50道经典SQL练习题 MYSQL8.0实现

小咪咪 2021-12-14 19:29 259阅读 0赞

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][_SQL][  
50道SQL练习题][50_SQL][  
50道SQL练习题及答案与详细分析][50_SQL 1]

#### 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 WHERE
     sc1.score > sc2.score AND sc1.student_id = sc2.student_id;

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

1.  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 WHERE
     sc1.student_id = sc2.student_id;

1.  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') sc1
     LEFT JOIN (SELECT student_id, score FROM score WHERE cource_id = '02') sc2 ON
     sc1.student_id = sc2.student_id;

1.  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') sc1
     RIGHT JOIN (SELECT student_id, score FROM score WHERE cource_id = '02') sc2 ON
     sc1.student_id = sc2.student_id WHERE sc1.student_id IS NULL;

SELECT * FROM score sc WHERE sc.cource_id = '02' AND
     sc.student_id NOT IN (SELECT student_id FROM score WHERE cource_id = '01');

1.  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) a
     INNER JOIN student s ON s.id = a.student_id
     WHERE a.avg_score >= 60;

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

1.  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;

1.  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) sc
     ON s.id = sc.id;

1.  查有成绩的学生信息

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

1.  查询「李」姓老师的数量

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

1.  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 WHERE
     sc.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 WHERE
     t.name = '张三' AND t.id = c.teacher_id AND c.id = sc.cource_id AND sc.student_id = s.id;

1.  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));

1.  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')) s3
     ON 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';

1.  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') AND
      sc.student_id <> '01' GROUP By sc.student_id HAVING
      COUNT(*) = (SELECT COUNT(*) FROM score WHERE student_id = '01')) sd
      ON sd.id = s.id;

1.  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')));

1.  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 score
      WHERE score < 60 GROUP BY student_id HAVING COUNT(*) >= 2) a
     ON a. student_id = s.id;

1.  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 ON
     sc.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;

1.  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) sc
     ON sc.student_id = s.id ORDER BY sc.avg_score DESC;

1.  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_rate
     FROM score, course WHERE course.id = score.cource_id GROUP BY cource_id
     ORDER BY s_count DESC, c_id ASC;

1.  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;

1.  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;

1.  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) total
          FROM score GROUP BY student_id ORDER BY total DESC) t,
          (SELECT @pre_total := NULL, @cur_rank := 0, @rank_counter := 1) r;

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

1.  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;

1.  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;

1.  查询每门课程被选修的学生数

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

1.  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) sc
     ON sc.student_id = s.id;

1.  查询男生、女生人数

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

1.  查询名字中含有「风」字的学生信息

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

1.  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 WHERE
     s1.name = s2.name AND s1.id <> s2.id GROUP By name;

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

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

1.  查询 1990 年出生的学生名单

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

1.  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;

1.  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;

1.  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;

1.  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;

1.  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;

1.  查询不及格的课程

SELECT * FROM score WHERE score < 60;

1.  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';

1.  求每门课程的学生人数

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

1.  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 AND
          c.teacher_id = t.id AND t.name = '张三'
    ORDER BY sc.score DESC LIMIT 1;

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

1.  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;

1.  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 sc2
           WHERE sc1.cource_id = sc2.cource_id
           AND 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) ranking
          FROM score) r
    WHERE r.ranking < 3;

1.  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;

1.  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;

1.  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);

1.  查询各学生的年龄，只按年份来算

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

1.  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;

1.  查询本周过生日的学生

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

1.  查询下周过生日的学生

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

1.  查询本月过生日的学生

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

1.  查询下月过生日的学生

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

[_SQL]: https://blog.csdn.net/flycat296/article/details/63681089
[50_SQL]: https://zhuanlan.zhihu.com/p/32137597
[50_SQL 1]: https://www.jianshu.com/p/476b52ee4f1b