50道经典SQL练习题 MYSQL8.0实现

使用MYSQL8.0完成了“50道经典SQL练习题”，答案可能有误，欢迎指正。
参考了多位网友的答案
超经典SQL练习题，做完这些你的ＳＱＬ就过关了
50道SQL练习题
50道SQL练习题及答案与详细分析

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准备数据

CREATE TABLE student(id VARCHAR(10),name VARCHAR(10),age DATETIME,sex VARCHAR(10)
);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' , '孙风' , '1990-05-20' , '男');
insert into student values('04' , '李云' , '1990-08-06' , '男');
insert into student values('05' , '周梅' , '1991-12-01' , '女');
insert into student values('06' , '吴兰' , '1992-03-01' , '女');
insert into student values('07' , '郑竹' , '1989-07-01' , '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' , '2017-12-30' , '女');
insert into student values('12' , '赵六' , '2017-01-01' , '女');
insert into student values('13' , '孙七' , '2018-01-01' , '女');create table course(id varchar(10),name varchar(10),teacher_id varchar(10)
);insert into course values('01' , '语文' , '02');
insert into course values('02' , '数学' , '01');
insert into course values('03' , '英语' , '03');create table teacher(id varchar(10),name varchar(10)
);insert into teacher values('01' , '张三');
insert into teacher values('02' , '李四');
insert into teacher values('03' , '王五');create table score(student_id varchar(10),cource_id varchar(10),score decimal(18,1)
);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);

题解

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 * FROM (SELECT student_id, score FROM score WHERE cource_id = '01') sc1,(SELECT student_id, score FROM score WHERE cource_id = '02') sc2 WHEREsc1.score > sc2.score AND sc1.student_id = sc2.student_id;

SELECT * FROM (SELECT student_id, score FROM score WHERE cource_id = '01') sc1INNER JOIN (SELECT student_id, score FROM score WHERE cource_id = '02') sc2 ONsc1.student_id = sc2.student_id WHERE sc1.score > sc2.score;

2. 查询同时存在" 01 “课程和” 02 "课程的情况

SELECT * FROM (SELECT student_id, score FROM score WHERE cource_id = '01') sc1,(SELECT student_id, score FROM score WHERE cource_id = '02') sc2 WHEREsc1.student_id = sc2.student_id;

SELECT * FROM (SELECT student_id, score FROM score WHERE cource_id = '01') sc1INNER JOIN (SELECT student_id, score FROM score WHERE cource_id = '02') sc2 ONsc1.student_id = sc2.student_id;

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 * FROM (SELECT student_id, score FROM score WHERE cource_id = '01') sc1LEFT JOIN (SELECT student_id, score FROM score WHERE cource_id = '02') sc2 ONsc1.student_id = sc2.student_id;

4. 查询不存在" 01 “课程但存在” 02 "课程的情况

SELECT * FROM (SELECT student_id, score FROM score WHERE cource_id = '01') sc1RIGHT JOIN (SELECT student_id, score FROM score WHERE cource_id = '02') sc2 ONsc1.student_id = sc2.student_id WHERE sc1.student_id IS NULL;

SELECT * FROM score sc WHERE sc.cource_id = '02' ANDsc.student_id NOT IN (SELECT student_id FROM score WHERE cource_id = '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 a.student_id, s.name, a.avg_score FROM(SELECT student_id, AVG(score) avg_score FROM score GROUP BY student_id) aINNER JOIN student s ON s.id = a.student_idWHERE a.avg_score >= 60;

SELECT a.student_id, s.name, a.avg_score FROM(SELECT student_id, AVG(score) avg_score FROM scoreGROUP BY student_id HAVING avg_score > 60) aINNER JOIN student s ON s.id = a.student_id;

6. 查询在 SC 表存在成绩的学生信息

SELECT * FROM student s INNER JOIN (SELECT DISTINCT student_id id FROM score) sc ON sc.id = s.id;

SELECT DISTINCT s.* FROM student s, score sc WHERE s.id = sc.student_id;

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.id, s.name, sc.count, sc.sum FROM student s LEFT JOIN(SELECT student_id id, COUNT(student_id) count, SUM(score) sum FROM score GROUP BY student_id) scON s.id = sc.id;

8. 查有成绩的学生信息

SELECT * FROM student s INNER JOIN(SELECT DISTINCT student_id id FROM score) scON s.id = sc.id;

SELECT * FROM student s WHERE EXISTS (SELECT 1 FROM score sc WHERE sc.student_id = s.id);

SELECT * FROM student s WHERE S.id IN (SELECT student_id FROM score);

9. 查询「李」姓老师的数量

SELECT COUNT(*) FROM teacher WHERE name LIKE '李%';

10. 查询学过「张三」老师授课的同学的信息

SELECT * FROM student s WHERE s.id IN(SELECT student_id FROM score sc WHEREsc.cource_id = (SELECT id FROM course WHERE teacher_id = (SELECT id FROM teacher WHERE name = '张三')));

SELECT s.* FROM student s, teacher t, course c, score sc WHEREt.name = '张三' AND t.id = c.teacher_id AND c.id = sc.cource_id AND sc.student_id = s.id;

11. 查询没有学全所有课程的同学的信息

SELECT s.* FROM student s WHERE s.id NOT IN(SELECT student_id FROM score GROUP BY student_id HAVING COUNT(*) = (SELECT COUNT(*) FROM course));

12. 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 INNER JOIN(SELECT s2.student_id student_id FROM score s2 WHERE s2.student_id <> '01' AND s2.cource_id IN(SELECT s1.cource_id FROM score s1 WHERE s1.student_id = '01')) s3ON s3.student_id = s.id;

SELECT student.* FROM student WHERE id IN(SELECT student_id FROM score sc WHERE sc.cource_id IN(SELECT cource_id FROM score sc1 WHERE sc1.student_id = '01')) AND id <> '01';

13. 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

SELECT s.* FROM student s INNER JOIN(SELECT student_id id, COUNT(*) FROM score sc WHERE sc.cource_id IN(SELECT cource_id FROM score WHERE student_id = '01') ANDsc.student_id <> '01' GROUP By sc.student_id HAVINGCOUNT(*) = (SELECT COUNT(*) FROM score WHERE student_id = '01')) sdON sd.id = s.id;

14. 查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT student.* FROM student WHERE id NOT IN(SELECT student_id FROM score WHERE cource_id IN(SELECT c.id FROM course c WHERE c.teacher_id = (SELECT t.id FROM teacher t WHERE t.name = '张三')));

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 s.id, s.name, avg_score FROM student s INNER JOIN(SELECT student_id, AVG(score) avg_score, COUNT(*) FROM scoreWHERE score < 60 GROUP BY student_id HAVING COUNT(*) >= 2) aON a. student_id = s.id;

16. 检索" 01 "课程分数小于 60，按分数降序排列的学生信息

SELECT s.* FROM student s INNER JOIN(SELECT * FROM score WHERE cource_id = '01' AND score < 60) sc ONsc.student_id = s.id ORDER BY sc.score DESC;

SELECT s.*, sc.* FROM student s, (SELECT * FROM score WHERE cource_id = '01' AND score < 60) sc  WHERE s.id = sc.student_id ORDER BY sc.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

SELECT s.*, sc.score, sc.avg_score FROM student s LEFT JOIN(SELECT student_id student_id, score score, AVG(score) avg_score FROM score GROUP BY student_id) scON sc.student_id = s.id ORDER BY sc.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，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

-- 选择性聚合
SELECT cource_id c_id, course.name name, COUNT(*) s_count, MAX(score) max_score,MIN(score) min_score, AVG(score) avg_score,CONCAT(SUM(CASE WHEN score >= 60 THEN 1 ELSE 0 END) / COUNT(*) * 100, '%') pass_rate,CONCAT(SUM(CASE WHEN score >= 70 AND score < 80 THEN 1 ELSE 0 END) / COUNT(*) * 100, '%') mid_rate,CONCAT(SUM(CASE WHEN score >= 80 AND score < 90 THEN 1 ELSE 0 END) / COUNT(*) * 100, '%') good_rate,CONCAT(SUM(CASE WHEN score >= 90 THEN 1 ELSE 0 END) / COUNT(*) * 100, '%') excellent_rateFROM score, course WHERE course.id = score.cource_id GROUP BY cource_idORDER BY s_count DESC, c_id ASC;

19. 按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

--- MYSQL 8.0 实现
SELECT cource_id, score, RANK() OVER(PARTITION BY cource_id ORDER BY score DESC) FROM score;

SELECT s.cource_id, s.score,
IF(@pre_course_id = s.cource_id,@rank_counter := @rank_counter + 1,@rank_counter := 1) temp1,
IF(@pre_course_id = s.cource_id,IF(@pre_score = s.score, @cur_rank, @cur_rank := @rank_counter),@cur_rank := 1) ranking,
@pre_score := s.score temp2,
@pre_course_id := s.cource_id temp3
FROM score s, (SELECT @cur_rank := 0, @pre_course_id := NULL, @pre_score := NULL, @rank_counter := 1)r
ORDER BY s.cource_id, s.score DESC;

20. 按各科成绩进行排序，并显示排名， Score 重复时合并名次

--- MYSQL 8.0 实现
SELECT cource_id, score, DENSE_RANK() OVER(PARTITION BY cource_id ORDER BY score DESC) FROM score;

SELECT s.cource_id, s.score,
IF(@pre_score = s.score, @cur_rank, @cur_rank := @cur_rank + 1) temp1,
@pre_score := s.score,
IF(@pre_course_id = s.cource_id, @cur_rank, @cur_rank := 1) ranking,
@pre_course_id := s.cource_id
FROM score s, (SELECT @cur_rank :=0, @pre_score = NULL, @pre_course_id := NULL) r
ORDER BY cource_id, score DESC;

SELECT s.cource_id, s.score,
IF(@pre_course_id = s.cource_id,IF(@pre_score = s.score, @cur_rank, @cur_rank := @cur_rank + 1),@cur_rank := 1) ranking,
@pre_score := s.score,
@pre_course_id := s.cource_id
FROM score s, (SELECT @cur_rank :=0, @pre_score = NULL, @pre_course_id := NULL) r
ORDER BY cource_id, score DESC;

21. 查询学生的总成绩，并进行排名，总分重复时保留名次空缺

SELECT student_id, SUM(score) total, RANK() OVER(ORDER BY SUM(score) DESC) ranking
FROM score GROUP BY student_id;

SELECT t.student_id, t.total,
IF(@pre_total = t.total, @cur_rank, @cur_rank := @rank_counter) ranking,
@pre_total := t.total temp1,
@rank_counter := @rank_counter + 1 temp2
FROM (SELECT student_id, SUM(score) totalFROM score GROUP BY student_id ORDER BY total DESC) t,(SELECT @pre_total := NULL, @cur_rank := 0, @rank_counter := 1) r;

22. 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 student_id, SUM(score) total, DENSE_RANK() OVER(ORDER BY SUM(score) DESC) ranking
FROM score GROUP BY student_id;

SELECT t.student_id, t.total,
IF(@pre_total = t.total, @cur_rank, @cur_rank := @cur_rank + 1) ranking,
@pre_total := t.total temp1
FROM (SELECT student_id, SUM(score) totalFROM score GROUP BY student_id ORDER BY total DESC) t,(SELECT @pre_total := NULL, @cur_rank := 0) r;

23. 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.id, c.name,
SUM(IF(sc.score >= 0 AND sc.score < 60, 1, 0)) / COUNT(*) '[0-60)',
SUM(IF(sc.score >= 60 AND sc.score < 70, 1, 0)) / COUNT(*) '[60-70)',
SUM(IF(sc.score >= 70 AND sc.score < 85, 1, 0)) / COUNT(*) '[70-85)',
SUM(IF(sc.score >= 85 AND sc.score <= 100, 1, 0)) / COUNT(*) '[80-100]'
FROM course c LEFT JOIN score sc ON c.id = sc.cource_id
GROUP BY sc.cource_id;

24. 查询各科成绩前三名的记录

--关联子查询
SELECT sc1.cource_id, sc1.student_id, sc1.score
FROM score sc1
WHERE (SELECT COUNT(*) FROM score sc2 WHERE sc1.cource_id = sc2.cource_id AND sc1.score < sc2.score) < 3
ORDER BY sc1.cource_id, sc1.score DESC;

25. 查询每门课程被选修的学生数

SELECT cource_id, COUNT(*) FROM score GROUP BY cource_id;

26. 查询出只选修两门课程的学生学号和姓名

SELECT s.* FROM student s INNER JOIN(SELECT student_id, COUNT(*) s_count FROM score GROUP BY student_id HAVING COUNT(*) = 2) scON sc.student_id = s.id;

27. 查询男生、女生人数

SELECT sex, COUNT(*) FROM student GROUP BY sex;

28. 查询名字中含有「风」字的学生信息

SELECT student.* FROM student WHERE name Like '%风%';

29. 查询同名同性学生名单，并统计同名人数

SELECT s1.name, COUNT(*) FROM student s1, student s2 WHEREs1.name = s2.name AND s1.id <> s2.id GROUP By name;

SELECT s1.name, COUNT(*) FROM student s1INNER JOIN student s2 ON s1.name = s2.name AND s1.id <> s2.idORDER BY name;

SELECT name, COUNT(*) FROM student GROUP BY name HAVING COUNT(*) >1;

30. 查询 1990 年出生的学生名单

SELECT * FROM student WHERE YEAR(age) = 1990;

31. 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 cource_id, AVG(score) avg_score
FROM score GROUP BY cource_id
ORDER BY avg_score DESC, cource_id;

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 s.id, s.name, AVG(sc.score) avg_score
FROM student s INNER JOIN score sc ON s.id = sc.student_id
GROUP BY sc.student_id HAVING avg_score >=85;

33. 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.name, sc.score FROM student s
LEFT JOIN score sc ON s.id = sc.student_id
WHERE sc.cource_id = (SELECT id FROM course WHERE name = '数学') AND sc.score < 60;

34. 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.id, s.name, sc.cource_id, sc.score FROM student s
LEFT JOIN score sc ON s.id = sc.student_id;

35. 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.name, c.name, sc.score
FROM score sc INNER JOIN student s ON s.id = sc.student_id
INNER JOIN course c ON c.id = sc.cource_id
WHERE sc.score >= 70;

36. 查询不及格的课程

SELECT * FROM score WHERE score < 60;

37. 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.id, s.name, sc.score FROM score sc
INNER JOIN student s ON sc.student_id = s.id
WHERE sc.score >= 80 AND sc.cource_id = '01';

38. 求每门课程的学生人数

SELECT cource_id, COUNT(*) num FROM score
GROUP BY(cource_id);

39. 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

SELECT s.id, s.name, sc.score FROM score sc
INNER JOIN student s ON sc.student_id = s.id
WHERE sc.cource_id = (SELECT id FROM course WHERE teacher_id =(SELECT id FROM teacher WHERE name = '张三'))
ORDER BY sc.score DESC LIMIT 1;

SELECT s.id, s.name, sc.score FROM score sc, student s, teacher t, course c
WHERE sc.student_id = s.id AND sc.cource_id = c.id ANDc.teacher_id = t.id AND t.name = '张三'
ORDER BY sc.score DESC LIMIT 1;

40. 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

SELECT s.id, s.name, sc.score FROM score sc, student s, teacher t, course c
WHERE sc.student_id = s.id AND sc.cource_id = c.id ANDc.teacher_id = t.id AND t.name = '张三' ANDsc.score = (SELECT MAX(ss.score)FROM (SELECT sc.scoreFROM score sc, student s, teacher t, course cWHERE sc.student_id = s.id AND sc.cource_id = c.id ANDc.teacher_id = t.id AND t.name = '张三') ss);

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

SELECT sc1.student_id, sc1.cource_id, sc1.score FROM score sc1
INNER JOIN score sc2 ON sc1.student_id = sc2.student_id
WHERE sc1.cource_id <> sc2.cource_id
GROUP BY student_id, cource_id;

42. 查询每门功成绩最好的前两名

SELECT sc1.cource_id, sc1.student_id, sc1.score FROM score sc1
WHERE (SELECT COUNT(*) FROM score sc2WHERE sc1.cource_id = sc2.cource_idAND sc1.score < sc2.score) < 2
ORDER BY cource_id;

---MYSQL8.0 实现
SELECT *
FROM (SELECT cource_id, student_id, score,ROW_NUMBER() OVER (PARTITION BY cource_id ORDER BY score DESC) rankingFROM score) r
WHERE r.ranking < 3;

43. 统计每门课程的学生选修人数（超过 5 人的课程才统计）。

SELECT cource_id, COUNT(*) student_num FROM score
GROUP BY(cource_id) HAVING COUNT(*) > 5;

44. 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(*) >= 2;

45. 查询选修了全部课程的学生信息

SELECT student_id FROM score GROUP BY student_id
HAVING COUNT(*) = (SELECT COUNT(*) FROM course);

46. 查询各学生的年龄，只按年份来算

SELECT name, (YEAR(CURRENT_DATE()) - YEAR(age)) age FROM student;

47. 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 name, TIMESTAMPDIFF(YEAR, age, CURDATE()) FROM student;

48. 查询本周过生日的学生

SELECT * FROM student WHERE WEEKOFYEAR(age) = WEEKOFYEAR(CURRENT_DATE());

49. 查询下周过生日的学生

SELECT * FROM student WHERE WEEKOFYEAR(age) = WEEKOFYEAR(DATE_ADD(CURRENT_DATE(), INTERVAL 7 DAY));

50. 查询本月过生日的学生

SELECT * FROM student WHERE MONTH(age) = MONTH(CURRENT_DATE());

51. 查询下月过生日的学生

SELECT * FROM student WHERE MONTH(age) = MONTH(DATE_ADD(CURRENT_DATE(), INTERVAL 1 MONTH));

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