最近在学习MYSQL 数据库，在此mark 一下做过的sql 相关练习

表结构如下：

teacher表

tid

tname

class表

cid

caption

course表

cid

cname

teacher_id

student表

sid

gender

class_id

sname

score表

sid

student_id

course_id

num

class :

 teacher :

 course :

student :

score :

根据以上图片建表并进行SQL 语句练习：  (本次练习使用的MySQL 版本为 8.0.21，系统为windows 10 系统)

1.自行创建测试数据库

/*Navicat Premium Data Transfer

Source Server : localhost

Source Server Type : MySQL

Source Server Version : 50624

Source Host : localhost

Source Database : sqlexam

Target Server Type : MySQL

Target Server Version : 50624

File Encoding : utf-8

Date: 2020年10月8日09:02:02*/

SETNAMES utf8;SET FOREIGN_KEY_CHECKS = 0;--------------------------------Table structure for `class`------------------------------

DROP TABLE IF EXISTS`class`;CREATE TABLE`class` (

`cid`int(11) NOT NULLAUTO_INCREMENT,

`caption`varchar(32) NOT NULL,PRIMARY KEY(`cid`)

) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;--------------------------------Records of `class`------------------------------

BEGIN;INSERT INTO `class` VALUES ('1', '三年二班'), ('2', '三年三班'), ('3', '一年二班'), ('4', '二年九班');COMMIT;--------------------------------Table structure for `course`------------------------------

DROP TABLE IF EXISTS`course`;CREATE TABLE`course` (

`cid`int(11) NOT NULLAUTO_INCREMENT,

`cname`varchar(32) NOT NULL,

`teacher_id`int(11) NOT NULL,PRIMARY KEY(`cid`),KEY`fk_course_teacher` (`teacher_id`),CONSTRAINT `fk_course_teacher` FOREIGN KEY (`teacher_id`) REFERENCES`teacher` (`tid`)

) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;--------------------------------Records of `course`------------------------------

BEGIN;INSERT INTO `course` VALUES ('1', '生物', '1'), ('2', '物理', '2'), ('3', '体育', '3'), ('4', '美术', '2');COMMIT;--------------------------------Table structure for `score`------------------------------

DROP TABLE IF EXISTS`score`;CREATE TABLE`score` (

`sid`int(11) NOT NULLAUTO_INCREMENT,

`student_id`int(11) NOT NULL,

`course_id`int(11) NOT NULL,

`num`int(11) NOT NULL,PRIMARY KEY(`sid`),KEY`fk_score_student` (`student_id`),KEY`fk_score_course` (`course_id`),CONSTRAINT `fk_score_course` FOREIGN KEY (`course_id`) REFERENCES`course` (`cid`),CONSTRAINT `fk_score_student` FOREIGN KEY (`student_id`) REFERENCES`student` (`sid`)

) ENGINE=InnoDB AUTO_INCREMENT=53 DEFAULT CHARSET=utf8;--------------------------------Records of `score`------------------------------

BEGIN;INSERT INTO `score` VALUES ('1', '1', '1', '10'), ('2', '1', '2', '9'), ('5', '1', '4', '66'), ('6', '2', '1', '8'), ('8', '2', '3', '68'), ('9', '2', '4', '99'), ('10', '3', '1', '77'), ('11', '3', '2', '66'), ('12', '3', '3', '87'), ('13', '3', '4', '99'), ('14', '4', '1', '79'), ('15', '4', '2', '11'), ('16', '4', '3', '67'), ('17', '4', '4', '100'), ('18', '5', '1', '79'), ('19', '5', '2', '11'), ('20', '5', '3', '67'), ('21', '5', '4', '100'), ('22', '6', '1', '9'), ('23', '6', '2', '100'), ('24', '6', '3', '67'), ('25', '6', '4', '100'), ('26', '7', '1', '9'), ('27', '7', '2', '100'), ('28', '7', '3', '67'), ('29', '7', '4', '88'), ('30', '8', '1', '9'), ('31', '8', '2', '100'), ('32', '8', '3', '67'), ('33', '8', '4', '88'), ('34', '9', '1', '91'), ('35', '9', '2', '88'), ('36', '9', '3', '67'), ('37', '9', '4', '22'), ('38', '10', '1', '90'), ('39', '10', '2', '77'), ('40', '10', '3', '43'), ('41', '10', '4', '87'), ('42', '11', '1', '90'), ('43', '11', '2', '77'), ('44', '11', '3', '43'), ('45', '11', '4', '87'), ('46', '12', '1', '90'), ('47', '12', '2', '77'), ('48', '12', '3', '43'), ('49', '12', '4', '87'), ('52', '13', '3', '87');COMMIT;--------------------------------Table structure for `student`------------------------------

DROP TABLE IF EXISTS`student`;CREATE TABLE`student` (

`sid`int(11) NOT NULLAUTO_INCREMENT,

`gender`char(1) NOT NULL,

`class_id`int(11) NOT NULL,

`sname`varchar(32) NOT NULL,PRIMARY KEY(`sid`),KEY`fk_class` (`class_id`),CONSTRAINT `fk_class` FOREIGN KEY (`class_id`) REFERENCES`class` (`cid`)

) ENGINE=InnoDB AUTO_INCREMENT=17 DEFAULT CHARSET=utf8;--------------------------------Records of `student`------------------------------

BEGIN;INSERT INTO `student` VALUES ('1', '男', '1', '理解'), ('2', '女', '1', '钢蛋'), ('3', '男', '1', '张三'), ('4', '男', '1', '张一'), ('5', '女', '1', '张二'), ('6', '男', '1', '张四'), ('7', '女', '2', '铁锤'), ('8', '男', '2', '李三'), ('9', '男', '2', '李一'), ('10', '女', '2', '李二'), ('11', '男', '2', '李四'), ('12', '女', '3', '如花'), ('13', '男', '3', '刘三'), ('14', '男', '3', '刘一'), ('15', '女', '3', '刘二'), ('16', '男', '3', '刘四');COMMIT;--------------------------------Table structure for `teacher`------------------------------

DROP TABLE IF EXISTS`teacher`;CREATE TABLE`teacher` (

`tid`int(11) NOT NULLAUTO_INCREMENT,

`tname`varchar(32) NOT NULL,PRIMARY KEY(`tid`)

) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;--------------------------------Records of `teacher`------------------------------

BEGIN;INSERT INTO `teacher` VALUES ('1', '张磊老师'), ('2', '李平老师'), ('3', '刘海燕老师'), ('4', '朱云海老师'), ('5', '李杰老师');COMMIT;SET FOREIGN_KEY_CHECKS = 1;

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2.查询 生物 课程比 物理 课程成绩高的所有学生的学号

--我的写法：

select A.student_id from score asAleft join score as B on A.student_id =B.student_idwhere A.course_id in (select cid from course where cname = '生物')and B.course_id in (select cid from course where cname = '物理')and A.num >B.num;--参考写法：

select A.student_id from(select score.sid, student_id,course.cname,num from score left join course on course.cid = score.course_id where course.cname = '生物') asAinner join(select score.sid,student_id,course.cname,num from score left join course on course.cid = score.course_id where course.cname = '物理') asBon A.student_id =B.student_idwhere A.num > B.num;

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3.查询平均成绩大于60分的同学的学号和平均成绩

select b.student_id,student.sname,B.avgnum from(select student_id ,AVG( num) as avgnum fromscoreGROUP BY student_id having AVG(num) > 60

ORDER BY avg(num) desc) asBLEFT JOIN student on B.student_id = student.sid ;

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4.查询所有同学的学号，姓名，选课数，总成绩

select student.sid,student.sname,Count(score.course_id),sum(score.num) fromstudentleft join score on score.student_id =student.sidGROUP BY student.sid;

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5.查询姓李的老师的个数

select '姓李的老师个数',count(1) from teacher where tname like '李%';

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6.查询没学过 叶平 老师课的同学的学号和姓名

select sa.sid,sa.sname from student assaWHERE sa.sid not in(select score.student_id fromscoreleft join student on score.student_id =student.sidleft join course on course.cid =score.course_idleft join teacher on teacher.tid =course.teacher_idwhere teacher.tname like '李平%');

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7.查询学过 001 并且也学过编号 002 课程的同学的学号和姓名

select score.student_id,student.sname fromscoreleft join student on student.sid =score.student_idwhere score.course_id = 1 or score.course_id = 2

GROUP BY score.student_id having count(score.course_id) > 1 ;

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8.查询学过 李平 老师所有课程的同学的学号和姓名

select A.student_id,student.sname from score asAleft join student on student.sid =A.student_idwhere A.course_id in(SELECT course.cid fromcourseleft join teacher on teacher.tid =course.teacher_idwhere teacher.tname like '李平%')group by A.student_id having count(A.course_id) =(SELECT count(course.cid) fromcourseleft join teacher on teacher.tid =course.teacher_idwhere teacher.tname like '李平%');

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9.查询课程编号 002 的成绩比课程编号 001 课程低的所有同学的学号和姓名

select A.student_id,student.sname from score asAinner join score as B on A.student_id =B.student_idleft join student on student.sid =A.student_idwhere A.course_id = 1 and B.course_id = 2 and A.num < B.num;

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10 查询有课程成绩小于 60 分的同学的学号和姓名

select student_id,sname fromscoreleft join student on student.sid =score.student_idwhere score.num < 60

group byscore.student_id;select sid,sname from student where sid in(select student_id from score where num < 60 GROUP BYstudent_id);select sid,sname from student where sid in(select distinct student_id from score where num < 60) ; --distinct 的效率不高，能不用就不用，用其他方法实现去重

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11.查询没有学全所有课程的同学的学号和姓名

--解答分析：--1.查出所有课程ID--2.查出所有学过所有课程的学生ID--3.从所有学生中去掉 上面查询出来的学生

--主要：这道题有一个题目理解的问题，就是"没有学全所有课程的学生"包不包括一门课没选的学生?--如果理解为包括，则是答案1的解法--如果理解为不包括，则是答案2、3 的解法

--答案一：

select sid,sname fromstudentwhere sid not in(select student_id fromscoreGROUP BY student_id having count(1) = ( select count(1) fromcourse)

);--答案二:

select student.sid,student.sname fromstudentleft JOIN score on score.student_id =student.sidGROUP BY student.sid having count(score.course_id) >= 1 and count(score.course_id) < ( select count(1) fromcourse) ;--答案三:

select student_id ,sname fromscoreleft join student on student.sid =score.student_idGROUP BY student_id HAVING count(1)

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12.查询至少有一门课与学号 001 的同学所学相同的同学的学号和姓名

--解答思路:--1.先查出001 的同学所学的课程ID--2.查score 表中student_id != 1 的学生的课程ID 在上面查出的范围内的数据

select student.sid,student.sname fromscoreleft join student on student.sid =score.student_idwhere student_id != 1 and score.course_id in (select B.course_id from score as B where B.student_id = 1)GROUP BY score.student_id;

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13.查询至少学过学号为 001 同学所选课程中任意一门课的其他同学学号和姓名

select student.sid,student.sname fromscoreleft join student on student.sid =score.student_idwhere student_id != 1 and course_id in (SELECT course_id from score where student_id = 1)GROUP BY score.student_id having count(score.course_id) >= (select count(1) from score where student_id = 1);

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14.查询 和 002 号同学学习的课程完全相同的其他同学的学号和姓名

select score.student_id,student.sname fromscoreleft join student on student.sid =score.student_idwhere student_id in(select student_id fromscorewhere student_id != 2

GROUP BY student_id having count(course_id ) = (select count(1) from score where student_id = 2))and course_id in (select course_id from score where student_id = 2)GROUP BY student_id having count(course_id) = (select count(1) from score where student_id = 2);

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15.删除学习 '叶平' 老师课的score表记录

delete fromscorewhere course_id in(select course.cid fromcourseleft join teacher on teacher.tid =course.teacher_idwhere tname like '叶平%');

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16.向score 表插入一些记录，这些记录要求符合以下条件：1.没有上过编号 002 课程的同学学号 2.插入 002 号课程的平均成绩

select sid,2,(select avg(num) from score where course_id = 2 ) fromstudentwhere sid not in (select student_id from score where course_id = 2);

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17、按平均成绩从低到高显示所有学生的 生物,物理,体育 三门的课程成绩，按如下形式显示： 学生ID,生物,物理,体育,有效课程数,有效平均分；

select A.sid as学生ID,

(SELECT num from score left JOIN course on course.cid = score.course_id where course.cname = '生物' and score.student_id = A.sid) as生物,

(SELECT num from score left JOIN course on course.cid = score.course_id where course.cname = '物理' and score.student_id = A.sid) as物理,

(SELECT num from score left JOIN course on course.cid = score.course_id where course.cname = '体育' and score.student_id = A.sid) as体育,

(SELECT count(1) from score where score.student_id =A.sidand num is not null and score.course_id in(select cid from course where cname in ('生物','物理','体育'))) as有效课程数,

(SELECT avg(num) from score where score.student_id =A.sidand num is not null and score.course_id in(select cid from course where cname in ('生物','物理','体育'))) as三门成绩有效平均分from student asAleft join score on score.student_id =A.sidGROUP BY A.sid;

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18.查询各科成绩最高分和最低分，以如下形式显示：课程ID，最高分，最低分

selectcourse.cidas课程ID ,max(score.num) as最高分,min(score.num) as最低分 .fromcourseleft JOIN score on score.course_id =course.cidGROUP BY course.cid;

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19.按各科平均成绩从低到高和及格率的百分比从高到低顺序

selectA.cidas课程ID,avg(num) as课程平均分,AVG(A.passed ) 及格率from (select course_id as cid,num,case when num >=60 then 1 else 0 end as passed from score ) asAGROUP BYA.cidORDER BY avg(num) asc ,AVG(A.passed ) desc;selectcourse_idas课程ID,avg(num) as课程平均分,AVG(case when num >=60 then 1 else 0 end ) as及格率fromscoreGROUP BYcourse_idORDER BY avg(num) asc ,AVG(case when num >=60 then 1 else 0 end ) desc;

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20.课程平均分从高到低显示(显示任课老师)

selectcourse.cidas课程ID,avg(score.num) as平均分,

teacher.tnameas授课老师fromcourseleft join score on score.course_id =course.cidleft join teacher on teacher.tid =course.teacher_idGROUP BYcourse.cidorder by avg(score.num) desc;

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21.查询个课程成绩前三名的记录(不考虑成绩并列情况)

selectA.course_id,

A.student_id,

A.numfrom score asAwhere num >= (select num from score where course_id = A.course_id ORDER BY num desc LIMIT 2,1)ORDER BY A.course_id asc ,A.num desc;SELECTscore.course_id,score.student_id,score.numfromscoreleft JOIN(selectA.course_id,

(select num from score where score.course_id = A.course_id ORDER BY num desc LIMIT 2,1) asnumfrom score asAGROUP BYcourse_id

)as B on score.course_id =B.course_idwhere score.num >=B.numORDER BY score.course_id asc ,score.num desc;

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22.查询每门课程被选修的学生数

SELECTcourse.cid ,count(score.course_id)fromcourseleft JOIN score on score.course_id =course.cidGROUP BY course.cid;

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23.查询出只选修了一门课程的全部学生的学号和姓名

select student.sid,student.sname fromscoreleft JOIN student on student.sid =score.student_idgroup by score.student_id having count(course_id) = 1;

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24.查询男生、女生人数

SELECT gender,count(1) from student group by gender;

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25.查询姓 张 的学生名单

select * from student where sname like '张%';

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26.查询同名同姓学生名单，并统计同名人数

select sname ,COUNT(1) from student GROUP BY sname ;

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27.查询每门课程的平均分，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列

select course.cid,avg( case when isnull(score.num) then 0 else score.num end) fromcourseLEFT JOIN score on score.course_id =course.cidGROUP BYcourse.cidORDER BY avg( case when isnull(score.num) then 0 else score.num END) asc ,score.course_id;

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28.查询平均成绩大于 85 的所有学生的学号和姓名

select student.sid,student.sname,avg(num) fromscoreleft JOIN student on student.sid =score.student_idGROUP BY student_id having avg(num) > 60

ORDER BY avg(num) desc;

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29 查询课程为 物理，且分数低于60 的学生的姓名和分数

SELECT course_id,student.sname,num fromscoreLEFT JOIN student on student.sid =score.student_idwhere course_id = (select cid from course where cname = "物理") and num < 60;

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30.查询课程编号为003 且课程成绩在80 分以上的学生的学号和姓名

SELECT course_id,student.sname,num fromscoreLEFT JOIN student on student.sid =score.student_idwhere course_id = 3 and num >80;

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31.求选了课程的学生人数

select count(1) from score GROUP BY student_id;

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32.查询选修 李平 老师所教授的学生中，成绩最高的学生姓名和成绩

select score.course_id,score.student_id,student.sname,score.num fromscoreleft join(select course_id ,max(num) as num from score where course_id in(select cid from course where teacher_id = (select tid FROM teacher where tname like '李平%'))group by course_id ) asAon score.course_id =A.course_idleft join student on student.sid =score.student_idwhere score.num = A.num;

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33.查询各个课程及相应的选修人数

select course.cid,course.cname,count(score.student_id) fromcourseleft JOIN score on score.course_id =course.cidGROUP BY course.cid;

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34.查询不同课程但成绩相同的学生的学号，课程号，学生成绩

--不要在意这道题的结果，因为这个查询需求根本就是个奇葩

select a1.student_id, a1.course_id,a2.student_id,a2.course_id ,a2.num from score as a1 ,score asa2where (a1.student_id = a2.student_id and a1.course_id < a2.course_id and a1.num = a2.num )or(a1.student_id!= a2.student_id and a1.course_id != a2.course_id and a1.num =a2.num)ORDER BY a1.course_id asc;

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35.查询每门课程成绩最好的前两名

selectA.course_id,

A.student_id,

A.numfrom score asAwhere num >= (select num from score where course_id = A.course_id ORDER BY num desc LIMIT 1,1)ORDER BY A.course_id asc ,A.num desc;SELECTscore.course_id,score.student_id,score.numfromscoreleft JOIN(selectA.course_id,

(select num from score where score.course_id = A.course_id ORDER BY num desc LIMIT 1,1) asnumfrom score asAGROUP BYcourse_id

)as B on score.course_id =B.course_idwhere score.num >=B.numORDER BY score.course_id asc ,score.num desc;

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36.检索至少选修两门课程的学生学号

select student_id fromscoreGROUP BY student_id having count(course_id) >=2;

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37.查询全部学生都选修的课程的课程号和课程名

select score.course_id,course.cname fromscoreleft JOIN course on course.cid =score.course_idgroup by course_id having count(student_id) = (select count(1) from student);

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38.查询没学过 叶平 老师讲授的任意一门课程的学生姓名

select sname fromstudentwhere sid not in(select student_id fromscorewhere course_id in(select cid fromcourseleft JOIN teacher on teacher.tid =course.teacher_idwhere tname like '李平%')

);

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39.查询两门以上不及格课程的学生的学号及平均成绩

select student_id,avg(num) fromscorewhere num <60

GROUP BY student_id having count(1)>=2 ;

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40.检索 004 课程分数小于60，按分数降序排列的同学学号

select student_id,num fromscorewhere course_id = 4 and num < 60

order by num desc;

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41删除 002 同学的 001 课程成绩

delete fromscorewhere student_id = 2 and course_id = 1;

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终于吐血整理完了，撒花  ✿✿ヽ(°▽°)ノ✿

这些题目参考自 武沛齐老师博客园： https://www.cnblogs.com/ wupeiqi

答案均是我解答出来的，如有错误，欢迎提出指正。转载请标明出处！