create database execise_new;
use execise_new;
-- 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(20) 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));
-- 插入数据
-- 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成绩高的学生的信息及学生分数
##错误做法（自己)
select st.*,t1.s_score,t2.s_score from Student as st,
(select sc.s_score ,sc.s_id from Score as sc where sc.c_id='01')as t1
,
(select sc.s_score ,sc.s_id from Score as sc where sc.c_id='02') as t2
where t1.s_score > t2.s_score ;
##正确做法
select a.*, b.s_score as 01_score, c.s_score as 02_score from
Student as a
join Score as b on a.s_id=b.s_id and b.c_id='01'
left join Score as c on a.s_id=c.s_id and c.c_id='02' or c.c_id=null
where b.s_score>c.s_score;
#2.查询"01"课程比"02"课程成绩低的学生的信息及课程分数
select a.*,b.s_score as 01_score,c.s_score as 02_score from
Student as a
join Score as b on( a.s_id=b.s_id and b.c_id='01')
left join Score as c on (a.s_id=c.s_id and c.c_id='02' or c.c_id=null)
where b.s_score<c.s_score;
#3.查询平均成绩大于等于60分的学生编号和学生姓名和平均成绩
##错误做法
select a.s_id,a.s_name,avg(b.s_score) as avgscore from
Student as a ,Score as b
group by b.s_id
having avg(b.s_score)>60;
##错误做法
select b.s_id,b.s_name,round(avg(a.s_score)) as avg_score from
Student as b,Score as a
group by b.s_id,b.s_name having round(avg(a.s_score),2)>=60;
#正确做法
select b.s_id,b.s_name,round(avg(a.s_score)) as avg_score from
Student as b
join Score as a on b.s_id=a.s_id #先连接再group by
group by b.s_id,b.s_name having round(avg(a.s_score),2)>=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)
select b.s_id,b.s_name,round(avg(a.s_score),2) as avg_score from
Student as b
left join Score as a on a.s_id=a.s_id
group by b.s_id,b.s_name having round(avg(a.s_score),2)<60
union
select a.s_id,a.s_name ,0 as avg_score from
Student as a
where a.s_id not in (select distinct s_id from Score);
##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
##自己的做法
select a.s_id,a.s_name,count(b.c_id) as count_num,sum(b.s_score) as sum_num
from Student as a
join Score as b on a.s_id=b.s_id
group by a.s_id
union
select a.s_id,a.s_name,0 as count_num,0 as sum_num from
Student as a
where a.s_id not in (select distinct s_id from Score);
#答案做法 用了left join,省了很多步骤
select a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) as sum_score from
student a
left join score b on a.s_id=b.s_id
GROUP BY a.s_id,a.s_name;
##6.查询‘李’姓老师的数量
#做对啦
select count(t_name) as count_num from
Teacher where t_name like '%李%';
##7.查询学过’张三‘老师授课的同学的信息
##自己的做法 （做对啦）
select a.* from Student as a
join Score as d on a.s_id=d.s_id
where d.c_id in
(
select c.c_id from Course as c where c.t_id in
(select b.t_id from Teacher as b where b.t_name='张三')
);
##答案做法
select a.* from
student a
join score b on a.s_id=b.s_id where b.c_id in(
select c_id from course where t_id =(
select 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
##自己做法 有重复记录
select a.* from Student as a
join Score as d on a.s_id=d.s_id
where d.c_id in
(
select c.c_id from Course as c where c.t_id in
(select b.t_id from Teacher as b where b.t_name!='张三')
);
##根据答案改变 注意not in
select * from Student c where c.s_id not in(
select a.s_id from Student as a
join Score as d on a.s_id=d.s_id
where d.c_id in
(
select c.c_id from Course as c where c.t_id in
(select b.t_id from Teacher as b where b.t_name='张三')
));
##答案做法 注意not in放在哪里
select * from
student c
where c.s_id not in(
select a.s_id from student a join score b on a.s_id=b.s_id where b.c_id in(
select c_id from course where t_id =(
select t_id from teacher where t_name = '张三')));
##9.查询学过01也学过02的课程的同学的信息
##正确答案 （自己没写出来）
select a.* from
Student as a,Score as b,Score as c
where a.s_id=b.s_id and a.s_id=c.s_id and b.c_id='01' and c.c_id='02';
##10.查询学过01但是没有学过02的同学的信息
##自己写的有小Bug
select a.* from
Student as a,Score as b,Score as c
where a.s_id=b.s_id and a.s_id=c.s_id and b.c_id='01' and c.c_id!='02';
##正确答案 注意in /not in的用法
select a.* from
Student as 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');
#根据上面的灵感 可以把第9问写成下面的形式
select a.* from
Student as a
where a.s_id in (select s_id from score where c_id='01')
and a.s_id 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
##自己的做法 （比答案更简单~~）
select a.* from
Student as a where a.s_id in(
select s_id from score group by s_id having count(c_id)<3);
##答案做法
select s.* from
student s where s.s_id in(
select s_id from score where s_id not in(
select a.s_id from score a
join score b on a.s_id = b.s_id and b.c_id='02'
join score c on a.s_id = c.s_id and c.c_id='03'
where a.c_id='01'));
##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
##自己做法
select a.* from
Student as a where a.s_id in(
select s_id from score group by s_id having count(c_id)>0);
##答案做法 注意in的用法 更general
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好的同学学习的课程完全相同的其他同学的信息
#自己的做法 并不对
select * from student where s_id =(
select distinct a.s_id from score a where a.c_id =(select a.c_id from score a where a.s_id='01')
);
#正确做法 最后的having过滤很重要
select a.* from student a where a.s_id in(
select distinct s_id from score where s_id!='01' and c_id in(select c_id from score where s_id='01')
group by s_id
having count(1)=(select count(1) from score where s_id='01'));
#14.查询没学过‘张三’老师教的任一门课程的学生姓名
#自己做法
select a.s_name from student as a
where a.s_id not in (
select distinct s_id from Score where c_id in(
select c_id from Course where t_id=
(select t_id from Teacher where t_name='张三')));
#答案做法
select a.s_name from student a where a.s_id not 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'))
group by s_id);
#15.查询两门及以上不及格课程的同学的学号姓名及平均成绩
#自己做法 (不对！不知道为什么）
select a.s_id,a.s_name,round(avg(b.s_score)) from
student as a ,Score as b
where a.s_id in(
select s_id from Score group by s_id
having s_score<60 and count(c_id)>=2);
select s_id from Score as b group by b.s_id
having count(b.s_score<60)>=2 );
#答案做法
select a.s_id,a.s_name,ROUND(AVG(b.s_score)) from
student a -- 注意这里的left join是因为avg用到了而且一定要用join
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;
#16.检索01分数小于60，按分数降序排列的学生信息
##自己的做法 没有排序
select a.* from Student as a where a.s_id in(
select s_id from Score where c_id='01' and s_score<=60)
;
##正确答案
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
## 自己做的 没有每一门的成绩
select a.*,b.c_id,b.s_score,round(avg(b.s_score),2) from
Student as a,Score as b
where a.s_id=b.s_id
group by a.s_id
order by round(avg(b.s_score),2) DESC;
##正确答案
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
##答案做法 自己不会
select a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),
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 as a left join course as 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（涉及函数先放着）
#只有单科成绩的
select a.s_id,a.c_id,
@i:=@i +1 as i_,
@k:=(case when @score=a.s_score then @k else @i end) as rank_,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score WHERE c_id='01' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s;
#20、查询学生的总成绩并进行排名（涉及函数先放着
##没运行出来 也没错啊！！！
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 a.c_id,b.c_name,round(avg(a.s_score),2) from score as a
join Course as b on a.c_id=b.c_id
group by a.c_id
order by round(avg(a.s_score),2) DESC;
##答案做法
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 涉及函数先放着
#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
##答案做法 自己反正不会23333
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 f on a.c_id = f.c_id;
#24.查询学生平均成绩及其名次
-- select声明变量时候必须:=
select a.s_id,
-- i的顺序一直变大
@i:=@i+1 as '不保留空缺排名',
-- 只有在前后两次排序值不同时才使用顺序号
@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 ROUND(AVG(s_score),2))a,
-- @k 表示最终的排名（相同值时序号相同）
-- @j 表示顺序排名
-- @avg_score上一次排序值
(select @avg_score:=0,@i:=0,@k:=0)b;
#25.查询各科成绩前第三名的记录
##正确答案解法一：
select * from score as t1
where (select count(*) from score as t2 where t1.c_id=t2.c_id and t2.s_score>t1.s_score )<=3
order by t1.c_id;
##正确答案 自己不会做 答案也没看懂 ！！！
-- 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,a.s_score DESC;
#26.查询每门课程被选修的学生数
-- 做对啦
select count(s_id) ,c_id from score 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
#我的做法 =答案做法
select a.s_id,a.s_name from Student as a
where a.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 count(s_sex),s_sex from student group by s_sex;
#29.查询名字中含有‘风’字的学生信息
select * from Student where s_name like '%风%';
#30.查询同名同姓的学生名单，并且统计同名人数
##自己的做法 少了group by
select a.s_sex,a.s_name from student as a
join student as b on a.s_name=b.s_name
and a.s_id!=b.s_id ;
##答案做法
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 * 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，按照课程编号升序排列alter
select c_id,round(avg(s_score),2) from Score
group by c_id
order by round(avg(s_score),2) DESC,c_id;
#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
#不特别声明就是自己和答案做的一样~~
select a.s_id,a.s_name ,round(avg(b.s_score),2)
from student as a
join score as b on a.s_id=b.s_id
group by b.s_id
having round(avg(b.s_score),2)>=85;
#34.查询课程名称为’数学‘，而且分数低于60的学生姓名和分数
#我的做法 --也对~
select a.s_name,b.s_score from student as a
left join score as b on b.s_id=a.s_id
left join course as c on c.c_id=b.c_id
where c.c_name='数学' and b.s_score<60;
#答案做法
select a.s_name,b.s_score from score b LEFT 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
##做错了
select a.s_id,a.s_name,b.s_score from student as a
join score as b on a.s_id=b.s_id
group by a.s_id,a.s_name;
##答案
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；
##我的做法 ：另一种思路
select a.s_name,b.c_name,c.s_score from student a
left join score as c on c.s_id=a.s_id
left join course as b on b.c_id=c.c_id
where c.s_score >70
group by a.s_name, b.c_name;
##答案做法
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.查询不及格课程
select a.s_id,b.c_name from score as a
join course as b on b.c_id=a.c_id
where a.s_score<60
group by a.s_id,b.c_name ;
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 ;
#38.查询课程编号为01且课程成绩在80分以上的学生的学号和姓名
#自己做法 也对
select a.s_id,a.s_name from student as a
where a.s_id in (
select s_id from score where c_id='01' and s_score>80);
#答案做法
select a.s_id,b.s_name from score a LEFT JOIN student b on a.s_id = b.s_id
where a.c_id = '01' and a.s_score>80;
#39.求每门课程的学生人数
select count(s_id),c_id from score
group by c_id;
#40.查询选走张三老师所教课程的学生中，成绩最高的学生信息
##自己做的 有bug
select a.* from student a
where a.s_id in(
select b.s_id from score as b where b.score in (select max(s_score) from score where c_id='02') and b.c_id =(
select c_id from course where t_id=(
select t_id from teacher where t_name='张三')) );
##答案做法
-- 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
##一开始自己没有加distinct
select distinct a.s_id,a.c_id,a.s_score from score as a
join score b on a.s_score=b.s_score and a.c_id!=b.c_id;
#42.查询每门成绩最好的前两名
##不会看答案
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，查询结果按人数降序排列，若人数相同，按课程号升序排列
#自己写的 也对
select c_id,count(s_id) from score
group by c_id
having count(s_id)>5
order by count(s_id),c_id;
#答案
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
#me
select s_id ,count(c_id)from score
group by s_id
having count(c_id)>=2;
#key
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
#me
select a.* from student as a where a.s_id in(
select b.s_id from score as b group by b.s_id having count(b.c_id)=3
);
#key(select count(*) from course)
select a.* from student as a where a.s_id in(
select b.s_id from score as b group by b.s_id
having count(b.c_id)=(select count(*) from course)
);
#46.查询各学生的年龄
#不会 看的答案
select s_birth,(
-- -- 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一
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
#自己写的 不对。。。。
select a.* from student as a
where (date_format(now(),'%m%d')-date_format(a.s_birth,'%m%d'))<=7;
#key
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(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(s_birth);
#50.查询下个月过生日的学生
select * from student where month(date_format(now(),'%Y%m%d'))+1=month(s_birth);MySQL练习题50
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