分享50道mysq练习题l

分享50道mysq练习题l

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_score60;

##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=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);