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【MySQL】练习题(带数据库和答案)

toyiye 2024-06-21 12:34 9 浏览 0 评论

数据库文件

将文本粘贴到txt文件后，将后缀名改为sql可以（SQLyog）直接导入

/*
SQLyog Ultimate v12.3.1 (64 bit)
MySQL - 5.5.27 : Database - lianxi
*********************************************************************
*/
/*!40101 SET NAMES utf8 */;
/*!40101 SET SQL_MODE=''*/;
/*!40014 SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0 */;
/*!40014 SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0 */;
/*!40101 SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='NO_AUTO_VALUE_ON_ZERO' */;
/*!40111 SET @OLD_SQL_NOTES=@@SQL_NOTES, SQL_NOTES=0 */;
CREATE DATABASE /*!32312 IF NOT EXISTS*/`lianxi` /*!40100 DEFAULT CHARACTER SET utf8 */;
USE `lianxi`;
/*Table structure for table `course` */
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
 `c_id` varchar(20) NOT NULL DEFAULT '',
 `c_name` varchar(20) NOT NULL DEFAULT '',
 `t_id` varchar(20) NOT NULL,
 PRIMARY KEY (`c_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
/*Data for the table `course` */
insert into `course`(`c_id`,`c_name`,`t_id`) values
('01','语文','02'),
('02','数学','01'),
('03','英语','03');
/*Table structure for table `score` */
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score` (
 `s_id` varchar(20) NOT NULL DEFAULT '',
 `c_id` varchar(20) NOT NULL DEFAULT '',
 `s_score` int(3) DEFAULT NULL,
 PRIMARY KEY (`s_id`,`c_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
/*Data for the table `score` */
insert into `score`(`s_id`,`c_id`,`s_score`) values
('01','01',80),
('01','02',90),
('01','03',99),
('02','01',70),
('02','02',60),
('02','03',80),
('03','01',80),
('03','02',80),
('03','03',80),
('04','01',50),
('04','02',30),
('04','03',20),
('05','01',76),
('05','02',87),
('06','01',31),
('06','03',34),
('07','02',89),
('07','03',98);
/*Table structure for table `student` */
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (
 `s_id` varchar(20) NOT NULL DEFAULT '',
 `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`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
/*Data for the table `student` */
insert into `student`(`s_id`,`s_name`,`s_birth`,`s_sex`) values
('01','赵雷','1990-01-01','男'),
('02','钱电','1990-12-21','男'),
('03','孙风','1990-05-20','男'),
('04','李云','1990-08-06','男'),
('05','周梅','1991-12-01','女'),
('06','吴兰','1992-03-01','女'),
('07','郑竹','1989-07-01','女'),
('08','王菊','1990-01-20','女');
/*Table structure for table `teacher` */
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (
 `t_id` varchar(20) NOT NULL DEFAULT '',
 `t_name` varchar(20) NOT NULL DEFAULT '',
 PRIMARY KEY (`t_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
/*Data for the table `teacher` */
insert into `teacher`(`t_id`,`t_name`) values
('01','张三'),
('02','李四'),
('03','王五');
/*!40101 SET SQL_MODE=@OLD_SQL_MODE */;
/*!40014 SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS */;
/*!40014 SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS */;
/*!40111 SET SQL_NOTES=@OLD_SQL_NOTES */;

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

select a.* ,b.s_score as 01_score,c.s_score as 02_score from

student a

join score b on a.s_id=b.s_id and b.c_id='01'

left join score 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

--也可以这样写

select a.*,b.s_score as 01_score,c.s_score as 02_score from student a,score b,score 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'

and 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 a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL

join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.s_score

-- 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 b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from

student b

join score a on b.s_id = a.s_id

GROUP BY b.s_id,b.s_name HAVING avg_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)

select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from

student b

left join score a on b.s_id = a.s_id

GROUP BY b.s_id,b.s_name HAVING avg_score <60

union

select a.s_id,a.s_name,0 as avg_score from

student 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

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

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

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

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 a.c_id from course a join teacher b on a.t_id = b.t_id where 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 a.* from

student a,score b,score 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、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息

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

--@wendiepei的写法

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)

--@k1051785839的写法

select *

from student

where s_id not in(

select s_id from score t1

group by s_id having count(*) =(select count(distinct c_id) 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 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

--@ouyang_1993的写法

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')

--@k1051785839的写法

SELECT

t3.*

FROM

(

SELECT

s_id,

group_concat(c_id ORDER BY c_id) group1

FROM

score

WHERE

s_id <> '01'

GROUP BY

s_id

) t1

INNER JOIN (

SELECT

group_concat(c_id ORDER BY c_id) group2

FROM

score

WHERE

s_id = '01'

GROUP BY

s_id

) t2 ON t1.group1 = t2.group2

INNER JOIN student t3 ON t1.s_id = t3.s_id

-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名

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')));

-- 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 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 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

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;

--@喝完这杯还有一箱的写法

SELECT a.s_id,MAX(CASE a.c_id WHEN '01' THEN a.s_score END ) 语文,

MAX(CASE a.c_id WHEN '02' THEN a.s_score END ) 数学,

MAX(CASE a.c_id WHEN '03' THEN a.s_score END ) 英语,

avg(a.s_score),b.s_name FROM Score a JOIN Student b ON a.s_id=b.s_id GROUP BY a.s_id ORDER BY 5 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

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 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

-- mysql没有rank函数

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

--@k1051785839的写法

(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

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.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 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)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,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 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

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 人数 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、查询同名同性学生名单，并统计同名人数

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 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 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,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

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 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；

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 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；

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(*) 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

-- 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

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

-- 牛逼的写法

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，查询结果按人数降序排列，若人数相同，按课程号升序排列

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(*) 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 * from student where s_id in(

select s_id from score GROUP BY s_id HAVING count(*)=(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_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(s_birth)

select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

-- 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、查询下月过生日的学生

select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)

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