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50道MySQL面试题，经典~
一、测试表数据
学生表：student [学号,学生姓名,出生年月,性别]
成绩表：score [学号,课程号,成绩]
课程表：course [课程号,课程名称,教师号]
教师表：teacher [教师号,教师姓名)
下面是表结构和数据，直接执行即可~
#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(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
`t_name` varchar(20) 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(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
`s_name` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
`s_birth` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
`s_sex` varchar(10) 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(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
`c_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
`s_score` int(0) 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', 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);
-- ----------------------------
-- Table structure for course
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
`c_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
`c_name` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
`t_id` varchar(20) 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');
二、50道SQL题（不含答案）
好了，请打开你亲爱的navicat，关闭百度、手机等一切阻碍你进步的绊脚石。开始你的表演~
#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
三、50道SQL题（含答案）
-- 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;
50道SQL面试题开撸
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）
#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;
2、查询平均成绩大于60分的学生的学号和平均成绩
#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 > 60 ORDER BY avg_score desc;
3、查询所有学生的学号、姓名、选课数、总成绩
#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),0) from student stu
LEFT JOIN score sc on stu.s_id = sc.s_id GROUP BY stu.s_id;
4、查询姓“猴”的老师的个数
select count(*) from teacher where t_name like '猴%';
5、查询没学过“数学老师-杰斯”老师课的学生的学号、姓名
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')
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
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';
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
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;
8、查询课程编号为“02”的总成绩
select SUM(sc.s_score) from score sc where c_id = '02';
9、查询所有，课程成绩小于60分的学生的学号、姓名
select stu.s_id,stu.s_name from student stu,score sc where stu.s_id = sc.s_id and sc.s_score < 60 GROUP BY stu.s_id;
10、查询没有学全所有课的学生的学号、姓名
SELECT stu2.s_id,stu2.s_name from student stu2 ,
(SELECT stu.s_id,stu.s_name,IFNULL(COUNT(c.c_id),0) 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);
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）
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;
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）
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'))
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
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') ;
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, 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 < 60 GROUP BY stu.s_id) a where stu2.s_id = a.s_id and a.count >= 2;
16、检索"01"课程分数小于60，按分数降序排列的学生信息
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 < 60 ORDER BY sc.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(难)
-- 全量展示
SELECT * from
(SELECT stu.s_id,stu.s_name,sc.c_id,IFNULL(sc.s_score,0) 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;
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
SELECT
c.c_id,c.c_name,MAX(sc.s_score),MIN(sc.s_score),AVG(sc.s_score),CONCAT(round(100 * sum(case when sc.s_score>=60 then 1.0 else 0.0 end)/count(*),2),'%') as '及格率'
from score sc, course c
where sc.c_id = c.c_id
GROUP BY c.c_id;
19、查询学生的总成绩并进行排名
SELECT
stu.s_id,stu.s_name,IFNULL(SUM(sc.s_score),0) 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
20、查询不同老师所教不同课程平均分,从高到低显示
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;
21、查询学生平均成绩及其名次
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;
22、按各科成绩进行排序，并显示排名(难)
#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;
23、查询每门功课成绩最好的前两名学生姓名
#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<=2;
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
#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 2 and 3;
25、查询各科成绩前三名的记录（不考虑成绩并列情况）
#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 1 and 3;
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
#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 >= 85) as '[100-85]',
(select count(*) from score sc2 where sc2.c_id=c.c_id and sc2.s_score BETWEEN 70 and 85) as '[85-70]',
(select count(*) from score sc3 where sc3.c_id=c.c_id and sc3.s_score BETWEEN 60 and 70) as '[70-60]',
(select count(*) from score sc4 where sc4.c_id=c.c_id and sc4.s_score < 60) as '[<60]'
from course c
GROUP BY c.c_id;
27、查询每门课程被选修的学生数
select c_id,count(c_id) count from score group by c_id;
28、查询出只有两门课程的全部学生的学号和姓名
#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=2;
29、查询男生、女生人数
select s_sex,count(s_sex) count_sex from student group by s_sex;
30、查询名字中含有"风"字的学生信息
select * from student where s_name like '%风%';
31、查询1990年出生的学生名单
select s_id,s_name from student where year(s_birth)='1990';
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
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>=85;
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
select
c_id, avg(s_score) avg_score
from score
group by c_id
order by avg_score,c_id desc;
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 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<60 and c.c_name='数学';
35、查询所有学生的课程及分数情况
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 ;
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 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>70;
37、查询不及格的课程并按课程号从大到小排列
select c_id from score where s_score < 60 GROUP BY c_id order by c_id desc;
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
select stu.s_id,stu.s_name
from student stu, score sc where stu.s_id=sc.s_id
and sc.s_score>80 and sc.c_id=03;
39、求每门课程的学生人数
select c_id, count(*) count from score group by c_id;
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
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 1;
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）
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;
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
select
c_id, count(*) count
from score
group by c_id
order by c_id asc;
43、检索至少选修两门课程的学生学号
select s_id,count(*) count from score group by s_id having count>=2;
44、查询选修了全部课程的学生信息
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 = 3;
45、查询各学生的年龄
SELECT * ,(YEAR(CURDATE()) - YEAR(s_birth)) AS age from student;
46、查询两门以上不及格课程的同学的学号及其平均成绩
select s_id,avg(s_score) avg
from score
where s_score < 60
group by s_id
having count(*) >= 2;
47、查询本月过生日的学生
select s_id from student where month(s_birth) = month(now());
48、查询下一个月过生日的学生
select s_id from student where month(s_birth) = month(now())+1;
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