一 介绍
本节主题
- 多表连接查询
- 复合条件连接查询
- 子查询
准备表
#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 |
+----+------------+--------+------+--------+
表department与employee
二 多表连接查询
#重点：外链接语法 SELECT 字段列表
FROM 表1 INNER|LEFT|RIGHT JOIN 表2
ON 表1.字段 = 表2.字段;
1 交叉连接：不适用任何匹配条件。生成笛卡尔积
mysql> select * from employee,department;
+----+------------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+----+------------+--------+------+--------+------+--------------+
| 1 | egon | male | 18 | 200 | 200 | 技术 |
| 1 | egon | male | 18 | 200 | 201 | 人力资源 |
| 1 | egon | male | 18 | 200 | 202 | 销售 |
| 1 | egon | male | 18 | 200 | 203 | 运营 |
| 2 | alex | female | 48 | 201 | 200 | 技术 |
| 2 | alex | female | 48 | 201 | 201 | 人力资源 |
| 2 | alex | female | 48 | 201 | 202 | 销售 |
| 2 | alex | female | 48 | 201 | 203 | 运营 |
| 3 | wupeiqi | male | 38 | 201 | 200 | 技术 |
| 3 | wupeiqi | male | 38 | 201 | 201 | 人力资源 |
| 3 | wupeiqi | male | 38 | 201 | 202 | 销售 |
| 3 | wupeiqi | male | 38 | 201 | 203 | 运营 |
| 4 | yuanhao | female | 28 | 202 | 200 | 技术 |
| 4 | yuanhao | female | 28 | 202 | 201 | 人力资源 |
| 4 | yuanhao | female | 28 | 202 | 202 | 销售 |
| 4 | yuanhao | female | 28 | 202 | 203 | 运营 |
| 5 | liwenzhou | male | 18 | 200 | 200 | 技术 |
| 5 | liwenzhou | male | 18 | 200 | 201 | 人力资源 |
| 5 | liwenzhou | male | 18 | 200 | 202 | 销售 |
| 5 | liwenzhou | male | 18 | 200 | 203 | 运营 |
| 6 | jingliyang | female | 18 | 204 | 200 | 技术 |
| 6 | jingliyang | female | 18 | 204 | 201 | 人力资源 |
| 6 | jingliyang | female | 18 | 204 | 202 | 销售 |
| 6 | jingliyang | female | 18 | 204 | 203 | 运营 |
+----+------------+--------+------+--------+------+--------------+
2 内连接：只连接匹配的行
#找两张表共有的部分，相当于利用条件从笛卡尔积结果中筛选出了正确的结果
#department没有204这个部门，因而employee表中关于204这条员工信息没有匹配出来
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id;
+----+-----------+------+--------+--------------+
| id | name | age | sex | name |
+----+-----------+------+--------+--------------+
| 1 | egon | 18 | male | 技术 |
| 2 | alex | 48 | female | 人力资源 |
| 3 | wupeiqi | 38 | male | 人力资源 |
| 4 | yuanhao | 28 | female | 销售 |
| 5 | liwenzhou | 18 | male | 技术 |
+----+-----------+------+--------+--------------+ #上述sql等同于
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;
3 外链接之左连接：优先显示左表全部记录
#以左表为准，即找出所有员工信息，当然包括没有部门的员工
#本质就是：在内连接的基础上增加左边有右边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id;
+----+------------+--------------+
| id | name | depart_name |
+----+------------+--------------+
| 1 | egon | 技术 |
| 5 | liwenzhou | 技术 |
| 2 | alex | 人力资源 |
| 3 | wupeiqi | 人力资源 |
| 4 | yuanhao | 销售 |
| 6 | jingliyang | NULL |
+----+------------+--------------+
4 外链接之右连接：优先显示右表全部记录
#以右表为准，即找出所有部门信息，包括没有员工的部门
#本质就是：在内连接的基础上增加右边有左边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id;
+------+-----------+--------------+
| id | name | depart_name |
+------+-----------+--------------+
| 1 | egon | 技术 |
| 2 | alex | 人力资源 |
| 3 | wupeiqi | 人力资源 |
| 4 | yuanhao | 销售 |
| 5 | liwenzhou | 技术 |
| NULL | NULL | 运营 |
+------+-----------+--------------+
5 全外连接：显示左右两个表全部记录
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
#注意：mysql不支持全外连接 full JOIN
#强调：mysql可以使用此种方式间接实现全外连接
select * from employee left join department on employee.dep_id = department.id
union
select * from employee right join department on employee.dep_id = department.id
;
#查看结果
+------+------------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+------+------------+--------+------+--------+------+--------------+
| 1 | egon | male | 18 | 200 | 200 | 技术 |
| 5 | liwenzhou | male | 18 | 200 | 200 | 技术 |
| 2 | alex | female | 48 | 201 | 201 | 人力资源 |
| 3 | wupeiqi | male | 38 | 201 | 201 | 人力资源 |
| 4 | yuanhao | female | 28 | 202 | 202 | 销售 |
| 6 | jingliyang | female | 18 | 204 | NULL | NULL |
| NULL | NULL | NULL | NULL | NULL | 203 | 运营 |
+------+------------+--------+------+--------+------+--------------+ #注意 union与union all的区别：union会去掉相同的纪录
三 符合条件连接查询
#示例1：以内连接的方式查询employee和department表，并且employee表中的age字段值必须大于25,即找出年龄大于25岁的员工以及员工所在的部门
select employee.name,department.name from employee inner join department
on employee.dep_id = department.id
where age > 25; #示例2：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.id,employee.name,employee.age,department.name from employee,department
where employee.dep_id = department.id
and age > 25
order by age asc;
四 子查询
#1：子查询是将一个查询语句嵌套在另一个查询语句中。
#2：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。
#3：子查询中可以包含：IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等关键字
#4：还可以包含比较运算符：= 、 !=、> 、<等
1 带IN关键字的子查询
#查询平均年龄在25岁以上的部门名
select id,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 in
(select id from department where name='技术'); #查看不足1人的部门名(子查询得到的是有人的部门id)
select name from department where id not in (select distinct dep_id from employee);
2 带比较运算符的子查询
#比较运算符：=、!=、>、>=、<、<=、<>
#查询大于所有人平均年龄的员工名与年龄
mysql> select name,age from emp where age > (select avg(age) from emp);
+---------+------+
| name | age |
+---------+------+
| alex | 48 |
| wupeiqi | 38 |
+---------+------+
2 rows in set (0.00 sec) #查询大于部门内平均年龄的员工名、年龄
select t1.name,t1.age from emp t1
inner join
(select dep_id,avg(age) avg_age from emp group by dep_id) t2
on t1.dep_id = t2.dep_id
where t1.age > t2.avg_age;
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。True或False
当返回True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
#department表中存在dept_id=203，Ture
mysql> select * from employee
-> where exists
-> (select id from department where id=200);
+----+------------+--------+------+--------+
| 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 |
+----+------------+--------+------+--------+ #department表中存在dept_id=205，False
mysql> select * from employee
-> where exists
-> (select id from department where id=204);
Empty set (0.00 sec)
练习：查询每个部门最新入职的那位员工
company.employee
员工id id int
姓名 emp_name varchar
性别 sex enum
年龄 age int
入职日期 hire_date date
岗位 post varchar
职位描述 post_comment varchar
薪水 salary double
办公室 office int
部门编号 depart_id int #创建表
create table employee(
id int not null unique auto_increment,
name varchar(20) not null,
sex enum('male','female') not null default 'male', #大部分是男的
age int(3) unsigned not null default 28,
hire_date date not null,
post varchar(50),
post_comment varchar(100),
salary double(15,2),
office int, #一个部门一个屋子
depart_id int
); #查看表结构
mysql> desc employee;
+--------------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------+-----------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(20) | NO | | NULL | |
| sex | enum('male','female') | NO | | male | |
| age | int(3) unsigned | NO | | 28 | |
| hire_date | date | NO | | NULL | |
| post | varchar(50) | YES | | NULL | |
| post_comment | varchar(100) | YES | | NULL | |
| salary | double(15,2) | YES | | NULL | |
| office | int(11) | YES | | NULL | |
| depart_id | int(11) | YES | | NULL | |
+--------------+-----------------------+------+-----+---------+----------------+ #插入记录
#三个部门：教学，销售，运营
insert into employee(name,sex,age,hire_date,post,salary,office,depart_id) values
('egon','male',18,'','老男孩驻沙河办事处外交大使',7300.33,401,1), #以下是教学部
('alex','male',78,'','teacher',1000000.31,401,1),
('wupeiqi','male',81,'','teacher',8300,401,1),
('yuanhao','male',73,'','teacher',3500,401,1),
('liwenzhou','male',28,'','teacher',2100,401,1),
('jingliyang','female',18,'','teacher',9000,401,1),
('jinxin','male',18,'','teacher',30000,401,1),
('成龙','male',48,'','teacher',10000,401,1), ('歪歪','female',48,'','sale',3000.13,402,2),#以下是销售部门
('丫丫','female',38,'','sale',2000.35,402,2),
('丁丁','female',18,'','sale',1000.37,402,2),
('星星','female',18,'','sale',3000.29,402,2),
('格格','female',28,'','sale',4000.33,402,2), ('张野','male',28,'','operation',10000.13,403,3), #以下是运营部门
('程咬金','male',18,'','operation',20000,403,3),
('程咬银','female',18,'','operation',19000,403,3),
('程咬铜','male',18,'','operation',18000,403,3),
('程咬铁','female',18,'','operation',17000,403,3)
; #ps：如果在windows系统中，插入中文字符，select的结果为空白，可以将所有字符编码统一设置成gbk
准备表和记录
SELECT
*
FROM
emp AS t1
INNER JOIN (
SELECT
post,
max(hire_date) max_date
FROM
emp
GROUP BY
post
) AS t2 ON t1.post = t2.post
WHERE
t1.hire_date = t2.max_date;
答案一（链表）
mysql> select (select t2.name from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) from emp as t1 group by post;
+---------------------------------------------------------------------------------------+
| (select t2.name from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) |
+---------------------------------------------------------------------------------------+
| 张野 |
| 格格 |
| alex |
| egon |
+---------------------------------------------------------------------------------------+
rows in set (0.00 sec) mysql> select (select t2.id from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) from emp as t1 group by post;
+-------------------------------------------------------------------------------------+
| (select t2.id from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) |
+-------------------------------------------------------------------------------------+
| 14 |
| 13 |
| 2 |
| 1 |
+-------------------------------------------------------------------------------------+
rows in set (0.00 sec) #正确答案
mysql> select t3.name,t3.post,t3.hire_date from emp as t3 where id in (select (select id from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) from emp as t1 group by post);
+--------+-----------------------------------------+------------+
| name | post | hire_date |
+--------+-----------------------------------------+------------+
| egon | 老男孩驻沙河办事处外交大使 | 2017-03-01 |
| alex | teacher | 2015-03-02 |
| 格格 | sale | 2017-01-27 |
| 张野 | operation | 2016-03-11 |
+--------+-----------------------------------------+------------+
rows in set (0.00 sec)
答案二（子查询）
答案一为正确答案，答案二中的limit 1有问题（每个部门可能有>1个为同一时间入职的新员工），我只是想用该例子来说明可以在select后使用子查询
可以基于上述方法解决：比如某网站在全国各个市都有站点，每个站点一条数据，想取每个省下最新的那一条市的网站质量信息
五 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 ('', '三年二班'), ('', '三年三班'), ('', '一年二班'), ('', '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 ('', '生物', ''), ('', '物理', ''), ('', '体育', ''), ('', '美术', '');
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 ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', '');
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 ('', '男', '', '理解'), ('', '女', '', '钢蛋'), ('', '男', '', '张三'), ('', '男', '', '张一'), ('', '女', '', '张二'), ('', '男', '', '张四'), ('', '女', '', '铁锤'), ('', '男', '', '李三'), ('', '男', '', '李一'), ('', '女', '', '李二'), ('', '男', '', '李四'), ('', '女', '', '如花'), ('', '男', '', '刘三'), ('', '男', '', '刘一'), ('', '女', '', '刘二'), ('', '男', '', '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 ('', '张磊老师'), ('', '李平老师'), ('', '刘海燕老师'), ('', '朱云海老师'), ('', '李杰老师');
COMMIT; SET FOREIGN_KEY_CHECKS = 1;
从init.sql文件中导入数据
#准备表、记录
mysql> create database db1;
mysql> use db1;
mysql> source /root/init.sql
表结构为
！！！重中之重：练习之前务必搞清楚sql逻辑查询语句的执行顺序
1、查询所有的课程的名称以及对应的任课老师姓名 2、查询学生表中男女生各有多少人 3、查询物理成绩等于100的学生的姓名 4、查询平均成绩大于八十分的同学的姓名和平均成绩 5、查询所有学生的学号，姓名，选课数，总成绩 6、 查询姓李老师的个数 7、 查询没有报李平老师课的学生姓名 8、 查询物理课程比生物课程高的学生的学号 9、 查询没有同时选修物理课程和体育课程的学生姓名 10、查询挂科超过两门(包括两门)的学生姓名和班级
、查询选修了所有课程的学生姓名 12、查询李平老师教的课程的所有成绩记录 13、查询全部学生都选修了的课程号和课程名 14、查询每门课程被选修的次数 15、查询之选修了一门课程的学生姓名和学号 16、查询所有学生考出的成绩并按从高到低排序（成绩去重） 17、查询平均成绩大于85的学生姓名和平均成绩 18、查询生物成绩不及格的学生姓名和对应生物分数 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 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 22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称； 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； 24、任课最多的老师中学生单科成绩最高的学生姓名
题目
#1、查询所有的课程的名称以及对应的任课老师姓名
SELECT
course.cname,
teacher.tname
FROM
course
INNER 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 性别,
count(1) 人数
FROM
student
GROUP BY
gender; #3、查询物理成绩等于100的学生的姓名
SELECT
student.sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
INNER JOIN course ON score.course_id = course.cid
WHERE
course.cname = '物理'
AND score.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
student.sname,
t1.avg_num
FROM
student
INNER JOIN (
SELECT
student_id,
avg(num) AS avg_num
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(注意：对于那些没有选修任何课程的学生也算在内)
SELECT
student.sid,
student.sname,
t1.course_num,
t1.total_num
FROM
student
LEFT JOIN (
SELECT
student_id,
COUNT(course_id) course_num,
sum(num) total_num
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(找出报名李平老师课程的学生，然后取反就可以)
SELECT
student.sname
FROM
student
WHERE
sid NOT IN (
SELECT DISTINCT
student_id
FROM
score
WHERE
course_id IN (
SELECT
course.cid
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid
WHERE
teacher.tname = '李平老师'
)
); #8、 查询物理课程比生物课程高的学生的学号(分别得到物理成绩表与生物成绩表，然后连表即可)
SELECT
t1.student_id
FROM
(
SELECT
student_id,
num
FROM
score
WHERE
course_id = (
SELECT
cid
FROM
course
WHERE
cname = '物理'
)
) AS t1
INNER JOIN (
SELECT
student_id,
num
FROM
score
WHERE
course_id = (
SELECT
cid
FROM
course
WHERE
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(没有同时选修指的是选修了一门的，思路是得到物理+体育课程的学生信息表，然后基于学生分组，统计count(课程)=1)
SELECT
student.sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
WHERE
cname = '物理'
OR cname = '体育'
)
GROUP BY
student_id
HAVING
COUNT(course_id) = 1
); #10、查询挂科超过两门(包括两门)的学生姓名和班级(求出<60的表，然后对学生进行分组，统计课程数目>=2)
SELECT
student.sname,
class.caption
FROM
student
INNER JOIN (
SELECT
student_id
FROM
score
WHERE
num < 60
GROUP BY
student_id
HAVING
count(course_id) >= 2
) AS t1
INNER JOIN class ON student.sid = t1.student_id
AND student.class_id = class.cid; #11、查询选修了所有课程的学生姓名(先从course表统计课程的总数，然后基于score表按照student_id分组，统计课程数据等于课程总数即可)
SELECT
student.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
*
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid
WHERE
teacher.tname = '李平老师'
); #13、查询全部学生都选修了的课程号和课程名(取所有学生数，然后基于score表的课程分组，找出count(student_id)等于学生数即可)
SELECT
cid,
cname
FROM
course
WHERE
cid IN (
SELECT
course_id
FROM
score
GROUP BY
course_id
HAVING
COUNT(student_id) = (
SELECT
COUNT(sid)
FROM
student
)
); #14、查询每门课程被选修的次数
SELECT
course_id,
COUNT(student_id)
FROM
score
GROUP BY
course_id; #15、查询之选修了一门课程的学生姓名和学号
SELECT
sid,
sname
FROM
student
WHERE
sid IN (
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,
t1.avg_num
FROM
student
INNER JOIN (
SELECT
student_id,
avg(num) 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
sname 姓名,
num 生物成绩
FROM
score
LEFT JOIN course ON score.course_id = course.cid
LEFT JOIN student ON score.student_id = student.sid
WHERE
course.cname = '生物'
AND score.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
sname
FROM
student
WHERE
sid = (
SELECT
student_id
FROM
score
WHERE
course_id IN (
SELECT
course.cid
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid
WHERE
teacher.tname = '李平老师'
)
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
#查看每门课程按照分数排序的信息，为下列查找正确与否提供依据
SELECT
*
FROM
score
ORDER BY
course_id,
num DESC; #表1：求出每门课程的课程course_id,与最高分数first_num
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id; #表2：去掉最高分，再按照课程分组，取得的最高分，就是第二高的分数second_num
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id; #将表1和表2联合到一起，得到一张表t3，包含课程course_id与该们课程的first_num与second_num
SELECT
t1.course_id,
t1.first_num,
t2.second_num
FROM
(
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t1
INNER JOIN (
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id
) AS t2 ON t1.course_id = t2.course_id; #查询前两名的学生（有可能出现并列第一或者并列第二的情况）
SELECT
score.student_id,
t3.course_id,
t3.first_num,
t3.second_num
FROM
score
INNER JOIN (
SELECT
t1.course_id,
t1.first_num,
t2.second_num
FROM
(
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t1
INNER JOIN (
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id
) AS t2 ON t1.course_id = t2.course_id
) AS t3 ON score.course_id = t3.course_id
WHERE
score.num >= t3.second_num
AND score.num <= t3.first_num; #排序后可以看的明显点
SELECT
score.student_id,
t3.course_id,
t3.first_num,
t3.second_num
FROM
score
INNER JOIN (
SELECT
t1.course_id,
t1.first_num,
t2.second_num
FROM
(
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t1
INNER JOIN (
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id
) AS t2 ON t1.course_id = t2.course_id
) AS t3 ON score.course_id = t3.course_id
WHERE
score.num >= t3.second_num
AND score.num <= t3.first_num
ORDER BY
course_id; #可以用以下命令验证上述查询的正确性
SELECT
*
FROM
score
ORDER BY
course_id,
num DESC; -- 21、查询不同课程但成绩相同的学号，课程号，成绩
-- 22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称；
-- 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；
-- 24、任课最多的老师中学生单科成绩最高的学生姓名
答案