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MySQL练习题50题（附带详细教程）

news 2025/8/20 7:19:15

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准备工作

create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(10));
create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));			create table Teacher(TId varchar(10),Tname varchar(10));			create table SC(SId varchar(10),CId varchar(10),score decimal(18,1));			insert into Student values('01' , '赵雷' , '1990-01-01' , '男'); insert into Student values('02' , '钱电' , '1990-12-21' , '男'); insert into Student values('03' , '孙⻛' , '1990-12-20' , '男'); insert into Student values('04' , '李云' , '1990-12-06' , '男'); insert into Student values('05' , '周梅' , '1991-12-01' , '⼥'); insert into Student values('06' , '吴兰' , '1992-01-01' , '⼥'); insert into Student values('07' , '郑⽵' , '1989-01-01' , '⼥'); insert into Student values('09' , '张三' , '2017-12-20' , '⼥'); insert into Student values('10' , '李四' , '2017-12-25' , '⼥'); insert into Student values('11' , '李四' , '2012-06-06' , '⼥'); insert into Student values('12' , '赵六' , '2013-06-13' , '⼥'); insert into Student values('13' , '孙七' , '2014-06-01' , '⼥');			insert into Course values('01' , '语⽂' , '02'); insert into Course values('02' , '数学' , '01'); insert into Course values('03' , '英语' , '03');		insert into Teacher values('01' , '张三'); insert into Teacher values('02' , '李四'); insert into Teacher values('03' , '王五');		insert into SC values('01' , '01' , 80); insert into SC values('01' , '02' , 90); insert into SC values('01' , '03' , 99); insert into SC values('02' , '01' , 70); insert into SC values('02' , '02' , 60); insert into SC values('02' , '03' , 80); insert into SC values('03' , '01' , 80); insert into SC values('03' , '02' , 80); insert into SC values('03' , '03' , 80); insert into SC values('04' , '01' , 50); insert into SC values('04' , '02' , 30); insert into SC values('04' , '03' , 20); insert into SC values('05' , '01' , 76); insert into SC values('05' , '02' , 87); insert into SC values('06' , '01' , 31); insert into SC values('06' , '03' , 34); insert into SC values('07' , '02' , 89); insert into SC values('07' , '03' , 98);

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 s.*, sc1.score AS score_01, sc2.score AS score_02
FROM Student s
-- 连接学生选“01”课程的成绩记录
JOIN SC sc1 ON s.SId = sc1.SId AND sc1.CId = '01'  
-- 连接学生选“02”课程的成绩记录
JOIN SC sc2 ON s.SId = sc2.SId AND sc2.CId = '02'  
WHERE sc1.score > sc2.score;

2. 查询同时存在 “01” 课程和 “02” 课程的情况

SELECT DISTINCT sc1.SId
FROM SC sc1
JOIN SC sc2 ON sc1.SId = sc2.SId
WHERE sc1.CId = '01' AND sc2.CId = '02';

3. 查询存在 “01” 课程但可能不存在 “02” 课程的情况（不存在时显示为 null ）

SELECT s.SId, s.Sname, sc1.score AS score_01, sc2.score AS score_02
FROM Student s
JOIN SC sc1 ON s.SId = sc1.SId AND sc1.CId = '01'
-- 左连接保证即使没有“02”课程成绩也能显示
LEFT JOIN SC sc2 ON s.SId = sc2.SId AND sc2.CId = '02';

4. 查询不存在 “01” 课程但存在 “02” 课程的情况

SELECT s.*, sc2.score AS score_02
FROM Student s
JOIN SC sc2 ON s.SId = sc2.SId AND sc2.CId = '02'
WHERE NOT EXISTS (SELECT 1 FROM SC sc1 WHERE sc1.SId = s.SId AND sc1.CId = '01'
);

5. 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 s.SId, s.Sname, AVG(sc.score) AS avg_score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname
HAVING AVG(sc.score) >= 60;

6. 查询在 SC 表存在成绩的学生信息

SELECT DISTINCT s.*
FROM Student s
JOIN SC sc ON s.SId = sc.SId;

7. 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（没成绩的显示为 null ）

SELECT s.SId, s.Sname, COUNT(sc.CId) AS course_count, COALESCE(SUM(sc.score), 0) AS total_score
FROM Student s
-- 左连接保证所有学生都能显示，包括没选课的
LEFT JOIN SC sc ON s.SId = sc.SId  
GROUP BY s.SId, s.Sname;

8. 查询 [李] 姓老师的数量

SELECT COUNT(*) AS teacher_count
FROM Teacher
WHERE Tname LIKE '李%';

9. 查询学过 [张三] 老师授课的同学的信息

SELECT DISTINCT s.*
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
JOIN Teacher t ON c.TId = t.TId
WHERE t.Tname = '张三';

10. 查询没有学全所有课程的同学的信息

-- 先查询出学全所有课程的学生，再取反
SELECT s.*
FROM Student s
WHERE s.SId NOT IN (SELECT sc.SIdFROM SC scGROUP BY sc.SIdHAVING COUNT(DISTINCT sc.CId) = (SELECT COUNT(*) FROM Course)
);

11. 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 DISTINCT s.*
FROM Student s
JOIN SC sc ON s.SId = sc.SId
WHERE sc.CId IN (SELECT CId FROM SC WHERE SId = '01'
) AND s.SId != '01';

12. 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

-- 先找出 01 同学学的课程集合，再找学的课程集合与之完全匹配的其他同学
SELECT s.*
FROM Student s
WHERE s.SId != '01'AND (SELECT GROUP_CONCAT(CId ORDER BY CId) FROM SC WHERE SId = s.SId) =(SELECT GROUP_CONCAT(CId ORDER BY CId) FROM SC WHERE SId = '01');

13. 查询没学过 “张三” 老师讲授的任一门课程的学生姓名

SELECT Sname
FROM Student
WHERE SId NOT IN (SELECT DISTINCT SC.SIdFROM SCJOIN Course ON SC.CId = Course.CIdJOIN Teacher ON Course.TId = Teacher.TIdWHERE Teacher.Tname = '张三'
);

14. 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 s.SId, s.Sname, AVG(sc.score) AS avg_score
FROM Student sJOIN SC sc ON s.SId = sc.SId
WHERE sc.score < 60
GROUP BY s.SId, s.Sname
HAVING COUNT(sc.CId) >= 2;

15. 检索 “01” 课程分数小于 60，按分数降序排列的学生信息

SELECT s.*, sc.score
FROM Student sJOIN SC sc ON s.SId = sc.SId
WHERE sc.CId = '01' AND sc.score < 60
ORDER BY sc.score DESC;

16. 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 s.SId,s.Sname,GROUP_CONCAT(CONCAT_WS(':', sc.CId, sc.score)ORDER BY sc.CIdSEPARATOR '; ') AS course_scores,AVG(sc.score) AS avg_score
FROM Student sLEFT JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname
ORDER BY avg_score DESC;

17. 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，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.CId,c.Cname,COUNT(sc.SId) AS student_count,MAX(sc.score) AS max_score,MIN(sc.score) AS min_score,AVG(sc.score) AS avg_score,-- 及格率CONCAT(ROUND(SUM(CASE WHEN sc.score >= 60 THEN 1 ELSE 0 END) / COUNT(sc.SId) * 100, 2), '%') AS pass_rate,-- 中等率CONCAT(ROUND(SUM(CASE WHEN sc.score >= 70 AND sc.score < 80 THEN 1 ELSE 0 END) / COUNT(sc.SId) * 100, 2), '%') AS medium_rate,-- 优良率CONCAT(ROUND(SUM(CASE WHEN sc.score >= 80 AND sc.score < 90 THEN 1 ELSE 0 END) / COUNT(sc.SId) * 100, 2), '%') AS good_rate,-- 优秀率CONCAT(ROUND(SUM(CASE WHEN sc.score >= 90 THEN 1 ELSE 0 END) / COUNT(sc.SId) * 100, 2), '%') AS excellent_rate
FROM Course cLEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname
ORDER BY student_count DESC, c.CId ASC;

18. 按各科平均成绩进行排序，并显示排名，Score 重复时保留名次空缺

-- 使用变量实现排名（MySQL 8.0 以下版本方式，8.0+ 可用窗口函数更简洁）
SET @rank := 0;
SET @prev_avg := NULL;SELECT sub.CId,sub.avg_score,CASEWHEN @prev_avg = sub.avg_score THEN @rankELSE @rank := @rank + 1END AS ranking,@prev_avg := sub.avg_score
FROM (SELECT c.CId, AVG(sc.score) AS avg_scoreFROM Course cLEFT JOIN SC sc ON c.CId = sc.CIdGROUP BY c.CIdORDER BY avg_score DESC) AS sub;

19. 按各科平均成绩进行排序，并显示排名，Score 重复时不保留名次空缺

-- MySQL 8.0+ 可用窗口函数 RANK()
SELECT c.CId,AVG(sc.score) AS avg_score,RANK() OVER (ORDER BY AVG(sc.score) DESC) AS ranking
FROM Course cLEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId;

20. 查询学生的总成绩，并进行排名，总分重复时保留名次空缺

-- MySQL 8.0 以下用变量，8.0+ 用窗口函数
SET @rank_total := 0;
SET @prev_total := NULL;SELECT sub.SId,sub.Sname,sub.total_score,CASEWHEN @prev_total = sub.total_score THEN @rank_totalELSE @rank_total := @rank_total + 1END AS ranking,@prev_total := sub.total_score
FROM (SELECT s.SId,s.Sname,SUM(sc.score) AS total_scoreFROM Student sLEFT JOIN SC sc ON s.SId = sc.SIdGROUP BY s.SId, s.SnameORDER BY total_score DESC) AS sub;

21. 查询学生的总成绩，并进行排名，总分重复时不保留名次空缺

SELECT s.SId,s.Sname,SUM(sc.score) AS total_score,DENSE_RANK() OVER (ORDER BY SUM(sc.score) DESC) AS ranking
FROM Student s
LEFT JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname
ORDER BY total_score DESC;

22. 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]，[60-0] 及所占百分比

SELECT c.CId,c.Cname,-- 统计[100-85]分数段人数SUM(CASE WHEN sc.score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS '100-85',-- 计算[100-85]分数段人数占比CONCAT(ROUND(SUM(CASE WHEN sc.score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) / COUNT(sc.SId) * 100, 2), '%') AS '100-85%',-- 统计[85-70]分数段人数SUM(CASE WHEN sc.score BETWEEN 70 AND 84 THEN 1 ELSE 0 END) AS '85-70',-- 计算[85-70]分数段人数占比CONCAT(ROUND(SUM(CASE WHEN sc.score BETWEEN 70 AND 84 THEN 1 ELSE 0 END) / COUNT(sc.SId) * 100, 2), '%') AS '85-70%',-- 统计[70-60]分数段人数SUM(CASE WHEN sc.score BETWEEN 60 AND 69 THEN 1 ELSE 0 END) AS '70-60',-- 计算[70-60]分数段人数占比CONCAT(ROUND(SUM(CASE WHEN sc.score BETWEEN 60 AND 69 THEN 1 ELSE 0 END) / COUNT(sc.SId) * 100, 2), '%') AS '70-60%',-- 统计[60-0]分数段人数SUM(CASE WHEN sc.score < 60 THEN 1 ELSE 0 END) AS '60-0',-- 计算[60-0]分数段人数占比CONCAT(ROUND(SUM(CASE WHEN sc.score < 60 THEN 1 ELSE 0 END) / COUNT(sc.SId) * 100, 2), '%') AS '60-0%'
FROM Course c
LEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname;

23. 查询各科成绩前三名的记录

-- 利用窗口函数 ROW_NUMBER 按课程分组并排序取前三名
WITH RankedScores AS (SELECT sc.CId,sc.SId,sc.score,ROW_NUMBER() OVER (PARTITION BY sc.CId ORDER BY sc.score DESC) AS rank_numFROM SC sc
)
SELECT rs.CId,c.Cname,s.Sname,rs.score
FROM RankedScores rs
JOIN Course c ON rs.CId = c.CId
JOIN Student s ON rs.SId = s.SId
WHERE rs.rank_num <= 3
ORDER BY rs.CId, rs.rank_num;

24. 查询每门课程被选修的学生数

SELECT c.CId,c.Cname,COUNT(DISTINCT sc.SId) AS student_count
FROM Course c
LEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname;

25. 查询出只选修两门课程的学生学号和姓名

SELECT s.SId,s.Sname
FROM Student s
JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname
HAVING COUNT(DISTINCT sc.CId) = 2;

26. 查询男生、女生人数

SELECT Ssex,COUNT(SId) AS student_count
FROM Student
GROUP BY Ssex;

27. 查询名字中含有「风」字的学生信息

SELECT *
FROM Student
WHERE Sname LIKE '%风%';

28. 查询同名同性学生名单，并统计同名人数

SELECT Sname,Ssex,COUNT(SId) AS name_count
FROM Student
GROUP BY Sname, Ssex
HAVING COUNT(SId) > 1;

29. 查询 1990 年出生的学生名单

SELECT *
FROM Student
WHERE YEAR(Sage) = 1990;

30. 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.CId,c.Cname,AVG(sc.score) AS avg_score
FROM Course c
LEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname
ORDER BY avg_score DESC, c.CId ASC;

31. 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 s.SId, s.Sname, AVG(sc.score) AS avg_score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname
HAVING AVG(sc.score) >= 85;

32. 查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

SELECT s.Sname, sc.score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
WHERE c.Cname = '数学' AND sc.score < 60;

33. 查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）

SELECT s.SId, s.Sname, c.Cname, sc.score
FROM Student s
-- 左连接课程表，保证所有学生都显示，不管是否选课
LEFT JOIN SC sc ON s.SId = sc.SId 
-- 左连接课程表，保证即使没选课也能关联到课程信息（没选课时课程名称等为 null ）
LEFT JOIN Course c ON sc.CId = c.CId;

34. 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 s.Sname, c.Cname, sc.score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
WHERE sc.score > 70;

35. 查询不及格的课程

SELECT c.CId, c.Cname, sc.SId, sc.score
FROM Course c
JOIN SC sc ON c.CId = sc.CId
WHERE sc.score < 60;

36. 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 s.SId, s.Sname
FROM Student s
JOIN SC sc ON s.SId = sc.SId
WHERE sc.CId = '01' AND sc.score > 80;

37. 求每门课程的学生人数

SELECT c.CId, c.Cname, COUNT(DISTINCT sc.SId) AS student_count
FROM Course c
LEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname;

38. 成绩不重复，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

-- 先找到张三老师授课的课程编号，再关联 SC 表找到对应课程成绩最高的学生
SELECT s.*, sc.score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
JOIN Teacher t ON c.TId = t.TId
WHERE t.Tname = '张三'AND sc.score = (SELECT MAX(score) FROM SC WHERE CId IN (SELECT CId FROM Course WHERE TId = (SELECT TId FROM Teacher WHERE Tname = '张三')));

39. 成绩有重复的情况下，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

-- 使用窗口函数 RANK() ，成绩相同排名相同，取排名第一的（包含重复成绩情况）
WITH RankedScores AS (SELECT s.*, sc.score,RANK() OVER (ORDER BY sc.score DESC) AS rnkFROM Student sJOIN SC sc ON s.SId = sc.SIdJOIN Course c ON sc.CId = c.CIdJOIN Teacher t ON c.TId = t.TIdWHERE t.Tname = '张三'
)
SELECT * 
FROM RankedScores 
WHERE rnk = 1;

40. 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 sc1.SId, sc1.CId, sc1.score
FROM SC sc1
JOIN SC sc2 ON sc1.score = sc2.score AND sc1.CId != sc2.CId AND sc1.SId = sc2.SId;

41. 查询每门课程成绩最好的前两名

WITH RankedScores AS (SELECT sc.CId,sc.SId,sc.score,-- 按课程分组，对成绩降序排名ROW_NUMBER() OVER (PARTITION BY sc.CId ORDER BY sc.score DESC) AS rank_numFROM SC sc
)
SELECT rs.CId,c.Cname,s.Sname,rs.score
FROM RankedScores rs
JOIN Course c ON rs.CId = c.CId
JOIN Student s ON rs.SId = s.SId
WHERE rs.rank_num <= 2
ORDER BY rs.CId, rs.rank_num;

42. 统计每门课程的学生选修人数（超过 5 人的课程才统计）

SELECT c.CId,c.Cname,COUNT(DISTINCT sc.SId) AS student_count
FROM Course c
LEFT JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname
HAVING COUNT(DISTINCT sc.SId) > 5;

43. 检索至少选修两门课程的学生学号

SELECT SId
FROM SC
GROUP BY SId
HAVING COUNT(DISTINCT CId) >= 2;

44. 查询选修了全部课程的学生信息

SELECT s.*
FROM Student s
WHERE (SELECT COUNT(DISTINCT CId) FROM SC WHERE SId = s.SId) = (SELECT COUNT(CId) FROM Course);

45. 查询各学生的年龄，只按年份来算

SELECT SId,Sname,-- 用当前年份减去出生年份得到年龄YEAR(NOW()) - YEAR(Sage) AS age
FROM Student;

46. 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 SId,Sname,-- 判断当前月日和出生月日，决定年龄计算方式CASE WHEN DATE_FORMAT(NOW(), '%m%d') < DATE_FORMAT(Sage, '%m%d') THEN YEAR(NOW()) - YEAR(Sage) - 1ELSE YEAR(NOW()) - YEAR(Sage) END AS age
FROM Student;

47. 查询本周过生日的学生

SELECT *
FROM Student
WHERE WEEKOFYEAR(Sage) = WEEKOFYEAR(NOW());

48. 查询下周过生日的学生

SELECT *
FROM Student
WHERE WEEKOFYEAR(Sage) = WEEKOFYEAR(NOW()) + 1;

49. 查询本月过生日的学生

SELECT *
FROM Student
WHERE MONTH(Sage) = MONTH(NOW());

50. 查询下月过生日的学生

SELECT *
FROM Student
WHERE MONTH(Sage) = MONTH(NOW()) + 1;

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