视频教程地址：link
基本知识：
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');
SELECT * FROM Student;
--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');
select * from Course;
--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');
select * from Teacher;
--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50题
1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
select * from Course;
SELECT * FROM Score;
方法一：
select
c.*,a.s_score s01,b.s_score s02
from score a, score b,student c
where a.c_id='01'
and b.c_id = '02'
and a.s_id = b.s_id
and c.s_id = a.s_id
and a.s_score>b.s_score;
方法二：
SELECT st.* ,s1.s_score as '01_score',s2.s_score as '02_score'
FROM student AS st,score AS s1,score AS S2
WHERE st.s_id=s1.s_id
AND st.s_id=s2.s_id
AND s1.c_id='01'
AND s2.c_id='02'
AND s1.s_score>s2.s_score;
#Q1:解题方法二：
select
s.*,t.s01,t.s02
from
(SELECT
a.s_id,
max(case when a.c_id='01'then a.s_score end) s01,
max(case when a.c_id ='02' then a.s_score end )s02
from
score a
group by
a.s_id) t,Student s
where t.s01>t.s02
and t.s_id = s.s_id;
2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
#Q1:解题方法二：
select
s.*,t.s01,t.s02
from
(SELECT
a.s_id,
max(case when a.c_id='01'then a.s_score end) s01,
max(case when a.c_id ='02' then a.s_score end )s02
from
score a
group by
a.s_id) t,Student s
where t.s01<t.s02
and t.s_id = s.s_id;
3、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
select avg(s_score),a.s_id,(select s_name from student s where s.s_id=a.s_id) s_name
from score a
group by s_id
having AVG(s_score)>60;
4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
(包括有成绩的和无成绩的)
我的写法：
select avg(s_score),a.s_id,(select s_name from student s where s.s_id=a.s_id) s_name
from score a
group by s_id
having AVG(s_score)<60;
方法二：
select * from score;
//这里行不通，因为score表中没有王菊的记录，如果有则可以用下列的方法。
SELECT student.s_id AS '学生编号',student.s_name AS '学生姓名',AVG(score.s_score)
AS '平均成绩' FROM student
RIGHT OUTER JOIN score ON student.s_id=score.s_id
GROUP BY student.s_id HAVING AVG(score.s_score)<60 ;
5、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
我的写法：
select a.s_id,b.s_name,sum(a.s_score) sum_s,count(a.c_id) cnt_s
from score a
inner join student b
on a.s_id = b.s_id
group by
a.s_id;
select * from course;
SELECT * FROM student;
SELECT * FROM score;
参考答案：
SELECT student.s_id AS '学生编号',student.s_name AS '学生姓名',COUNT(score.c_id)
AS '选课总数', SUM(score.s_score) FROM student
LEFT JOIN score ON student.s_id=score.s_id
GROUP BY student.s_id,student.s_name;
6、查询"李"姓老师的数量
我的写法：
select count(*) from teacher where t_name like "%李%";
------
select * from Teacher;
SELECT COUNT(*) FROM Teacher WHERE t_name LIKE '李%';
select count(t_id) from teacher where t_name like '李%';
7、查询学过"张三"老师授课的同学的信息
我的写法：
select c.*
from course a, score b, student c, teacher d
where d.t_id=a.t_id
and a.c_id = b.c_id
and b.s_id = c.s_id
and d.t_name = '张三';
SELECT * FROM Course;
SELECT student.* FROM student
JOIN score ON student.s_id=score.s_id
WHERE score.c_id in (select course.c_id FROM Course WHERE Course.t_id IN (
SELECT Teacher.t_id FROM Teacher WHERE t_name='张三'));
8、查询没学过"张三"老师授课的同学的信息
我的写法：
select * from student where s_id not in (select b.s_id
from course a, score b, teacher d
where d.t_id=a.t_id
and a.c_id = b.c_id
and d.t_name = '张三');
为什么不可以这样？
SELECT student.* FROM student
LEFT OUTER JOIN score ON student.s_id=score.s_id
WHERE score.c_id NOT IN (SELECT Course.c_id FROM Course WHERE Course.t_id in
SELECT Teacher.t_id FROM Teacher WHERE Teacher.t_name='张三'));
方法二：
SELECT * FROM student
WHERE student.s_id not in (SELECT score.s_id FROM score
JOIN course ON score.c_id=course.c_id IN (
SELECT Teacher.t_id FROM Teacher WHERE Teacher.t_name='张三'));
9、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.
SELECT * FROM score;
我的写法：自链接
select s.*
FROM student s ,score a, score b
where a.c_id='01' and b.c_id='02'
and a.s_id = b.s_id and b.s_id = s.s_id;
方法一：
SELECT student.* FROM student
JOIN Score as s1 ON student.s_id=s1.s_id AND s1.c_id='01'
JOIN Score as s2 ON student.s_id=s2.s_id AND s2.c_id='02';
10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
我的写法：
select s.* FROM
(select s_id,
max(case when c_id='01' then s_score end) s01,
max(case when c_id='02' then s_score end) s02
from score a
group by a.s_id) t, student s
where t.s_id = s.s_id
and t.s01 is not null
and t.s02 is null;
作者的方法：
方法一：
SELECT student.*
FROM student
WHERE student.s_id in (SELECT s_id FROM score where c_id='01') AND student.s_id not in
(SELECT s_id FROM score where c_id='02');
方法二：
select a.* from
student a
where a.s_id in (select s_id from score where c_id='01' ) and a.s_id not in(select s_id from score where c_id='02')
11、查询没有学全所有课程的同学的信息
我:
SELECT a.*,count(b.c_id)
from student a
left join score b
on a.s_id = b.s_id
group by
a.s_id
having count(b.c_id) <(select count(c_id) from course);
---
SELECT student.* FROM student
left JOIN Score ON student.s_id= score.s_id
GROUP BY student.s_id HAVING count(score.c_id)<(SELECT count(*) FROM course);
方法二：
select s.* from student s
left join Score s1 on s1.s_id=s.s_id
group by s.s_id having count(s1.c_id)<(select count(*) from course)
12、查询至少有一门课与学号为"01"的同学所学相同的同学的信息
select * from score;
我写：
SELECT a.*
from student a
left JOIN score b on a.s_id = b.s_id
where c_id in (select c_id from score where s_id='01');
#01所学,一步，一步解题
select c_id from score where s_id='01';
#12
SELECT distinct a.*
from student a
left JOIN score b
on a.s_id = b.s_id
where c_id in (select c_id from score where s_id='01');
#01所学
select c_id from score where s_id='01';
方法一：
select * from student where s_id in(
select distinct a.s_id from score a where a.c_id in(select a.c_id from score a where a.s_id='01')
);
13、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
SELECT * FROM student;
SELECT * FROM score;
方法一：
SELECT student.* FROM student
WHERE student.s_id IN (SELECT score.s_id FROM score GROUP BY score.s_id
HAVING(COUNT(score.s_id)=(SELECT COUNT(score.s_id)
from score WHERE score.s_id='01')))
AND student.s_id NOT IN (SELECT score.s_id FROM score
WHERE score.c_id IN (SELECT DISTINCT score.c_id FROM score
WHERE score.c_id NOT IN (
SELECT score.c_id FROM score WHERE score.s_id='01')
)GROUP BY score.s_id
)
AND student.s_id NOT IN ('01');
--@ouyang_1993的分析:
SELECT
Student.*
FROM
Student
WHERE
s_id IN (SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(s_id) = (
#下面的语句是找到'01'同学学习的课程数
SELECT COUNT(c_id) FROM Score WHERE s_id = '01'
)
)
AND s_id NOT IN (
#下面的语句是找到学过‘01’同学没学过的课程，有哪些同学。并排除他们
SELECT s_id FROM Score
WHERE c_id IN(
#下面的语句是找到‘01’同学没学过的课程
SELECT DISTINCT c_id FROM Score
WHERE c_id NOT IN (
#下面的语句是找出‘01’同学学习的课程
SELECT c_id FROM Score WHERE s_id = '01'
)
) GROUP BY s_id
) #下面的条件是排除01同学
AND s_id NOT IN ('01')
14、查询没学过"张三"老师讲授的任一门课程的学生姓名
SELECT * FROM score;
方法一：
SELECT student.s_name FROM student
WHERE student.s_id NOT IN
(SELECT score.s_id FROM score WHERE score.c_id IN
(SELECT course.c_id FROM course WHERE course.t_id IN (
SELECT teacher.t_id FROM Teacher WHERE teacher.t_name='张三')
)
);
15、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩
SELECT *FROM score;
我的写法：
select avg(b.s_score) avg_s,a.s_id,a.s_name
from student a
left join score b
on a.s_id = b.s_id
group by a.s_id;
再查询2门以上不及格
select avg(b.s_score) avg_s,a.s_id,a.s_name
from student a
left join score b
on a.s_id = b.s_id
group by a.s_id
having
sum(case when b.s_score >= 60 then 0 else 1 end)>=2;
SELECT student.s_id,student.s_name ,AVG(score.s_score)FROM student
LEFT JOIN score ON student.s_id= score.s_id
WHERE student.s_id in (
SELECT score.s_id FROM score WHERE score.s_score<60 GROUP BY score.s_id HAVING COUNT(score.s_id)>=2)
GROUP BY student.s_id,student.s_name ;
分析：
如果同时存在where，group by，的时候的执行顺序应该是这样的：
1，首先where后面添加条件把数据进行了过滤，返回一个结果集
2，然后group by将上面返回的结果集进行分组，返回一个结果集
3，然后having将上面分组好了的结果集进行再次过滤，返回最后的结果集
4，select开始查询结果集
注意：
where:过滤表中数据的条件
group by:如何将上面过滤出的数据分组
having:对上面已经分组的数据进行过滤的条件
COUNT(常量) 和 COUNT(*)表示的是直接查询符合条件的数据库表的行数。
而COUNT(列名)表示的是查询符合条件的列的值不为NULL的行数。
最后，就是我们一直还没提到的COUNT(字段)，他的查询就比较简单粗暴了，就是进行全表扫描，然后判断指定字段的值是不是为NULL，不为NULL则累加。
相比COUNT(*)，COUNT(字段)多了一个步骤就是判断所查询的字段是否为NULL，所以他的性能要比COUNT(*)慢。
作者的方法：
select a.s_id,a.s_name,ROUND(AVG(b.s_score)) from
student a
left join score b on a.s_id = b.s_id
where a.s_id in(
select s_id from score where s_score<60 GROUP BY s_id having count(1)>=2)
GROUP BY a.s_id,a.s_name
16、检索"01"课程分数小于60，按分数降序排列的学生信息
SELECT student.* FROM student
WHERE student.s_id IN (
SELECT score.s_id FROM score WHERE score.s_score<60 AND score.c_id='01')
方法一：
SELECT student.*,score.c_id,score.s_score
FROM student,score
WHERE student.s_id=score.s_id
AND score.c_id='01'
AND score.s_score<60
ORDER BY score.s_score DESC;
方法二：
SELECT student.* ,score.c_id,score.s_score
FROM student
JOIN score ON student.s_id=score.s_id
AND score.c_id='01'
AND score.s_score<60
ORDER BY score.s_score DESC;
比较一下这里：
SELECT student.* ,score.c_id,score.s_score
FROM student
LEFT JOIN score ON student.s_id=score.s_id
AND score.c_id='01'
AND score.s_score<60
ORDER BY score.s_score DESC;
作者的写法：
select a.*,b.c_id,b.s_score from
student a,score b
where a.s_id = b.s_id and b.c_id='01' and b.s_score<60 ORDER BY b.s_score DESC;
17、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩
方法一：(有问题)
SELECT student.*,score.c_id,score.s_score,AVG(score.s_score)
FROM student,score
WHERE student.s_id=score.s_id
GROUP BY score.s_id
ORDER BY AVG(score.s_score) DESC;
改进：
SELECT score1.s_id AS 学生编号,
(SELECT score2.s_score FROM score AS score2 WHERE score1.s_id=score2.s_id AND score2.c_id='01') AS '语文' ,
(SELECT score2.s_score FROM score AS score2 WHERE score1.s_id=score2.s_id AND score2.c_id='02') AS '数学' ,
(SELECT score2.s_score FROM score AS score2 WHERE score1.s_id=score2.s_id AND score2.c_id='03') AS '英语' ,
AVG(score1.s_score) AS '平均成绩'
FROM score AS score1
GROUP BY score1.s_id
ORDER BY AVG(score1.s_score) DESC;
也可以：
SELECT score1.s_id AS 学生编号,
(SELECT s_score FROM score WHERE score1.s_id=s_id AND c_id='01') AS '语文',
(SELECT s_score FROM score WHERE score1.s_id=s_id AND c_id='02') AS '数学',
(SELECT s_score FROM score WHERE score1.s_id=s_id AND c_id='03') AS '英语',
AVG(score1.s_score) AS '平均成绩'
FROM score AS score1
GROUP BY score1.s_id
ORDER BY AVG(score1.s_score) DESC;
方法二：（有问题）
SELECT student.*,score.c_id,score.s_score,AVG(score.s_score)
FROM student
INNER JOIN score
ON student.s_id=score.s_id
GROUP BY score.s_id
ORDER BY AVG(score.s_score) DESC;
作者的写法：
select a.s_id,(select s_score from score where s_id=a.s_id and c_id='01') as 语文,
(select s_score from score where s_id=a.s_id and c_id='02') as 数学,
(select s_score from score where s_id=a.s_id and c_id='03') as 英语,
round(avg(s_score),2) as 平均分 from score a GROUP BY a.s_id ORDER BY 平均分 DESC;
18.查询各科成绩最高分、最低分和平均分：以如下形式显示：
课程ID，课程name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率
及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90
方法一：
SELECT score.c_id AS '课程ID',
course.c_name AS '课程name',
MAX(score.s_score) AS '最高分',
MIN(score.s_score) AS '最低分',
ROUND(AVG(score.s_score),3) AS '平均成绩',
ROUND(100*(SUM(CASE WHEN score.s_score>=60 THEN 1 ELSE 0 END)/SUM(CASE WHEN score.s_score THEN 1 ELSE 0 END)),3) AS '及格率',
ROUND(100*(SUM(CASE WHEN score.s_score>=70 AND score.s_score<=80 THEN 1 ELSE 0 END)/SUM(CASE WHEN score.s_score THEN 1 ELSE 0 END)),3) AS '中等率',
ROUND(100*(SUM(CASE WHEN score.s_score>=80 AND score.s_score<=90 THEN 1 ELSE 0 END)/SUM(CASE WHEN score.s_score THEN 1 ELSE 0 END)),3) AS '优良率',
ROUND(100*(SUM(CASE WHEN score.s_score>=90 THEN 1 ELSE 0 END)/SUM(CASE WHEN score.s_score THEN 1 ELSE 0 END)),3) AS '优秀率'
FROM score
LEFT JOIN course ON score.c_id=course.c_id
GROUP BY score.c_id,course.c_name;
作者的写法：
select a.c_id,b.c_name,MAX(s_score),MIN(s_score),
ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,
ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率,
ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率,
ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率
from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name
19、按各科成绩进行排序，并显示排名。（还没解决）
mysql没有rank函数
关于排名的写法如下：
https://blog.csdn.net/yanhanhui1/article/details/114532447?spm=1001.2101.3001.6661.1&utm_
medium=distribute.pc_relevant_t0.none-task-blog-2%7Edefault%7EBlogCommendFromBaidu%7Edefault-1.
highlightwordscore&depth_1-utm_source=distribute.pc_relevant_t0.none-task-blog-2%7Edefault%7EBlogCommendFromBaidu%7Edefault-1.highlightwordscore
作者的写法：
select a.s_id,a.c_id,
@i:=@i +1 as i保留排名,
@k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
) a,(select @k:=0,@i:=0,@score:=0) s
我的理解写法：(有问题，并没有把各科分出来在排名)
SELECT tt.s_id,
tt.c_id,
@i :=@i +1 AS 'i保留排名',
@k :=(case when @score =tt.s_score then @k else @i end) AS 'rank不保留排名',
@score :=tt.s_score AS score
FROM (SELECT score.s_id,score.c_id,score.s_score FROM score GROUP BY score.s_id,score.c_id,score.s_score ORDER BY score.s_score DESC) AS tt,
(SELECT @k :=0,@i :=0,@score :=0) AS paishengbiao2;
注：tt,c为派生表，且必须有别名，不能为字符串。
另一种方法：
(select * from (select
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='01') rank
FROM score t1 where t1.c_id='01'
order by t1.s_score desc) t1)
union
(select * from (select
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='02') rank
FROM score t1 where t1.c_id='02'
order by t1.s_score desc) t2)
union
(select * from (select
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='03') rank
FROM score t1 where t1.c_id='03'
order by t1.s_score desc) t3)
20、查询学生的总成绩并进行排名。
方法一：（有问题，没有显示排名，只是按总成绩从高往低排序而已）
SELECT score.s_id AS '学生编号',SUM(score.s_score) AS '总成绩'
FROM score
GROUP BY score.s_id
ORDER BY SUM(score.s_score) DESC;
方法二：通过派生表及定义临时变量来进行排名。
SELECT temp2.s_id,
@i := @i+1 AS '保留排名',
@k := (case when @score = temp2.sum1 then @k else @i end ) AS '不保留排名',
@score :=temp2.sum1 AS '总成绩'
FROM
(SELECT @i :=0,@k :=0,@score :=0) AS temp1,
(SELECT score.s_id,SUM(score.s_score) as sum1 FROM score GROUP BY score.s_id ORDER BY SUM(score.s_score) DESC) AS temp2;
方法
作者的写法：
select a.s_id,
@i:=@i+1 as i,
@k:=(case when @score=a.sum_score then @k else @i end) as rank,
@score:=a.sum_score as score
from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,
(select @k:=0,@i:=0,@score:=0)s
21、查询不同老师所教不同课程平均分从高到低显示
SELECT teacher.t_name,teacher.t_id,course.c_id,ROUND(AVG(score.s_score),2) avg_score
FROM course
left JOIN score ON score.c_id=course.c_id
left JOIN teacher ON teacher.t_id=course.t_id
GROUP BY course.c_id,course.t_id,teacher.t_name ORDER BY avg_score desc;
作者的写法：
select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course a
left join score b on a.c_id=b.c_id
left join teacher c on a.t_id=c.t_id
GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;
22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩
select * from score order by c_id,s_score desc;
方法一：
SELECT student.* ,temp1.c_id,temp1.s_score,temp1.排名
FROM (
SELECT score1.c_id,score1.s_score,score1.s_id,@i :=@i+1 as 排名 FROM score score1,
(SELECT @i :=0) as temp2 WHERE score1.c_id='01' ORDER BY score1.s_score DESC
) AS temp1
left join student on temp1.s_id=student.s_id
where temp1.排名 BETWEEN 2 AND 3
UNION
SELECT student.* ,temp3.c_id,temp3.s_score,temp3.排名
FROM (
SELECT score2.c_id,score2.s_score,score2.s_id,@j :=@j+1 as 排名 FROM score score2,
(SELECT @j :=0) as temp4 WHERE score2.c_id='02' ORDER BY score2.s_score DESC
) AS temp3
left join student on temp3.s_id=student.s_id
where temp3.排名 BETWEEN 2 AND 3
UNION
SELECT student.*,temp5.c_id,temp5.s_score,temp5.排名
FROM (
SELECT score3.s_id,score3.c_id,score3.s_score,@k :=@k+1 AS 排名 FROM score as score3,
(select @k :=0) AS temp6 where score3.c_id='03' ORDER BY score3.s_score DESC
) AS temp5
LEFT JOIN student ON student.s_id = temp5.s_id
WHERE 排名 BETWEEN 2 AND 3;
作者的写法：
select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01'
ORDER BY a.s_score DESC
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3
UNION
select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02'
ORDER BY a.s_score DESC
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3
UNION
select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03'
ORDER BY a.s_score DESC
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3;
23、统计各科成绩各分数段人数：课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比
SELECT DISTINCT f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比
FROM score a
LEFT JOIN (SELECT c_id,SUM(CASE WHEN s_score>85 AND s_score<=100 THEN 1 ELSE 0 END) AS `85-100`,
ROUND(100*(SUM(CASE WHEN s_score>85 AND s_score<=100 THEN 1 ELSE 0 END)/COUNT(*)),2) AS 百分比
FROM score GROUP BY c_id
) AS b ON a.c_id=b.c_id
LEFT JOIN (SELECT c_id,SUM(CASE WHEN s_score>70 AND s_score<=85 THEN 1 ELSE 0 END) AS `70-85`,
ROUND(100*(SUM(CASE WHEN s_score>70 AND s_score<=85 THEN 1 ELSE 0 END)/COUNT(*)),2) AS 百分比
FROM score GROUP BY c_id
) AS c on a.c_id = c.c_id
LEFT JOIN (SELECT c_id,SUM(CASE WHEN s_score>60 AND s_score<=70 THEN 1 ELSE 0 END) AS `60-70`,
ROUND(100*(SUM(CASE WHEN s_score>60 AND s_score<=70 THEN 1 ELSE 0 END)/COUNT(*)),2) AS 百分比
FROM score GROUP BY c_id
) AS d ON a.c_id=d.c_id
LEFT JOIN (SELECT c_id,SUM(CASE WHEN s_score>=0 AND s_score<=60 THEN 1 ELSE 0 END) AS `0-60`,
ROUND(100*(SUM(CASE WHEN s_score>=0 AND s_score<=60 THEN 1 ELSE 0 END)/COUNT(*)),2) AS 百分比
FROM score GROUP BY c_id
) AS e ON a.c_id = e.c_id
LEFT JOIN course AS f ON a.c_id=f.c_id;
注意：别用冒号'' 要用``
作者的写法：
select distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score a
left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,
ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)b on a.c_id=b.c_id
left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,
ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)c on a.c_id=c.c_id
left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,
ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)d on a.c_id=d.c_id
left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,
ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)e on a.c_id=e.c_id
left join course f on a.c_id = f.c_id
24、查询学生平均成绩及其名次
SELECT a.s_id,
@i :=@i+1 AS '不保留排名',
@k :=(case WHEN @avg_score=a.avg_s THEN @k else @i end) AS '不保留排名',
@avg_score :=a.avg_s AS '平均分'
FROM (SELECT s_id,ROUND(AVG(s_score),2) AS avg_s FROM score GROUP BY s_id ORDER BY avg_s DESC) AS a,
(SELECT @i :=0,@k=0,@avg_score) AS b;
作者的写法：
select a.s_id,
@i:=@i+1 as '不保留空缺排名',
@k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',
@avg_score:=avg_s as '平均分'
from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id ORDER BY avg_s DESC)a,(select @avg_score:=0,@i:=0,@k:=0)b;
25、查询各科成绩前三名的记录
1.选出b表比a表成绩大的所有组
2.选出比当前id成绩大的 小于三个的
SELECT a.s_id,a.c_id,a.s_score FROM score a
LEFT JOIN score b ON a.c_id=b.c_id AND a.s_score<b.s_score
GROUP BY a.s_id,a.c_id,a.s_score HAVING(COUNT(b.s_id))<3
ORDER BY a.c_id ASC,a.s_score DESC;
作者的写法：
select a.s_id,a.c_id,a.s_score from score a
left join score b on a.c_id = b.c_id and a.s_score<b.s_score
group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3
ORDER BY a.c_id,a.s_score DESC
26、查询每门课程被选修的学生数
SELECT c_id,count(s_id) FROM score group by c_id
select c_id,count(s_id) from score a GROUP BY c_id
27、查询出只有两门课程的全部学生的学号和姓名
方法一：
SELECT a.s_id,a.s_name FROM student a
where a.s_id in (SELECT s_id FROM score GROUP BY s_id HAVING(count(c_id))=2);
方法二：
SELECT a.s_id,a.s_name FROM student AS a
JOIN score AS b ON a.s_id=b.s_id
GROUP BY b.s_id HAVING(COUNT(b.c_id)=2);
作者的写法：
select s_id,s_name from student where s_id in(
select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2);
28、查询男生、女生人数
SELECT s_sex,count(s_sex) AS '总人数' FROM student GROUP BY s_sex;
作者的写法：
select s_sex,COUNT(s_sex) as 人数 from student GROUP BY s_sex;
29、查询名字中含有"风"字的学生信息。
SELECT a.* FROM student AS a WHERE a.s_name LIKE '%风%';
作者的写法：
select * from student where s_name like '%风%';
30、查询同名同性学生名单，并统计同名人数
select a.s_name,a.s_sex,count(*) FROM student AS a
JOIN student AS b ON a.s_id !=b.s_id AND a.s_name=b.s_name AND a.s_sex=b.s_sex
GROUP BY a.s_name,a.s_sex;
作者的写法：
select a.s_name,a.s_sex,count(*) from student a JOIN
student b on a.s_id !=b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex
GROUP BY a.s_name,a.s_sex;
31、查询1990年出生的学生名单
select student.* from student where student.s_birth LIKE '1990%';
select s_name from student where s_birth like '1990%'
32、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列
SELECT score.c_id,avg(score.s_score) AS avg_score FROM score group by score.c_id
order by avg_score DESC,score.c_id ASC;
作者的写法：
select c_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC
33、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩
select a.s_id,a.s_name,AVG(b.s_score) FROM student AS a
LEFT JOIN score AS b ON a.s_id=b.s_id
GROUP BY b.s_id HAVING avg(b.s_score)>=85;
作者的写法：
select a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a
left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85
34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
方法一：
select a.s_name,b.s_score FROM student AS a
JOIN score b ON a.s_id=b.s_id AND b.s_score<60
JOIN course c ON b.c_id=c.c_id AND c.c_name='数学';
方法二：
SELECT a.s_name,b.s_score FROM score b JOIN student AS a ON a.s_id=b.s_id
WHERE b.c_id IN (SELECT c_id FROM course WHERE c_name='数学')
AND b.s_score<60;
作者的写法：
select a.s_name,b.s_score from score b join student a on a.s_id=b.s_id where b.c_id=(
select c_id from course where c_name ='数学') and b.s_score<60
35、查询所有学生的课程及分数情况；
SELECT a.s_id,a.s_name,
SUM(CASE c.c_name WHEN '语文' THEN b.s_score ELSE 0 END) as '语文',
SUM(CASE c.c_name WHEN '数学' THEN b.s_score ELSE 0 END) as '数学',
SUM(CASE c.c_name WHEN '英语' THEN b.s_score ELSE 0 END) as '英语',
SUM(b.s_score) AS '总分'
FROM student AS a
LEFT JOIN score AS b ON a.s_id=b.s_id
LEFT JOIN course AS c ON b.c_id= c.c_id
group by a.s_id,a.s_name;
作者的写法：
select a.s_id,a.s_name,
SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文',
SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',
SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语',
SUM(b.s_score) as '总分'
from student a left join score b on a.s_id = b.s_id
left join course c on b.c_id = c.c_id
GROUP BY a.s_id,a.s_name
36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数；
方法一：
SELECT a.s_id,a.s_name,b.c_id,c.c_name,b.s_score FROM student AS a
JOIN score AS b ON a.s_id=b.s_id
JOIN course AS c ON c.c_id=b.c_id
WHERE b.s_score>70
GROUP BY a.s_id,b.c_id,c.c_id;
方法二：
SELECT a.s_id,a.s_name,b.c_id,c.c_name,b.s_score FROM student AS a
left JOIN score AS b ON a.s_id=b.s_id AND b.s_score>70
JOIN course AS c ON c.c_id=b.c_id
作者的写法：
select a.s_name,b.c_name,c.s_score from course b left join score c on b.c_id = c.c_id
left join student a on a.s_id=c.s_id where c.s_score>=70
37、查询不及格的课程
select * from score order by s_score asc;
方法一：
select a.s_id,a.s_name,b.c_id,c.c_name,b.s_score FROM student AS a
JOIN score AS b ON a.s_id=b.s_id
JOIN course AS c ON b.c_id=c.c_id
WHERE b.s_score<60;
方法二：
select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_id
where a.s_score<60
39、求每门课程的学生人数
SELECT b.c_id,count(b.c_id) FROM score as b group by b.c_id ;
作者的写法：
select count(*) from score GROUP BY c_id;
40、查询选修"张三"老师所授课程的学生中，成绩最高的学生信息及其成绩
方法一：（有点问题，如果有相同分数的学生就不能查出来了）
SELECT a.* ,b.s_score,b.c_id,c.c_name FROM student as a
JOIN score as b ON a.s_id =b.s_id
JOIN course as c ON b.c_id = c.c_id
JOIN teacher as d ON c.t_id=d.t_id
WHERE t_name='张三'
ORDER BY b.s_score DESC
LIMIT 0,1;
方法二：
第一步：查询张三老师的id;
SELECT c_id FROM course AS c,teacher AS d WHERE c.t_id=d.t_id AND t_name='张三';
第二步：查询最高分（有可能有相同的分数）
SELECT MAX(s_score) FROM score WHERE c_id='02';
第三步：查询信息;
SELECT a.*,b.s_score,b.c_id,c.c_name FROM student AS a
LEFT JOIN score AS b ON a.s_id=b.s_id
LEFT JOIN course AS c ON b.c_id=c.c_id
WHERE b.c_id=(SELECT c_id FROM course AS c,teacher AS d WHERE c.t_id=d.t_id AND d.t_name='张三')
AND b.s_score IN (SELECT MAX(s_score) FROM score WHERE c_id='02');
作者的写法：
-- 查询老师id
select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'
-- 查询最高分（可能有相同分数）
select MAX(s_score) from score where c_id='02'
-- 查询信息
select a.*,b.s_score,b.c_id,c.c_name from student a
LEFT JOIN score b on a.s_id = b.s_id
LEFT JOIN course c on b.c_id=c.c_id
where b.c_id =(select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三')
and b.s_score in (select MAX(s_score) from score where c_id='02');
41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩。
方法一：
SELECT DISTINCT b.s_id,b.c_id,b.s_score FROM score AS a
JOIN score AS b ON a.c_id !=b.c_id AND a.s_score=b.s_score;
方法二：
SELECT DISTINCT a.s_id,a.c_id,a.s_score FROM score AS a,score AS b WHERE a.c_id !=b.c_id AND a.s_score=b.s_score;
作者的写法：
select DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score
42、查询每门功成绩最好的前两名
SELECT a.s_id,a.c_id,a.s_score FROM score AS a
WHERE (SELECT COUNT(*) FROM score AS b WHERE b.c_id=a.c_id AND b.s_score>=a.s_score)<=2 ORDER BY a.c_id;
作者的写法：
select a.s_id,a.c_id,a.s_score from score a
where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id;
43、统计每门课程的学生选修人数（超过5人的课程才统计）。要求输出课程号和选修人数。
SELECT c_id AS 课程编号,count(c_id) FROM score GROUP BY c_id HAVING COUNT(c_id)>5;
作者的写法：
select c_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC
44、检索至少选修两门课程的学生学号。
SELECT s_id,count(s_id) FROM score GROUP BY s_id HAVING COUNT(s_id)>=2 ORDER BY s_id ASC,count(s_id) ASC;
作者的写法：
select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2
45、查询选修了全部课程的学生信息
SELECT student.* FROM student WHERE student.s_id IN (
SELECT s_id FROM score GROUP BY s_id HAVING COUNT(s_id)=3)
作者的写法：
select * from student where s_id in(
select s_id from score GROUP BY s_id HAVING count(s_id)=(select count(*) FROM course ));
46、查询各学生的年龄
按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一
SELECT s_id,s_name,(DATE_FORMAT(NOW(),"%Y")-DATE_FORMAT(s_birth,"%Y")-
(case WHEN DATE_FORMAT(NOW(),"%m%d")<DATE_FORMAT(s_birth,"%m%d") THEN 1 ELSE 0 END)) as age
FROM student;
作者的写法：
select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') -
(case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as age
from student;
47、查询本周过生日的学生
SELECT *FROM student WHERE WEEK(DATE_FORMAT(NOW(),"%Y%m%d"))=WEEK(DATE_FORMAT(s_birth,"%Y%m%d"));
作者的写法：
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
48、查询下周过生日的学生
SELECT * FROM student WHERE (WEEK(DATE_FORMAT(NOW(),"%Y%m%d"))+1)=WEEK(DATE_FORMAT(s_birth,"%Y%m%d"));
作者的写法：
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)
49、查询本月过生日的学生
SELECT * FROM student WHERE MONTH(DATE_FORMAT(NOW(),"%Y%m%d"))=MONTH(DATE_FORMAT(s_birth,"%Y%m%d"));
作者的写法：
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth);
50、查询下月过生日的学生
SELECT * FROM student WHERE (MONTH(DATE_FORMAT(NOW(),"%Y%m%d"))+1)=MONTH(DATE_FORMAT(s_birth,"%Y%m%d"));
作者的写法：
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)