘4‘, ‘22‘), (‘38‘, ‘10‘, ‘1‘, ‘90‘), (‘39‘, ‘10‘, ‘2‘, ‘77‘), (‘40‘, ‘10‘, ‘3‘, ‘43‘), (‘41‘, ‘10‘, ‘4‘, ‘87‘), (‘42‘, ‘11‘, ‘1‘, ‘90‘), (‘43‘, ‘11‘, ‘2‘, ‘77‘), (‘44‘, ‘11‘, ‘3‘, ‘43‘), (‘45‘, ‘11‘, ‘4‘, ‘87‘), (‘46‘, ‘12‘, ‘1‘, ‘90‘), (‘47‘, ‘12‘, ‘2‘, ‘77‘), (‘48‘, ‘12‘, ‘3‘, ‘43‘), (‘49‘, ‘12‘, ‘4‘, ‘87‘), (‘52‘, ‘13‘, ‘3‘, ‘87‘);
COMMIT;

-- ----------------------------
--  Table structure for `student`
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (
  `sid` int(11) NOT NULL AUTO_INCREMENT,
  `gender` char(1) NOT NULL,
  `class_id` int(11) NOT NULL,
  `sname` varchar(32) NOT NULL,
  PRIMARY KEY (`sid`),
  KEY `fk_class` (`class_id`),
  CONSTRAINT `fk_class` FOREIGN KEY (`class_id`) REFERENCES `class` (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=17 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `student`
-- ----------------------------
BEGIN;
INSERT INTO `student` VALUES (‘1‘, ‘男‘, ‘1‘, ‘理解‘), (‘2‘, ‘女‘, ‘1‘, ‘钢蛋‘), (‘3‘, ‘男‘, ‘1‘, ‘张三‘), (‘4‘, ‘男‘, ‘1‘, ‘张一‘), (‘5‘, ‘女‘, ‘1‘, ‘张二‘), (‘6‘, ‘男‘, ‘1‘, ‘张四‘), (‘7‘, ‘女‘, ‘2‘, ‘铁锤‘), (‘8‘, ‘男‘, ‘2‘, ‘李三‘), (‘9‘, ‘男‘, ‘2‘, ‘李一‘), (‘10‘, ‘女‘, ‘2‘, ‘李二‘), (‘11‘, ‘男‘, ‘2‘, ‘李四‘), (‘12‘, ‘女‘, ‘3‘, ‘如花‘), (‘13‘, ‘男‘, ‘3‘, ‘刘三‘), (‘14‘, ‘男‘, ‘3‘, ‘刘一‘), (‘15‘, ‘女‘, ‘3‘, ‘刘二‘), (‘16‘, ‘男‘, ‘3‘, ‘刘四‘);
COMMIT;

-- ----------------------------
--  Table structure for `teacher`
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (
  `tid` int(11) NOT NULL AUTO_INCREMENT,
  `tname` varchar(32) NOT NULL,
  PRIMARY KEY (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `teacher`
-- ----------------------------
BEGIN;
INSERT INTO `teacher` VALUES (‘1‘, ‘张磊老师‘), (‘2‘, ‘李平老师‘), (‘3‘, ‘刘海燕老师‘), (‘4‘, ‘朱云海老师‘), (‘5‘, ‘李杰老师‘);
COMMIT;

SET FOREIGN_KEY_CHECKS = 1;

准备表

 

use day49
source c:\\init.sql


1、查询所有的课程的名称以及对应的任课老师姓名
select cname,tname from course left join teacher on course.teacher_id=teacher.tid;
2、查询学生表中男女生各有多少人
select gender,count(sid) from student group by gender;
3、查询物理成绩等于100的学生的姓名
#子查询的方式
select sname from student where sid in
(
select student_id from score
    where course_id = (select cid from course where cname=‘物理‘) and num=100
);
#连表的方式
select sname from student inner join
(
select student_id from score
    where course_id = (select cid from course where cname=‘物理‘) and num=100
) as a
on a.student_id=student.sid
;

4、查询平均成绩大于八十分的同学的姓名和平均成绩
select student.sname,t1.平均成绩 from student inner join
(select student_id,avg(num) 平均成绩 from score group by student_id having avg(num) > 80) as t1
on student.sid=t1.student_id;
5、查询所有学生的学号，姓名，选课数，总成绩
select student.sid,sname 学生名,选课数,总成绩 from student left join
(select student_id,count(course_id) 选课数,sum(num) 总成绩 from score group by student_id) as t1
on student.sid=t1.student_id
;
6、 查询姓李老师的个数
select count(tid) from teacher where tname like ‘李%‘;

7、 查询没有报李平老师课的学生姓名
select sname from student where sid not in (
select distinct student_id from score where course_id in (
select cid from course where teacher_id=(select tid from teacher where tname=‘李平老师‘)
)
);
8、查询物理课程比生物课程高的学生的学号
select t1.student_id from
(select student_id,num from score inner join course on score.course_id=course.cid
 where course.cname=‘物理‘) as t1
 inner join
(select student_id,num from score inner join course on score.course_id=course.cid
where course.cname=‘生物‘) as t2
on t1.student_id=t2.student_id
where t1.num > t2.num
;
9、 查询没有同时选修物理课程和体育课程的学生姓名
select sname from student where sid in (
select student_id from score inner join course
on course.cname in (‘物理‘,‘体育‘) and course.cid=score.course_id
group by student_id having count(course_id) !=2
);
10、查询挂科超过两门(包括两门)的学生姓名和班级名字
select t2.sname,class.caption from
(select sname,class_id from student inner join (
select student_id from score
where num < 60 group by student_id having count(course_id) >=2
) as t1
on student.sid=t1.student_id) as t2
inner join class
on class.cid = t2.class_id
;
11 、查询选修了所有课程的学生姓名
select sname from student inner join
(
select student_id from score group by student_id having count(course_id) = (select count(cid) from course)
) t1
on t1.student_id = student.sid
;
12、查询李平老师教的课程的所有成绩记录
select student_id,course_id,num from score inner join
(
select cid from course inner join teacher on teacher.tname=‘李平老师‘ and teacher.tid=course.teacher_id
) as t1
on t1.cid=score.course_id
;

13、查询全部学生都选修了的课程号和课程名
select course.cid,course.cname from course inner join
(
select course_id from score group by course_id
having count(student_id) = (select count(sid) from student)
) t1
on t1.course_id=course.cid
;

14、查询每门课程被选修的次数
select course.cname,选修人数 from course inner join
(
select course_id,count(student_id) as 选修人数 from score group by course_id
) as t1
on t1.course_id=course.cid
;
15、查询之选修了一门课程的学生姓名和学号
select sid,sname from student inner join
(
select student_id from score group by student_id having count(course_id)=1
) t1
on t1.student_id = student.sid
;
16、查询所有学生考出的成绩并按从高到低排序（成绩去重）
select distinct num from score order by num desc;

17、查询平均成绩大于85的学生姓名和平均成绩
select student.sname,avg_num from student inner join
(
select student_id,avg(num) as avg_num from score group by student_id having avg(num) > 85
) t1
on student.sid=t1.student_id
;

18、查询生物成绩不及格的学生姓名和对应生物分数
select student.sname,t1.num from student inner join
(
select student_id,num from score
where course_id=(select cid from course where cname=‘生物‘) and num < 60
) t1
on t1.student_id=student.sid
;
19、查询在所有选修了李平老师课程的学生中，这些课程(李平老师的课程，不是所有课程)平均成绩最高的学生姓名

select sname from student where sid =
(
select student_id from score where course_id in
(
select cid from course inner join teacher on teacher.tname=‘李平老师‘ and course.teacher_id=teacher.tid
)
group by student_id
order by avg(num) desc
limit 1
);

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