Day45：MySQL（多表的表记录的查询）
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一、外键约束
1、创建外键
--- 每一个班主任会对应多个学生 , 而每个学生只能对应一个班主任
----主表
CREATE TABLE ClassCharger(
id TINYINT PRIMARY KEY auto_increment,
name VARCHAR (20),
age INT ,
is_marriged boolean -- show create table ClassCharger: tinyint(1)
);
INSERT INTO ClassCharger (name,age,is_marriged) VALUES ("冰冰",12,0),
("丹丹",14,0),
("歪歪",22,0),
("姗姗",20,0),
("小雨",21,0);
----子表
CREATE TABLE Student(
id INT PRIMARY KEY auto_increment,
name VARCHAR (20),
charger_id TINYINT, --切记:作为外键一定要和关联主键的数据类型保持一致
-- [ADD CONSTRAINT charger_fk_stu]FOREIGN KEY (charger_id) REFERENCES ClassCharger(id)
) ENGINE=INNODB;
INSERT INTO Student(name,charger_id) VALUES ("alvin1",2),
("alvin2",4),
("alvin3",1),
("alvin4",3),
("alvin5",1),
("alvin6",3),
("alvin7",2);
DELETE FROM ClassCharger WHERE name="冰冰";
INSERT student (name,charger_id) VALUES ("yuan",1);
-- 删除居然成功,可是 alvin3显示还是有班主任id=1的冰冰的;
-----------增加外键和删除外键---------
ALTER TABLE student ADD CONSTRAINT abc
FOREIGN KEY(charger_id)
REFERENCES classcharger(id);
ALTER TABLE student DROP FOREIGN KEY abc;
2、 INNODB支持的ON语句
--外键约束对子表的含义: 如果在父表中找不到候选键,则不允许在子表上进行insert/update
--外键约束对父表的含义: 在父表上进行update/delete以更新或删除在子表中有一条或多条对
-- 应匹配行的候选键时,父表的行为取决于：在定义子表的外键时指定的
-- on update/on delete子句
-----------------innodb支持的四种方式---------------------------------------
-----cascade方式 在父表上update/delete记录时，同步update/delete掉子表的匹配记录
-----外键的级联删除：如果父表中的记录被删除，则子表中对应的记录自动被删除--------
FOREIGN KEY (charger_id) REFERENCES ClassCharger(id)
ON DELETE CASCADE
------set null方式 在父表上update/delete记录时，将子表上匹配记录的列设为null
-- 要注意子表的外键列不能为not null
FOREIGN KEY (charger_id) REFERENCES ClassCharger(id)
ON DELETE SET NULL
------Restrict方式 :拒绝对父表进行删除更新操作(了解)
------No action方式 在mysql中同Restrict,如果子表中有匹配的记录,则不允许对父表对应候选键
-- 进行update/delete操作（了解）
二、多表查询
-- 准备两张表 -- company.employee -- company.department create table employee( emp_id int auto_increment primary key not null, emp_name varchar(50), age int, dept_id int ); insert into employee(emp_name,age,dept_id) values (\'A\',19,200), (\'B\',26,201), (\'C\',30,201), (\'D\',24,202), (\'E\',20,200), (\'F\',38,204); create table department( dept_id int, dept_name varchar(100) ); insert into department values (200,\'人事部\'), (201,\'技术部\'), (202,\'销售部\'), (203,\'财政部\'); mysql> select * from employee; +--------+----------+------+---------+ | emp_id | emp_name | age | dept_id | +--------+----------+------+---------+ | 1 | A | 19 | 200 | | 2 | B | 26 | 201 | | 3 | C | 30 | 201 | | 4 | D | 24 | 202 | | 5 | E | 20 | 200 | | 6 | F | 38 | 204 | +--------+----------+------+---------+ rows in set (0.00 sec) mysql> select * from department; +---------+-----------+ | dept_id | dept_name | +---------+-----------+ | 200 | 人事部 | | 201 | 技术部 | | 202 | 销售部 | | 203 | 财政部 | +---------+-----------+ rows in set (0.01 sec)
1、多表查询之连接查询
1.笛卡尔积查询
mysql> SELECT * FROM employee,department; -- select employee.emp_id,employee.emp_name,employee.age, -- department.dept_name from employee,department; +--------+----------+------+---------+---------+-----------+ | emp_id | emp_name | age | dept_id | dept_id | dept_name | +--------+----------+------+---------+---------+-----------+ | 1 | A | 19 | 200 | 200 | 人事部 | | 1 | A | 19 | 200 | 201 | 技术部 | | 1 | A | 19 | 200 | 202 | 销售部 | | 1 | A | 19 | 200 | 203 | 财政部 | | 2 | B | 26 | 201 | 200 | 人事部 | | 2 | B | 26 | 201 | 201 | 技术部 | | 2 | B | 26 | 201 | 202 | 销售部 | | 2 | B | 26 | 201 | 203 | 财政部 | | 3 | C | 30 | 201 | 200 | 人事部 | | 3 | C | 30 | 201 | 201 | 技术部 | | 3 | C | 30 | 201 | 202 | 销售部 | | 3 | C | 30 | 201 | 203 | 财政部 | | 4 | D | 24 | 202 | 200 | 人事部 | | 4 | D | 24 | 202 | 201 | 技术部 | | 4 | D | 24 | 202 | 202 | 销售部 | | 4 | D | 24 | 202 | 203 | 财政部 | | 5 | E | 20 | 200 | 200 | 人事部 | | 5 | E | 20 | 200 | 201 | 技术部 | | 5 | E | 20 | 200 | 202 | 销售部 | | 5 | E | 20 | 200 | 203 | 财政部 | | 6 | F | 38 | 204 | 200 | 人事部 | | 6 | F | 38 | 204 | 201 | 技术部 | | 6 | F | 38 | 204 | 202 | 销售部 | | 6 | F | 38 | 204 | 203 | 财政部 | +--------+----------+------+---------+---------+-----------+
2.内连接
-- 查询两张表中都有的关联数据,相当于利用条件从笛卡尔积结果中筛选出了正确的结果。
select * from employee,department where employee.dept_id = department.dept_id;
--select * from employee inner join department on employee.dept_id = department.dept_id;
+--------+----------+------+---------+---------+-----------+
| emp_id | emp_name | age | dept_id | dept_id | dept_name |
+--------+----------+------+---------+---------+-----------+
| 1 | A | 19 | 200 | 200 | 人事部 |
| 2 | B | 26 | 201 | 201 | 技术部 |
| 3 | C | 30 | 201 | 201 | 技术部 |
| 4 | D | 24 | 202 | 202 | 销售部 |
| 5 | E | 20 | 200 | 200 | 人事部 |
+--------+----------+------+---------+---------+-----------+
3.外连接
--（1）左外连接：在内连接的基础上增加左边有右边没有的结果
select * from employee left join department on employee.dept_id = department.dept_id;
+--------+----------+------+---------+---------+-----------+
| emp_id | emp_name | age | dept_id | dept_id | dept_name |
+--------+----------+------+---------+---------+-----------+
| 1 | A | 19 | 200 | 200 | 人事部 |
| 5 | E | 20 | 200 | 200 | 人事部 |
| 2 | B | 26 | 201 | 201 | 技术部 |
| 3 | C | 30 | 201 | 201 | 技术部 |
| 4 | D | 24 | 202 | 202 | 销售部 |
| 6 | F | 38 | 204 | NULL | NULL |
+--------+----------+------+---------+---------+-----------+
--（2）右外连接：在内连接的基础上增加右边有左边没有的结果
select * from employee RIGHT JOIN department on employee.dept_id = department.dept_id;
+--------+----------+------+---------+---------+-----------+
| emp_id | emp_name | age | dept_id | dept_id | dept_name |
+--------+----------+------+---------+---------+-----------+
| 1 | A | 19 | 200 | 200 | 人事部 |
| 2 | B | 26 | 201 | 201 | 技术部 |
| 3 | C | 30 | 201 | 201 | 技术部 |
| 4 | D | 24 | 202 | 202 | 销售部 |
| 5 | E | 20 | 200 | 200 | 人事部 |
| NULL | NULL | NULL | NULL | 203 | 财政部 |
+--------+----------+------+---------+---------+-----------+
--（3）全外连接：在内连接的基础上增加左边有右边没有的和右边有左边没有的结果
-- mysql不支持全外连接 full JOIN
-- mysql可以使用此种方式间接实现全外连接
select * from employee RIGHT JOIN department on employee.dept_id = department.dept_id
UNION
select * from employee LEFT JOIN department on employee.dept_id = department.dept_id;
+--------+----------+------+---------+---------+-----------+
| emp_id | emp_name | age | dept_id | dept_id | dept_name |
+--------+----------+------+---------+---------+-----------+
| 1 | A | 19 | 200 | 200 | 人事部 |
| 2 | B | 26 | 201 | 201 | 技术部 |
| 3 | C | 30 | 201 | 201 | 技术部 |
| 4 | D | 24 | 202 | 202 | 销售部 |
| 5 | E | 20 | 200 | 200 | 人事部 |
| NULL | NULL | NULL | NULL | 203 | 财政部 |
| 6 | F | 38 | 204 | NULL | NULL |
+--------+----------+------+---------+---------+-----------+
-- 注意 union与union all的区别：union会去掉相同的纪录
2、多表查询之复合条件连接查询
-- 查询员工年龄大于等于25岁的部门
SELECT DISTINCT department.dept_name
FROM employee,department
WHERE employee.dept_id = department.dept_id
AND age>25;
--以内连接的方式查询employee和department表，并且以age字段的升序方式显示
select employee.emp_id,employee.emp_name,employee.age,department.dept_name
from employee,department
where employee.dept_id = department.dept_id
order by age asc;
3、多表查询之子查询
-- 子查询是将一个查询语句嵌套在另一个查询语句中。
-- 内层查询语句的查询结果，可以为外层查询语句提供查询条件。
-- 子查询中可以包含：IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等关键字
-- 还可以包含比较运算符：= 、 !=、> 、<等
-- 1. 带IN关键字的子查询
---查询employee表，但dept_id必须在department表中出现过
select * from employee
where dept_id IN
(select dept_id from department);
+--------+----------+------+---------+
| emp_id | emp_name | age | dept_id |
+--------+----------+------+---------+
| 1 | A | 19 | 200 |
| 2 | B | 26 | 201 |
| 3 | C | 30 | 201 |
| 4 | D | 24 | 202 |
| 5 | E | 20 | 200 |
+--------+----------+------+---------+
5 rows in set (0.01 sec)
-- 2. 带比较运算符的子查询
-- =、!=、>、>=、<、<=、<>
-- 查询员工年龄大于等于25岁的部门
select dept_id,dept_name from department
where dept_id IN
(select DISTINCT dept_id from employee where age>=25);
-- 3. 带EXISTS关键字的子查询
-- EXISTS关字键字表示存在。在使用EXISTS关键字时，内层查询语句不返回查询的记录。
-- 而是返回一个真假值。Ture或False
-- 当返回Ture时，外层查询语句将进行查询；当返回值为False时，外层查询语句不进行查询
select * from employee
WHERE EXISTS
(SELECT dept_name from department where dept_id=203);
--department表中存在dept_id=203，Ture
select * from employee
WHERE EXISTS
(SELECT dept_name from department where dept_id=205);
-- Empty set (0.00 sec)
ps: create table t1(select * from t2);
三、课后练习
练习一：
1、将所有的课程的名称以及对应的任课老师姓名打印出来
SELECT cid,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(gender) 人数 FROM student GROUP BY gender;
3、查询物理成绩等于100的学生的姓名
SELECT sid,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);
4、查询平均成绩大于八十分的同学的姓名和平均成绩
SELECT sname,AVG(num) FROM student,score WHERE student.sid=score.student_id GROUP BY student_id HAVING AVG(num)>80;
5、查询所有学生的学号，姓名，选课数，总成绩
SELECT student_id 学号 ,sname 姓名,COUNT(course_id) 选课数,SUM(num) 总成绩 FROM student,score WHERE student.sid=student_id GROUP BY student_id;
6、查询姓李老师的个数
SELECT COUNT(tname) 姓李老师个数 from teacher WHERE tname like \'李%\';
7、查询没有报李平老师课的学生姓名
SELECT sid,sname FROM student WHERE sid NOT IN (SELECT distinct student_id from score LEFT JOIN student on student_id=student.sid WHERE course_id in (SELECT cid from course,teacher WHERE teacher_id=tid and tname=\'李平老师\'));
mysql> select sid,sname from student where sid not in (SELECT distinct student_id from score where course_id in (SELECT cid from course,teacher WHERE teacher_id=tid and 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 A.student_id FROM
(SELECT * FROM score WHERE course_id=(SELECT cid FROM course WHERE cname=\'物理\'))A
INNER JOIN
(SELECT * FROM score WHERE course_id=(SELECT cid FROM course WHERE cname=\'生物\')) B ON A.student_id = B.student_id WHERE A.num>B.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 sid,sname from student WHERE sid NOT IN (SELECT student_id from score WHERE course_id in (2,3) GROUP BY student_id HAVING COUNT(student_id)=2);
10、查询挂科超过两门(包括两门)的学生姓名和班级
SELECT sname,caption FROM student,class WHERE class_id=cid and sid IN (SELECT student_id FROM score WHERE num<60 GROUP BY student_id HAVING COUNT(student_id)>=2);
11 、查询选修了所有课程的学生姓名
SELECT sname FROM student WHERE sid in (SELECT student_id FROM score GROUP BY student_id HAVING COUNT(course_id)=(SELECT COUNT(cid) FROM course));
12、查询李平老师教的课程的所有成绩记录
SELECT sid,sname,cname,num FROM student RIGHT JOIN (SELECT student_id,cname,num FROM score,course WHERE course_id=cid AND course_id in (SELECT cid from course,teacher WHERE teacher_id=tid and tname=\'李平老师\')) ls on ls.student_id=student.sid;
13、查询选课学生都选修了的课程号和课程名
SELECT cid,cname FROM course WHERE cid =(SELECT course_id from score GROUP BY course_id HAVING COUNT(student_id)=(select count(distinct student_id) from score))
练习二：
14、查询每门课程被选修的次数
SELECT course_id,cname,COUNT(course_id) 被选课次数 FROM score,course WHERE cid=course_id GROUP BY course_id;
15、查询只选修了一门课程的学生姓名和学号
SELECT sid,sname FROM student WHERE sid =(SELECT student_id FROM score GROUP BY student_id HAVING COUNT(course_id)=1);
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 sname,AVG(num) FROM score,student WHERE student_id =student.sid GROUP BY student_id HAVING AVG(num)>85;
18、查询生物成绩不及格的学生姓名和对应生物分数
SELECT sname,num FROM score,student WHERE student_id=student.sid AND course_id=(SELECT cid FROM course WHERE cname=\'生物\') and num<60;
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 student_id,sname,AVG(num) FROM score,student WHERE student_id=student.sid AND course_id in (SELECT cid FROM course WHERE teacher_id =( SELECT tid FROM teacher WHERE tname like \'李平%\')) GROUP BY student_id ORDER BY AVG(num) DESC LIMIT 1;
20、查询每门课程成绩最好的前两名学生姓名
21、查询不同课程但成绩相同的学号，课程号，成绩
SELECT * FROM score WHERE student_id IN (SELECT student_id FROM score GROUP BY num,student_id HAVING COUNT(student_id)>=2);
22、查询没学过“李平”老师课程的学生姓名以及选修的课程名称；
SELECT * from score WHERE student_id 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 like \'李平%\')));
23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名；
SELECT sid,sname FROM student WHERE sid IN (SELECT DISTINCT student_id FROM score WHERE course_id IN (SELECT course_id from score WHERE student_id=1) AND student_id!=1);
24、任课最多的老师中学生单科成绩最高的学生姓名
SELECT sname FROM student WHERE sid IN (SELECT student_id FROM score,(SELECT course_id,MAX(num) 最大值 FROM score WHERE course_id in (SELECT cid FROM course WHERE teacher_id=(SELECT teacher_id FROM course GROUP BY teacher_id ORDER BY COUNT(cid) DESC LIMIT 1)) GROUP BY course_id HAVING max(num)) A WHERE A.最大值=score.num AND score.course_id=A.course_id);
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