【MySQL】手撕SQL50题-优快云博客

本文链接：https://blog.youkuaiyun.com/zc_stats/article/details/103372124

写SQL应该明确知道程序的执行顺序：

SQL执行顺序：from-->[join on]-->where-->group by-->having-->select-->order by 。

其中，MySQL8.0已经支持窗口函数，这类函数在解决排序，TOP n等问题上很方便.http://www.mysqltutorial.org/mysql-window-functions/。

加入窗口函数后执行顺序：Note that window functions are performed on the result set after all JOIN, WHERE, GROUP BY, and HAVING clauses and before the ORDER BY, LIMIT and SELECT DISTINCT.（Oracle这些就不用说了早就支持。）

题目主要是四张表：course, score, student, teacher。具体表的内容主键啥的，等俺刷完之后再补吧

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 student.*, t.score1, t.score2 FROM student RIGHT JOIN (
(SELECT t1.s_id, t1.s_score AS score1, t2.s_score AS score2 FROM (
 (SELECT s_id, s_score FROM score WHERE c_id='01') as t1 ,
     (SELECT s_id, s_score FROM score WHERE c_id='02') as t2
      ) WHERE t1.s_id = t2.s_id and t1.s_score > t2.s_score
      )
       as t)
ON student.s_id=t.s_id;

# null 不能进行数字比较会直接过滤
1.1 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 t1.s_id, t1.s_score AS score1, t2.s_score AS score2 FROM (
    (SELECT s_id, s_score FROM score WHERE c_id='01') as t1 LEFT JOIN
        (SELECT s_id, s_score FROM score WHERE c_id='02') as t2
        ON t1.s_id = t2.s_id
    );

1.2 查询同时存在01和02课程的情况

SELECT t1.s_id ,t1.s_score AS score1, t2.s_score AS score2 FROM (
    (SELECT s_id, s_score FROM score WHERE c_id='01') as t1 JOIN
    (SELECT s_id, s_score FROM score WHERE c_id='02') as t2
    ON t1.s_id=t2.s_id and t1.s_score > t2.s_score
    );

1.3 查询选择了02课程但没有01课程的情况

SELECT * FROM score WHERE s_id NOT IN
        (SELECT s_id FROM score WHERE c_id='01') AND c_id='02';

2.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 t.s_id, s_name, t.avg_score FROM student
    JOIN
    (SELECT s_id, avg(s_score) AS avg_score FROM score GROUP BY s_id HAVING avg_score >=60) AS t
    ON student.s_id=t.s_id;

3.查询在score 表存在成绩的学生信息

    SELECT student.* FROM student
    JOIN
    (SELECT s_id, avg(s_score) FROM score GROUP BY s_id HAVING avg(s_score) IS NOT NULL) AS t
    ON student.s_id=t.s_id;
    DISTINCT
    SELECT  student.* FROM student, score WHERE student.s_id=score.s_id;

4.查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和

SELECT student.s_id, student.s_name, t.course_counts, t.score_sum FROM
    student JOIN
    (SELECT s_id, COUNT(*) AS course_counts, SUM(s_score) AS score_sum
    FROM score GROUP BY s_id) as t
    ON student.s_id=t.s_id;

5.查询「李」姓老师的数量

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

6.查询学过「张三」老师授课的同学的信息

    SELECT * FROM student WHERE s_id IN
    (SELECT s_id FROM score WHERE c_id IN
    (SELECT c_id FROM course JOIN teacher ON course.t_id=teacher.t_id WHERE teacher.t_name='张三'));

7.查询没有学全所有课程的同学的信息

SELECT * FROM student WHERE s_id IN (
        SELECT s_id FROM score GROUP BY s_id HAVING COUNT(*) < (SELECT COUNT(DISTINCT c_id) FROM course)
    );

 SELECT student.*, COUNT(*) FROM student
    RIGHT JOIN score
    ON student.s_id=score.s_id
    GROUP BY score.s_id
    HAVING COUNT(*) < (SELECT COUNT(DISTINCT c_id) FROM score);

8.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 * FROM student WHERE s_id IN
    (SELECT s_id FROM score WHERE c_id IN
        (SELECT DISTINCT c_id FROM score WHERE s_id='01')) AND s_id!='01';

9.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 * FROM student WHERE s_id IN
    (SELECT s_id FROM score GROUP BY s_id HAVING COUNT(*)=(SELECT COUNT(*) FROM score WHERE s_id='01'))
    AND s_id IN
    (SELECT s_id FROM score WHERE c_id IN (SELECT c_id FROM score WHERE s_id='01'));

10.查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT s_name FROM student WHERE s_id NOT IN (
    SELECT s_id FROM score WHERE c_id IN (SELECT c_id FROM teacher JOIN course ON teacher.t_id=course.t_id WHERE t_name='张三')
    );

SELECT s_name FROM student WHERE s_id NOT IN (
    SELECT s_id FROM score LEFT JOIN course ON score.c_id=course.c_id
    LEFT JOIN teacher ON teacher.t_id=course.t_id WHERE t_name='张三'
    );

11.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.s_id, s.s_name, avg(s_score), c.不及格数 FROM student AS s JOIN
    (SELECT a.不及格数, b.* FROM
    (SELECT s_id, count(*) AS '不及格数' FROM score WHERE s_score < 60 GROUP BY s_id HAVING COUNT(*) >= 2) AS a
     LEFT JOIN
     score AS b ON a.s_id=b.s_id
     ) AS c
     ON s.s_id=c.s_id GROUP BY s.s_id;

12.检索" 01 "课程分数小于 60，按分数降序排列的学生信息

SELECT student.*, t.s_score FROM student JOIN
    (SELECT s_id, s_score FROM score WHERE c_id='01' AND s_score < 60) AS t ON student.s_id=t.s_id
    ORDER BY t.s_score DESC;

13.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 * FROM score LEFT JOIN
    (SELECT s_id, AVG(s_score) AS '平均成绩' FROM score GROUP BY s_id) AS t
    ON score.s_id = t.s_id
    ORDER BY t.平均成绩 DESC;

14.查询各科成绩最高分、最低分和平均分,以如下形式显示：
    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 course.c_name,t.*, t.及格人数/t.选修人数 AS '及格率' FROM course JOIN
    (
    SELECT
     c_id,
      MAX(s_score) AS '最高分',
      MIN(s_score) AS '最低分',
      AVG(s_score) AS '平均分',
      COUNT(*) AS '选修人数',
      SUM(CASE WHEN s_score >=60 THEN 1 ELSE 0 END) AS '及格人数',
      SUM(CASE WHEN s_score >=70 AND s_score <80 THEN 1 ELSE 0 END) AS '中等人数',
      SUM(CASE WHEN s_score >=80 AND s_score <90 THEN 1 ELSE 0 END) AS '优良人数',
      SUM(CASE WHEN s_score >=90 THEN 1 ELSE 0 END) AS '优秀人数'
     FROM score GROUP BY c_id) as t
     ON course.c_id=t.c_id ORDER BY t.选修人数 DESC, t.c_id;

15.按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

SELECT c_id, s_score,
    RANK() OVER(PARTITION BY score.c_id ORDER BY score.s_score DESC) AS '排名'
    FROM score ;

15.1 按各科成绩进行行排序，并显示排名， Score 重复时合并名次

SELECT c_id, s_score,
    DENSE_RANK() OVER(PARTITION BY c_id ORDER BY s_score DESC) AS '排名'
    FROM score

16.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.s_id, student.s_name, t1.总成绩, t1.排名 FROM student JOIN
    (SELECT t.*, RANK() OVER(PARTITION BY s_id ORDER BY '总成绩' DESC) AS '排名' FROM score JOIN
    (SELECT s_id, SUM(s_score) AS '总成绩' FROM score GROUP BY s_id) AS t
    ON score.s_id=t.s_id
    ) AS t1
    ON student.s_id=t1.s_id;

16.1 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 t.*, DENSE_RANK() OVER(PARTITION BY s_id ORDER BY '总成绩' DESC) AS '排名' FROM score JOIN
    (SELECT s_id, SUM(s_score) AS '总成绩' FROM score GROUP BY s_id) AS t
    ON score.s_id=t.s_id;

17.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 course.c_name, t.* FROM course JOIN
    (SELECT
    c_id,
    COUNT(*) AS '人数',
    SUM(CASE WHEN s_score<=60 THEN 1 ELSE 0 END) / COUNT(*) AS '[0-60]',
    SUM(CASE WHEN s_score>60 AND s_score<=70 THEN 1 ELSE 0 END) / COUNT(*) AS '[60-70]',
    SUM(CASE WHEN s_score>70 AND s_score<=80 THEN 1 ELSE 0 END) / COUNT(*) AS '[80-90]',
    SUM(CASE WHEN s_score>80 AND s_score<=100 THEN 1 ELSE 0 END) / COUNT(*) AS '[90-100]'
    FROM score GROUP BY c_id) AS t
    ON course.c_id=t.c_id;

18.查询各科成绩前三名的记录

SELECT * FROM
    (SELECT *, ROW_NUMBER() OVER(PARTITION BY c_id ORDER BY s_score DESC) AS '排名'
     FROM score) AS t
     WHERE t.排名 >= 3;

19.查询每门课程被选修的学生数

SELECT c_name, t.* FROM course JOIN
    (SELECT c_id, COUNT(*) AS '选修人数' FROM score GROUP BY c_id) AS t
    ON course.c_id=t.c_id;

20.查询出只选修两门课程的学生学号和姓名

SELECT student.s_id, student.s_name FROM student JOIN
    (SELECT s_id FROM score GROUP BY s_id HAVING COUNT(c_id)=2) AS t
    ON student.s_id=t.s_id;

21. 查询男生、女生人数

SELECT s_sex, COUNT(*) AS '人数' FROM student GROUP BY s_sex;

22. 查询名字中含有「风」字的学生信息

SELECT * FROM student WHERE s_name like '%风%';

23.查询同名同性学生名单，并统计同名人数 （这题有点问题）

SELECT * FROM
    (SELECT s1.* FROM student AS s1, student AS s2
    WHERE s1.s_id!=s2.s_id AND s1.s_name=s2.s_name AND s1.s_sex=s2.s_sex) AS t1
    LEFT JOIN
    (SELECT s11.s_id, COUNT(*) AS '同名人数' FROM student AS s11, student AS s22 WHERE s11.s_id!=s22.s_id AND s11.s_name=s22.s_name
    GROUP BY s11.s_name) AS t2
    ON t1.s_id=t2.s_id;

24.查询 1990 年出生的学生名单

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

25.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，按课程编号降序排列

    SELECT c_id, AVG(s_score) FROM score GROUP BY c_id ORDER BY AVG(s_score) DESC, c_id DESC;

26.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_name, t.* FROM student
    JOIN
    (SELECT s_id, AVG(s_score) FROM score GROUP BY s_id HAVING AVG(s_score) >= 85) AS t
    ON student.s_id=t.s_id;

27.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_score, s_name FROM student
    LEFT JOIN score ON student.s_id=score.s_id
    LEFT JOIN course ON score.c_id=course.c_id
    WHERE course.c_name='数学' AND score.s_score<60;

28.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 * FROM score
    RIGHT JOIN course ON score.c_id=course.c_id;

29.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 score.*, c_name FROM score
    LEFT JOIN course
    ON score.c_id=course.c_id
    WHERE s_id NOT IN
    (SELECT DISTINCT s_id FROM score WHERE s_score <= 70);

30.查询不及格的课程

    SELECT DISTINCT course.c_name, course.c_id FROM course
    JOIN
    (SELECT c_id FROM score WHERE s_score < 60) AS t
    ON course.c_id=t.c_id;

31.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 student.s_id, s_name, t.s_score FROM student JOIN
    (SELECT * FROM score WHERE c_id='01' AND s_score>80) AS t
    ON student.s_id=t.s_id;

32.求每门课程的学生人数

    SELECT c_name, t.c_id, t.人数 FROM course JOIN
    (SELECT c_id, COUNT(*) AS '人数' FROM score GROUP BY c_id) AS t
    on course.c_id=t.c_id;

33.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 student.*,score.* FROM teacher JOIN course ON teacher.t_id=course.t_id
    RIGHT JOIN score ON course.c_id=score.c_id JOIN student ON score.s_id=student.s_id
    WHERE teacher.t_name='张三' ORDER BY score.s_score DESC LIMIT 1;

34.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

    SELECT t.*, RANK() OVER(PARTITION BY c_id ORDER BY t.s_score DESC) AS rank
    FROM
    (SELECT student.*,score.* FROM teacher JOIN course ON teacher.t_id=course.t_id
    RIGHT JOIN score ON course.c_id=score.c_id JOIN student ON score.s_id=student.s_id
    WHERE teacher.t_name='张三') AS t
    WHERE rank=1;

35.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 t1.*, t2.* FROM score AS t1, score AS t2 WHERE t1.s_score=t2.s_score AND t1.c_id!=t2.c_id;

36. 查询每门成绩最好的前两名

    SELECT *, ROW_NUMBER() OVER(PARTITION BY c_id ORDER BY s_score DESC) AS rank
     FROM score
    WHERE rank >= 2;

37. 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_id,COUNT(*) FROM score GROUP BY c_id HAVING COUNT(*) > 5;

38.检索至少选修两门课程的学生学号

SELECT s_id FROM score GROUP BY s_id HAVING COUNT(*) >=  2;

39.查询选修了全部课程的学生信息

    SELECT * FROM student JOIN
    (SELECT * FROM score GROUP BY s_id HAVING COUNT(*) = (SELECT COUNT(*) FROM course)) AS t
    ON student.s_id=t.s_id;

40.查询各学生的年龄，只按年份来算

 SELECT *, 2019-YEAR(s_birth) AS '年龄' FROM student;