50道MySQL面试题，经典~

#来源公众号：【码农编程进阶笔记】
-- ----------------------------
-- Table structure for teacher
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher`  (
  `t_id` varchar() CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
  `t_name` varchar() CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  PRIMARY KEY (`t_id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Dynamic;

-- ----------------------------
-- Records of teacher
-- ----------------------------
INSERT INTO `teacher` VALUES ('01', '数学老师-杰斯');
INSERT INTO `teacher` VALUES ('02', '语文老师-盲僧');
INSERT INTO `teacher` VALUES ('03', '英语老师-菲欧娜');

-- ----------------------------
-- Table structure for student
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student`  (
  `s_id` varchar() CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
  `s_name` varchar() CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  `s_birth` varchar() CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  `s_sex` varchar() CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  PRIMARY KEY (`s_id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Dynamic;

-- ----------------------------
-- Records of student
-- ----------------------------
INSERT INTO `student` VALUES ('01', '流浪法师-瑞兹', '1990-01-01', '男');
INSERT INTO `student` VALUES ('02', '探险家-EZ', '1990-12-21', '男');
INSERT INTO `student` VALUES ('03', '疾风剑豪-亚瑟', '1990-05-20', '男');
INSERT INTO `student` VALUES ('04', '迅捷斥候-提莫', '1990-08-06', '男');
INSERT INTO `student` VALUES ('05', '黑暗之女-安妮', '1991-12-01', '女');
INSERT INTO `student` VALUES ('06', '战争女神-希维尔', '1992-03-01', '女');
INSERT INTO `student` VALUES ('07', '光辉女郎-拉克丝', '1989-07-01', '女');
INSERT INTO `student` VALUES ('08', '放逐之刃-锐雯', '1990-01-20', '女');

-- ----------------------------
-- Table structure for score
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score`  (
  `s_id` varchar() CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
  `c_id` varchar() CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
  `s_score` int() NULL DEFAULT NULL,
  PRIMARY KEY (`s_id`, `c_id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Dynamic;

-- ----------------------------
-- Records of score
-- ----------------------------
INSERT INTO `score` VALUES ('01', '01', );
INSERT INTO `score` VALUES ('01', '02', );
INSERT INTO `score` VALUES ('01', '03', );
INSERT INTO `score` VALUES ('02', '01', );
INSERT INTO `score` VALUES ('02', '02', );
INSERT INTO `score` VALUES ('02', '03', );
INSERT INTO `score` VALUES ('03', '01', );
INSERT INTO `score` VALUES ('03', '02', );
INSERT INTO `score` VALUES ('03', '03', );
INSERT INTO `score` VALUES ('04', '01', );
INSERT INTO `score` VALUES ('04', '02', );
INSERT INTO `score` VALUES ('04', '03', );
INSERT INTO `score` VALUES ('05', '01', );
INSERT INTO `score` VALUES ('05', '02', );
INSERT INTO `score` VALUES ('06', '01', );
INSERT INTO `score` VALUES ('06', '03', );
INSERT INTO `score` VALUES ('07', '02', );
INSERT INTO `score` VALUES ('07', '03', );

-- ----------------------------
-- Table structure for course
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course`  (
  `c_id` varchar() CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
  `c_name` varchar() CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  `t_id` varchar() CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
  PRIMARY KEY (`c_id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Dynamic;

-- ----------------------------
-- Records of course
-- ----------------------------
INSERT INTO `course` VALUES ('01', '语文', '02');
INSERT INTO `course` VALUES ('02', '数学', '01');
INSERT INTO `course` VALUES ('03', '英语', '03');

#来源公众号：【码农编程进阶笔记】
-- 50道SQL面试题
-- 1、查询课程编号为“01”的课程比“02”的课程成绩高的所有学生的学号（难）

-- 2、查询平均成绩大于60分的学生的学号和平均成绩

-- 3、查询所有学生的学号、姓名、选课数、总成绩

-- 4、查询姓“猴”的老师的个数

-- 5、查询没学过“数学老师-杰斯”老师课的学生的学号、姓名

-- 6、查询学过“数学老师-杰斯”老师所教的所有课的同学的学号、姓名

-- 7、查询学过编号为“01”的课程并且也学过编号为“02”的课程的学生的学号、姓名

-- 8、查询课程编号为“02”的总成绩

-- 9、查询所有，课程成绩小于60分的学生的学号、姓名

-- 10、查询没有学全所有课的学生的学号、姓名

-- 11、查询至少有一门课与学号为“01”的学生所学课程相同的学生的学号和姓名 （难）

-- 12、查询和“01”号同学所学课程完全相同的其他同学的学号（难）

-- 13、查询没学过"数学老师-杰斯"老师讲授的任一门课程的学生姓名

-- 14、空

-- 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

-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息

-- 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(难)

-- 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

-- 19、查询学生的总成绩并进行排名

-- 20、查询不同老师所教不同课程平均分,从高到低显示

-- 21、查询学生平均成绩及其名次

-- 22、按各科成绩进行排序，并显示排名(难)

-- 23、查询每门功课成绩最好的前两名学生姓名

-- 24、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

-- 25、查询各科成绩前三名的记录（不考虑成绩并列情况）

-- 26、使用分段[100-85],[85-70],[70-60],[<60]来统计各科成绩，分别统计各分数段人数：课程ID和课程名称

-- 27、查询每门课程被选修的学生数

-- 28、查询出只有两门课程的全部学生的学号和姓名

-- 29、查询男生、女生人数

-- 30、查询名字中含有"风"字的学生信息

-- 31、查询1990年出生的学生名单

-- 32、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

-- 33、查询每门课程的平均成绩，结果按平均成绩升序排序，平均成绩相同时，按课程号降序排列

-- 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数

-- 35、查询所有学生的课程及分数情况

-- 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

-- 37、查询不及格的课程并按课程号从大到小排列

-- 38、查询课程编号为03且课程成绩在80分以上的学生的学号和姓名

-- 39、求每门课程的学生人数

-- 40、查询选修“数学老师-杰斯”老师所授课程的学生中成绩最高的学生姓名及其成绩

-- 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 （难）

-- 42、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

-- 43、检索至少选修两门课程的学生学号

-- 44、查询选修了全部课程的学生信息

-- 45、查询各学生的年龄

-- 46、查询两门以上不及格课程的同学的学号及其平均成绩

-- 47、查询本月过生日的学生

-- 48、查询下一个月过生日的学生

-- 先全数据关联一下看看
select * from student stu,score sc,course c,teacher t 
where stu.s_id=sc.s_id and sc.c_id =c.c_id and c.t_id = t.t_id;

#来源公众号：【码农编程进阶笔记】

SELECT a.s_id from 
(select * from score sc1 where sc1.c_id = '01') a , 
(select * from score sc2 where sc2.c_id = '02') b 
where a.s_id = b.s_id and a.s_score > b.s_score;

#来源公众号：【码农编程进阶笔记】

select a.s_id,a.avg_score from 
(select stu.s_id,stu.s_name,AVG(sc.s_score) as avg_score from student stu, score sc where stu.s_id = sc.s_id GROUP BY stu.s_id) a 
where a.avg_score >  ORDER BY avg_score desc;  

#来源公众号：【码农编程进阶笔记】

-- 我这里MySQL8执行sql时是出现了sql_mode=only_full_group_by错误，执行下行sql进行配置即可解决；
-- set @@global.sql_mode = 'STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_ENGINE_SUBSTITUTION'
select stu.s_id,stu.s_name,count(sc.c_id),IFNULL(SUM(sc.s_score),) from student stu 
LEFT JOIN score sc on stu.s_id = sc.s_id GROUP BY stu.s_id;

select count(*) from teacher where t_name like '猴%';

SELECT * from student a where a.s_id not in
 (select stu.s_id from student stu,score sc,course c,teacher t 
where stu.s_id=sc.s_id and sc.c_id =c.c_id and c.t_id = t.t_id and t.t_name ='数学老师-杰斯')

select stu.s_id,stu.s_name from 
student stu,score sc,course c,teacher t 
where stu.s_id=sc.s_id and sc.c_id =c.c_id and c.t_id = t.t_id and t.t_name ='数学老师-杰斯';

SELECT a.s_id,a.s_name from 
(SELECT stu1.s_id,stu1.s_name from student stu1, score sc1 where stu1.s_id = sc1.s_id and sc1.c_id ='01') a,  
(SELECT stu2.s_id,stu2.s_name from student stu2, score sc2 where stu2.s_id = sc2.s_id and sc2.c_id ='02') b
where a.s_id = b.s_id;

select SUM(sc.s_score) from score sc where c_id = '02';

select stu.s_id,stu.s_name from student stu,score sc where stu.s_id = sc.s_id and sc.s_score <  GROUP BY stu.s_id;

SELECT stu2.s_id,stu2.s_name from student stu2 ,
(SELECT stu.s_id,stu.s_name,IFNULL(COUNT(c.c_id),) as c_count from student stu 
LEFT JOIN score sc on stu.s_id = sc.s_id JOIN course c on sc.c_id = c.c_id GROUP BY stu.s_id) a
where stu2.s_id = a.s_id and a.c_count = (SELECT count(*) from course);

SELECT a.s_id,a.s_name from 
(SELECT stu1.s_id,stu1.s_name,sc1.c_id from student stu1,score sc1 where stu1.s_id = sc1.s_id) a,
(SELECT sc2.c_id from student stu2,score sc2 where stu2.s_id = sc2.s_id and stu2.s_id = '01') b
where a.c_id = b.c_id GROUP BY a.s_id;

select s_id,s_name from student 
where s_id in (
select s_id from score
where s_id <> '01'
group by s_id
having count(*) = (select count(*) from score where s_id = '01'))
and s_id not in(select distinct s_id from score where c_id not in (select c_id from score where s_id = '01'))

SELECT s_name from student stu2 where stu2.s_id not in
(select DISTINCT stu.s_id from student stu,score sc,course c,teacher t 
where stu.s_id=sc.s_id and sc.c_id =c.c_id and c.t_id = t.t_id and t.t_name = '数学老师-杰斯') ;

SELECT a.s_id, a.s_name, avg from student stu2,
(SELECT stu.s_id, stu.s_name, count(*) as count, AVG(sc.s_score) as avg from student stu, score sc 
where stu.s_id = sc.s_id and sc.s_score <  GROUP BY stu.s_id) a where stu2.s_id = a.s_id and a.count >= ;

SELECT stu.s_id, stu.s_name,sc.s_score from student stu, score sc where stu.s_id = sc.s_id and sc.c_id = '01' and sc.s_score <  ORDER BY sc.s_score desc;

-- 全量展示
SELECT * from 
(SELECT stu.s_id,stu.s_name,sc.c_id,IFNULL(sc.s_score,) from student stu LEFT JOIN score sc on stu.s_id = sc.s_id ) a,
(SELECT sc1.s_id,AVG(sc1.s_score) as avg from score sc1 GROUP BY sc1.s_id) b
where a.s_id = b.s_id ORDER BY avg desc;

-- 横排展示
select s_id,s_name,
(select s_score from score where score.s_id=student.s_id and c_id='01') as '语文',
(select s_score from score where score.s_id=student.s_id and c_id='02') as '数学',
(select s_score from score where score.s_id=student.s_id and c_id='03') as '英语',
(select avg(s_score) from score where score.s_id=student.s_id) avg_score
from student 
order by avg_score desc;

SELECT 
c.c_id,c.c_name,MAX(sc.s_score),MIN(sc.s_score),AVG(sc.s_score),CONCAT(round( * sum(case when sc.s_score>= then 1.0 else 0.0 end)/count(*),),'%') as '及格率' 
from score sc, course c 
where sc.c_id = c.c_id 
GROUP BY c.c_id;

SELECT 
stu.s_id,stu.s_name,IFNULL(SUM(sc.s_score),) as totalCount 
from student stu LEFT JOIN score sc on stu.s_id = sc.s_id GROUP BY stu.s_id 
ORDER BY totalCount desc

SELECT
t.t_name,c.c_name,AVG(sc.s_score) as avg 
from score sc, course c, teacher t where sc.c_id = c.c_id and c.t_id = t.t_id
GROUP BY t.t_id 
ORDER BY avg desc;

SELECT 
stu.s_id,stu.s_name,AVG(sc.s_score) as avg,rank() over(ORDER BY AVG(sc.s_score) desc) as `no` 
from student stu 
LEFT JOIN score sc on stu.s_id = sc.s_id 
GROUP BY stu.s_id 
ORDER BY avg desc;

#来源公众号：【码农编程进阶笔记】

select 
c_id,s_id,s_score,rank() over(partition by c_id 
order by s_score desc) num_row 
from score;

#来源公众号：【码农编程进阶笔记】

select a.s_id,a.s_name,b.num_row 
from student a 
join (select s_id, c_id, row_number() over(partition by c_id order by s_score desc) num_row from score) b on a.s_id=b.s_id 
where b.num_row<=;

#来源公众号：【码农编程进阶笔记】

select a.s_id,a.s_name,b.c_id,b.s_score,b.num_row 
from student a 
,(select 
      s_id,
      c_id,
      s_score,
      row_number() over(partition by c_id order by s_score desc) num_row from score) b 
where a.s_id=b.s_id 
and b.num_row between  and ;

#来源公众号：【码农编程进阶笔记】

select a.s_id,a.s_name,b.c_id,b.s_score,b.num_row 
from student a 
,(select 
      s_id,
      c_id,
      s_score,
      row_number() over(partition by c_id order by s_score desc) num_row from score) b 
where a.s_id=b.s_id 
and b.num_row between  and ;

#来源公众号：【码农编程进阶笔记】
SELECT c.c_id,c.c_name,
(select count(*) from score sc1 where sc1.c_id=c.c_id and sc1.s_score >= ) as '[100-85]',
(select count(*) from score sc2 where sc2.c_id=c.c_id and sc2.s_score BETWEEN  and ) as '[85-70]',
(select count(*) from score sc3 where sc3.c_id=c.c_id and sc3.s_score BETWEEN  and ) as '[70-60]',
(select count(*) from score sc4 where sc4.c_id=c.c_id and sc4.s_score < ) as '[<60]'
from course c 
GROUP BY c.c_id;

select c_id,count(c_id) count from score group by c_id;

#来源公众号：【码农编程进阶笔记】

select a.s_id,a.s_name,count(b.s_id) count 
from student a 
join score b on a.s_id=b.s_id 
group by s_id 
having count=;

select s_sex,count(s_sex) count_sex from student group by s_sex;

select * from student where s_name like '%风%';

select s_id,s_name from student where year(s_birth)='1990';

select a.s_id,a.s_name,avg(b.s_score) avg
from student a 
join score b on a.s_id=b.s_id 
group by b.s_id 
having avg>=;

select 
c_id, avg(s_score) avg_score 
from score 
group by c_id 
order by avg_score,c_id desc;

select sc.c_id,c.c_name,stu.s_id,stu.s_name,sc.s_score
from score sc 
join course c on sc.c_id=c.c_id 
join student stu on sc.s_id=stu.s_id 
where sc.s_score< and c.c_name='数学';

select stu.s_id,stu.s_name,c.c_id,c.c_name,sc.s_score 
from student stu 
left join score sc on stu.s_id=sc.s_id 
left join course c on sc.c_id=c.c_id ;

select c.s_name,b.c_name,a.s_score 
from score a 
join course b on a.c_id=b.c_id 
join student c on a.s_id=c.s_id 
where a.s_score>;

select c_id from score where s_score <  GROUP BY c_id order by c_id desc;

select stu.s_id,stu.s_name 
from student stu, score sc where stu.s_id=sc.s_id 
and sc.s_score> and sc.c_id=;

select c_id, count(*) count from score group by c_id;

select stu.s_id,stu.s_name,sc.s_score from
student stu,score sc,course c,teacher t where stu.s_id=sc.s_id and sc.c_id =c.c_id and c.t_id = t.t_id 
and t.t_name='数学老师-杰斯' 
order by sc.s_score desc 
limit ;

select a.s_id,a.c_id,a.s_score 
from score a, score b where a.s_score=b.s_score 
and a.c_id <> b.c_id and a.s_id=b.s_id 
group by a.s_id,a.c_id,a.s_score;

select 
c_id, count(*) count 
from score 
group by c_id 
order by c_id asc;

select s_id,count(*) count from score group by s_id having count>=;

select 
stu.s_id,stu.s_name,count(*) count 
from student stu,score sc 
where stu.s_id = sc.s_id
group by sc.s_id 
having count = ;

SELECT * ,(YEAR(CURDATE()) - YEAR(s_birth)) AS age from student; 

select s_id,avg(s_score) avg
from score 
where s_score <  
group by s_id 
having count(*) >= ;

select s_id from student where month(s_birth) = month(now());

select s_id from student where month(s_birth) = month(now())+;