数据库：MySQL（多表的表记录的查询）（三）

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 | +--------+----------+------+---------+ 6 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 | +---------+-----------+ 4 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'))AINNER 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 FROMcourse 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);

转载于:https://www.cnblogs.com/Vee-Wang/p/7246807.html