参照已有博客表及问题，但是SQL由自己练习写出，SQL不熟悉，写出的性能可能会很低
参考链接https://blog.csdn.net/qq_41936662/article/details/80393172
表名和字段
–1.学生表
Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别
–2.课程表
Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号
–3.教师表
Teacher(t_id,t_name) –教师编号,教师姓名
–4.成绩表
Score(s_id,c_id,s_score) –学生编号,课程编号,分数
测试数据
--建表
--学生表
CREATE TABLE `Student`(
`s_id` VARCHAR(20),
`s_name` VARCHAR(20) NOT NULL DEFAULT '',
`s_birth` VARCHAR(20) NOT NULL DEFAULT '',
`s_sex` VARCHAR(10) NOT NULL DEFAULT '',
PRIMARY KEY(`s_id`)
);
--课程表
CREATE TABLE `Course`(
`c_id` VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT '',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
);
--教师表
CREATE TABLE `Teacher`(
`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(`t_id`)
);
--成绩表
CREATE TABLE `Score`(
`s_id` VARCHAR(20),
`c_id` VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
);
--插入学生表测试数据
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
--课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
--教师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
--成绩表测试数据
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);
练习题和sql语句
1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
思路：1.查出"01"课程比"02"课程成绩高的学生的临时表
SELECT a.s_id,a.c_id,a.s_score FROM score a LEFT JOIN score b ON a.s_id = b.s_id AND a.c_id='01' AND b.c_id='02' WHERE a.s_score>b.s_score
2.将临时表结合student表查询
SELECT
s1.s_id,
s.s_name,
s.s_birth,
s.s_sex,
s1.01_score,
s1.02_score
FROM
(
SELECT
a.s_id,
a.c_id,
a.s_score AS 01_score,
b.s_score AS 02_score
FROM
score a
LEFT JOIN score b ON a.s_id = b.s_id
AND a.c_id = '01'
AND b.c_id = '02'
WHERE
a.s_score > b.s_score
) AS s1
LEFT JOIN student s ON s1.s_id = s.s_id
2.查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
思路：1.查出所有同学的平均分
SELECT s_id,SUM(s_score)/COUNT(s_id) as avg FROM score GROUP BY s_id
2.查询平均成绩大于60分的同学
SELECT * FROM (SELECT s_id,SUM(s_score)/COUNT(s_id) as avg FROM score GROUP BY s_id ) as tempscore
WHERE tempscore.avg>60
3.结合student表查询
SELECT s.s_id,s.s_name,s1.avg FROM student s
RIGHT JOIN (SELECT * FROM (SELECT s_id,SUM(s_score)/COUNT(s_id) as avg FROM score GROUP BY s_id ) as tempscore
WHERE tempscore.avg>60) as s1 ON s.s_id=s1.s_id
3.查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩
思路：1.查询每个同学选课总数
SELECT s_id,COUNT(c_id) FROM score GROUP BY s_id
2.查学生编号，姓名和选择总数
SELECT student.s_id,student.s_name,b.count FROM student LEFT JOIN (SELECT s_id,COUNT(c_id) as count FROM score GROUP BY s_id) as b on b.s_id=student.s_id
3.查询所有课程总成绩
SELECT s_id,SUM(s_score) FROM score GROUP BY s_id
4.结合
SELECT a.s_id,a.s_name,a.count,b.sum FROM (SELECT student.s_id as s_id,student.s_name AS s_name,b.count as count FROM student LEFT JOIN (SELECT s_id,COUNT(c_id) as count FROM score GROUP BY s_id) as b on b.s_id=student.s_id) as a
LEFT JOIN (SELECT s_id as s_id,SUM(s_score) as sum FROM score GROUP BY s_id) AS b
ON a.s_id=b.s_id
4.查询学过"张三"老师授课的同学的信息
思路：1.查出张三老师的tid
SELECT t_id FROM teacher WHERE t_name='张三'
2.根据t_id查出张三老师所授课
SELECT c_id FROM course WHERE t_id=(SELECT t_id FROM teacher WHERE t_name='张三')
3.取score表查询此门课有成绩的同学s_id
SELECT s_id FROM score WHERE c_id=(SELECT c_id FROM course WHERE t_id=(SELECT t_id FROM teacher WHERE t_name='张三'))
4.取student表中学生的信息
SELECT * FROM student WHERE s_id IN (SELECT s_id FROM score WHERE c_id=(SELECT c_id FROM course WHERE t_id=(SELECT t_id FROM teacher WHERE t_name='张三')))
5.查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息
思路：1.学过01的学生
SELECT s_id FROM score WHERE c_id='01'
2.学过02的学生
SELECT s_id FROM score WHERE c_id='02'
3.求交集
SELECT a.s_id FROM (SELECT s_id FROM score WHERE c_id='01')AS a
JOIN (SELECT s_id FROM score WHERE c_id='02')as b ON a.s_id = b.s_id
4.查询学生信息
SELECT * FROM student WHERE s_id in (SELECT a.s_id FROM (SELECT s_id FROM score WHERE c_id='01')AS a
JOIN (SELECT s_id FROM score WHERE c_id='02')as b ON a.s_id = b.s_id)
6.查询和"01"号的同学学习的课程完全相同的其他同学的信息
思路：1.查询01号同学学的课程
SELECT c_id FROM score WHERE s_id='01'
SELECT COUNT(c_id )FROM score WHERE s_id='01'
2.查询每个同学学的总数
SELECT COUNT(s_id) FROM score GROUP BY s_id
2.查询学习01号同学所学课程的学生(所学的IN 01号同学，并且总数=01号同学的总数)
SELECT s_id FROM score WHERE c_id IN (SELECT c_id FROM score WHERE s_id='01')
SELECT s_id,count(s_id) FROM (SELECT s_id FROM score WHERE c_id IN (SELECT c_id FROM score WHERE s_id='01')) as a GROUP BY s_id
SELECT * FROM (SELECT s_id,count(s_id) FROM (SELECT s_id FROM score WHERE c_id IN (SELECT c_id FROM score WHERE s_id='01')) as a GROUP BY s_id) AS b
SELECT s_id FROM (SELECT * FROM (SELECT s_id,count(s_id) as count FROM (SELECT s_id FROM score WHERE c_id IN (SELECT c_id FROM score WHERE s_id='01')) as a GROUP BY s_id) AS b) AS c WHERE c.count=(SELECT COUNT(c_id )FROM score WHERE s_id='01')
综上：
SELECT * FROM student WHERE s_id IN (SELECT s_id FROM (SELECT * FROM (SELECT s_id,count(s_id) as count FROM (SELECT s_id FROM score WHERE c_id IN (SELECT c_id FROM score WHERE s_id='01')) as a GROUP BY s_id) AS b) AS c WHERE c.count=(SELECT COUNT(c_id )FROM score WHERE s_id='01'))啊啊啊~太菜，想不到简单的写法，先记住这个套路，慢慢熟悉吧