MySQL 练习

最近在学习MYSQL 数据库，在此mark 一下做过的sql 相关练习

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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

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class :  teacher :  course :   

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 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
*/

SET NAMES utf8;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------
--  Table structure for `class`
-- ----------------------------
DROP TABLE IF EXISTS `class`;
CREATE TABLE `class` (
  `cid` int(11) NOT NULL AUTO_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 NULL AUTO_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 NULL AUTO_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 NULL AUTO_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 NULL AUTO_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;

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 as A
left join score as B on A.student_id = B.student_id
where 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') as A
inner 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') as B
on A.student_id = B.student_id
where A.num > B.num;

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  from score 
    GROUP BY student_id having AVG(num) > 60 
    ORDER BY avg(num) desc) as B
    LEFT JOIN student on B.student_id = student.sid ;

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) from student 
    left join score on score.student_id = student.sid
    GROUP BY student.sid;

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%';

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 as sa 
WHERE sa.sid not in (
select score.student_id from score 
left join student on score.student_id = student.sid 
left join course on course.cid = score.course_id
left join teacher on teacher.tid = course.teacher_id
where teacher.tname like '李平%'
);

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 from score
left join student on student.sid = score.student_id
where score.course_id = 1 or score.course_id = 2 
GROUP BY score.student_id having count(score.course_id) > 1 ;

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 as A
left join student on student.sid = A.student_id
where A.course_id in 
    (SELECT course.cid  from course 
    left join teacher on teacher.tid = course.teacher_id 
    where teacher.tname like '李平%') 
group by A.student_id having count(A.course_id) = 
            (SELECT count(course.cid)  from course 
            left join teacher on teacher.tid = course.teacher_id 
            where teacher.tname like '李平%');

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 as A
inner join score as B on A.student_id = B.student_id
left join student on student.sid = A.student_id
where A.course_id = 1 and B.course_id = 2 and A.num < B.num;

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 from score 
left join student on student.sid = score.student_id
where score.num < 60
group by score.student_id;

select sid,sname from student where sid in (
select student_id  from score where num < 60 GROUP BY student_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

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 from student 
where sid not in(
    select student_id from score 
    GROUP BY student_id having count(1) = ( select count(1) from course) 
);

-- 答案二:
select student.sid,student.sname from student 
left JOIN score on score.student_id = student.sid
GROUP BY student.sid having count(score.course_id) >= 1 and count(score.course_id) < ( select count(1) from course) ;

-- 答案三:
select student_id ,sname from score
left join student on student.sid = score.student_id 
GROUP BY student_id HAVING count(1) <(select count(1) from course);

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 from score 
left join student on student.sid = score.student_id
where 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;

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 from score
left join student on student.sid = score.student_id
where 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);

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  from score
left join student on student.sid = score.student_id
where    student_id in 
    (select student_id from score
        where 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);

15.删除学习 '叶平' 老师课的score表记录

delete from score 
where course_id in 
    (select course.cid from course
        left join teacher  on teacher.tid = course.teacher_id
        where tname like '叶平%'
    );

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 ) from student 
where sid not in (select student_id from score where course_id = 2);

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.sid 
            and 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.sid 
            and 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 as A 
left join score on score.student_id = A.sid
GROUP BY A.sid;

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

select 
    course.cid as 课程ID ,
    max(score.num) as 最高分,
    min(score.num) as 最低分  .
from course 
left JOIN score on score.course_id = course.cid
GROUP BY course.cid;

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

select
    A.cid as 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 ) as A
GROUP BY A.cid
ORDER BY avg(num) asc ,AVG(A.passed ) desc;

select
    course_id as 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
from  score 
GROUP BY course_id
ORDER BY avg(num) asc ,AVG(case when num >=60 then 1 else 0 end ) desc;

20.课程平均分从高到低显示（显示任课老师）

select 
    course.cid as 课程ID,
    avg(score.num) as  平均分,
    teacher.tname as 授课老师
from course 
left join score on score.course_id = course.cid
left join teacher on teacher.tid = course.teacher_id
GROUP BY course.cid
order by avg(score.num) desc;

21.查询个课程成绩前三名的记录（不考虑成绩并列情况）

select 
    A.course_id,
    A.student_id,
    A.num
from score as A
where 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;


SELECT score.course_id,score.student_id,score.num 
from score 
left JOIN(
        select 
            A.course_id,
            (select num from score where score.course_id = A.course_id ORDER BY num desc LIMIT 2,1) as num
        from score as A 
        GROUP BY course_id
    ) as B on score.course_id = B.course_id
where score.num >= B.num
ORDER BY score.course_id asc ,score.num desc;

22.查询每门课程被选修的学生数

SELECT
    course.cid ,
    count(score.course_id)
from course
left JOIN score on score.course_id = course.cid
GROUP BY course.cid;

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 from score 
left JOIN student on student.sid = score.student_id
group by score.student_id having count(course_id) = 1;

24.查询男生、女生人数

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

25.查询姓 张 的学生名单

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

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 ;

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) from course 
LEFT JOIN score on score.course_id = course.cid
GROUP BY course.cid
ORDER BY avg( case when isnull(score.num) then 0 else score.num END) asc ,score.course_id;

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) from score
left JOIN student on student.sid = score.student_id
GROUP BY student_id having avg(num) > 60
ORDER BY avg(num) desc;

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 from score 
LEFT JOIN student on student.sid = score.student_id
where course_id = (select cid from course where cname = "物理") and num < 60;

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 from score 
LEFT JOIN student on student.sid = score.student_id
where course_id = 3 and num >80;

31.求选了课程的学生人数

select count(1) from score GROUP BY student_id;

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 from  score
left 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 ) as A
    on score.course_id = A.course_id
left join student on student.sid = score.student_id
where score.num = A.num;

33.查询各个课程及相应的选修人数

select course.cid,course.cname,count(score.student_id) from course
left JOIN score on score.course_id = course.cid
GROUP BY course.cid;

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 as a2
where (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;

35.查询每门课程成绩最好的前两名

select 
    A.course_id,
    A.student_id,
    A.num
from score as A
where 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;

SELECT score.course_id,score.student_id,score.num 
from score 
left JOIN(
        select 
            A.course_id,
            (select num from score where score.course_id = A.course_id ORDER BY num desc LIMIT 1,1) as num
        from score as A 
        GROUP BY course_id
    ) as B on score.course_id = B.course_id
where score.num >= B.num
ORDER BY score.course_id asc ,score.num desc;

36.检索至少选修两门课程的学生学号

select student_id from score
GROUP BY student_id having count(course_id) >=2;

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 from score
left JOIN course on course.cid = score.course_id
group by course_id having count(student_id) = (select count(1) from student); 

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 from student 
where sid not in 
(    select student_id from score 
        where course_id in 
            (select cid from course 
                    left JOIN teacher on teacher.tid = course.teacher_id 
                    where tname like '李平%')
 );

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)  from score 
where num <60
GROUP BY student_id having count(1)>=2 ;

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 from score 
where course_id = 4  and num < 60
order by num desc;

41删除 002 同学的 001 课程成绩

delete from score 
where student_id = 2 and course_id = 1;

---

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！