MySQL经典练习题

本文参考CSDN博主「csdnluolei」的MySQL经典练习50题，链接如下：

https://blog.csdn.net/csdnluolei/article/details/83507312?spm=1001.2014.3001.5506

练习题涉及到四个表：学生信息表student、成绩表score、课程信息表course、教师信息表teacher。四个表的详细字段如下文所示，表格的建表语句和插入数据语句请前往 CSDN博主「csdnluolei」的“MySQL经典练习50题”文章复制粘贴。

学生信息表student：

成绩表score

课程信息表course

教师信息表teacher

练习题目及答案

1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数

#方法一：先使用自连接查找出课程1分数比课程2高的学生s_id和课程分数，然后再和学生信息表做关联查询
SELECT s.*,sc.c1_score,sc.c2_score 
FROMstudent s INNER JOIN (SELECT --使用自连接查找出课程1分数比课程2高的学生s_id和课程分数s1.`s_id` AS s_id,s1.`s_score` AS c1_score,s2.`s_score` AS c2_score FROMscore s1 INNER JOIN score s2 ON s1.`s_id` = s2.`s_id` AND s1.`c_id` = 01 AND s2.`c_id` = 02 AND s1.`s_score` > s2.`s_score`) sc ON s.s_id = sc.s_id #方法二：使用三表连接
SELECT s.*,s1.`s_score` AS c1_score,s2.`s_score` AS c2_score 
FROMstudent s LEFT JOIN score s1 ON s.`s_id` = s1.`s_id` AND s1.c_id = '01' LEFT JOIN score s2 ON s.`s_id` = s2.`s_id` AND s2.c_id = '02' 
WHERE s1.`s_score` > s2.`s_score` ;

2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
参考题目1，只需把条件筛选中的>变成<即可

3、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

#方法一：先在成绩表筛选出平均分大于60的学生编号，然后再和学生信息表连接
SELECT s.`s_id`,s.`s_name`,sc.avg_s 
FROMstudent s INNER JOIN (SELECT s_id,AVG(s_score) AS avg_s FROMscore GROUP BY s_id HAVING avg_s > 60) sc ON s.`s_id` = sc.s_id;
#方法二：先把成绩表和学生信息表连接，再分组统计学生的平均分，最后筛选
SELECT s.`s_id`,s.`s_name`,AVG(sc.s_score) AS avg_s 
FROMstudent s INNER JOIN score sc ON s.`s_id` = sc.`s_id` 
GROUP BY s.`s_id`,s.`s_name` 
HAVING avg_s > 60 

– 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩 (包括有成绩的和无成绩的)

--方法一：使用联合查询，第一个查询语句查询有成绩的同学信息，第二个查询语句查询没有成绩的同学信息
SELECT s.`s_id`,s.`s_name`,AVG(sc.s_score) AS avg_s 
FROMstudent s INNER JOIN score sc ON s.`s_id` = sc.`s_id` 
GROUP BY s.`s_id`,s.`s_name` 
HAVING avg_s <60 
UNION 
SELECTs.`s_id`,s.`s_name`,NULL AS avg_s 
FROMstudent s
WHERE s.`s_id` NOT IN (SELECT DISTINCT s_id FROM score);--方法二：使用左连接SELECT s.`s_id`,s.`s_name`,AVG(sc.`s_score`) AS avg_s 
FROMstudent s LEFT JOIN score sc ON s.`s_id` = sc.`s_id` 
GROUP BY s.`s_id` 
HAVING avg_s < 60 OR s.`s_id` NOT IN (SELECT DISTINCT s_id FROMscore) ;

– 5、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,并从高到低排序

SELECT st.`s_id`,st.`s_name`,COUNT(*) AS 选课总数,SUM(sc.`s_score`) AS 总成绩 
FROMstudent st LEFT JOIN score sc ON st.`s_id` = sc.s_id 
GROUP BY st.`s_id` 
ORDER BY 总成绩 DESC;

– 6、查询"李"姓老师的数量

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

– 7、查询学过"张三"老师授课的同学的信息

#方法一：使用子查询
SELECT st.`s_id`,st.`s_name`,st.`s_birth`,st.`s_sex` 
FROMstudent st 
WHERE s_id IN (SELECT --查询学过张三老师开设课程的学生的编号ids_id FROMscore WHERE c_id IN (SELECT --查询张三老师开设课程的编号c_id FROMcourse WHERE t_id IN (SELECT--查询张三老师的教师编号idt_id FROMteacher WHERE t_name = '张三'))) ;
#方法二：使用连接查询+子查询
SELECT st.* 
FROMstudent st INNER JOIN score sc ON st.`s_id` = sc.`s_id` 
WHERE sc.c_id IN (SELECT c_id FROMcourse WHERE t_id IN (SELECT t_id FROMteacher WHERE t_name = '张三')) ;

– 8、查询没学过"张三"老师授课的同学的信息

#使用子查询
SELECT st.`s_id`,st.`s_name`,st.`s_birth`,st.`s_sex` 
FROMstudent st 
WHERE s_id NOT IN (SELECT --查询学过张三老师开设课程的学生的编号ids_id FROMscore WHERE c_id IN (SELECT --查询张三老师开设课程的编号c_id FROMcourse WHERE t_id IN (SELECT --查询张三老师的教师编号idt_id FROMteacher WHERE t_name = '张三'))) ;

• -9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息,及两门课程成绩

#三表连接
SELECT st.*,sc1.s_score AS c1_s,sc2.s_score AS c2_s 
FROMstudent st INNER JOIN score sc1 ON st.`s_id` = sc1.`s_id` AND sc1.`c_id` = '01' INNER JOIN score sc2 ON st.`s_id` = sc2.`s_id` AND sc2.`c_id` = '02' 

– 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息

#方法一：使用连接查询
SELECT st.* 
FROMstudent st INNER JOIN score sc1 ON st.`s_id` = sc1.`s_id` AND sc1.`c_id` = '01' AND st.s_id NOT IN (SELECT --查询学过课程编号为02的学生编号ids_id FROMscore WHERE `c_id` = '02') ;
#方法二：使用子查询
SELECT * 
FROMstudent st 
WHERE st.`s_id` IN (SELECT s_id FROMscore WHERE c_id = '01') 
AND st.`s_id` NOT IN (SELECT s_id FROMscore WHERE c_id = '02')

– 11、查询没有学全所有课程的同学的信息

#方法一：
SELECT * 
FROMstudent 
WHERE s_id IN (SELECT --提取出符合条件的学生编号idc.`s_id` FROM(SELECT--查询每个学生学的课程个数，并筛选出没有学完全部课程的同学st.`s_id`,COUNT(sc.s_score) AS num FROMstudent st LEFT JOIN score sc ON st.`s_id` = sc.s_id GROUP BY st.`s_id` HAVING num <> (SELECT COUNT(DISTINCT c_id) FROMscore)) c);
#方法二：使用子查询
SELECT * 
FROMstudent 
WHERE s_id NOT IN (SELECT--查询学完全部课程的学生的编号 s_id FROMscore GROUP BY s_id HAVING COUNT(s_score) = 3) 

– 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

#方法一：使用子查询
SELECT * FROM student WHERE s_id IN
--查询至少有一门课程与学号为01的同学所学相同的同学的学号
(SELECT DISTINCT s_id FROM score WHERE c_id IN 
--查询学号为01的同学所学的课程
(SELECT c_id FROM score WHERE s_id='01'));
#方法二：使用连接查询
SELECT 
st.`s_id`,st.`s_name`,st.`s_birth`,st.`s_sex`
FROM student st
INNER JOIN score sc
ON st.`s_id`=sc.`s_id`
WHERE sc.`c_id` IN(SELECT c_id FROM score WHERE s_id='01')
AND st.`s_id`<>'01'
--group by有去重的作用
GROUP BY st.`s_id`,st.`s_name`,st.`s_birth`,st.`s_sex`;

– 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

#方法一：
SELECT * 
FROMstudent st 
WHERE st.`s_id` IN (SELECT s1.s_id FROMscore s1 JOIN score s2 ON s1.s_id = s2.s_id AND s2.c_id = '02' JOIN score s3 ON s1.s_id = s3.s_id AND s3.c_id = '03' WHERE s1.c_id = '01') ;#方法二：SELECT * 
FROMstudent st 
WHERE st.`s_id` IN (SELECT s_id FROM score WHERE c_id IN(SELECT c_id FROM score WHERE s_id='01') GROUP BY s_idHAVING COUNT(1)=(SELECT COUNT(1) FROM score WHERE s_id='01')) ;

– 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 st.`s_id`,st.`s_name`,COUNT(1) 课程数,AVG(sc.s_score) AS avg_s 
FROMscore sc LEFT JOIN student st ON sc.s_id = st.`s_id` 
WHERE sc.s_score < 60 
GROUP BY st.`s_id` 
HAVING COUNT(1) >= 2 ;

– 16、检索"01"课程分数小于60，按分数降序排列的学生信息及01分数

SELECT st.*,sc.s_score 
FROMstudent st RIGHT JOIN score sc ON st.`s_id` = sc.s_id 
WHERE 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

#方法一：
SELECT s_id,
(SELECT s_score FROM score WHERE s_id=a.s_id AND c_id='01') AS c1_score,
(SELECT s_score FROM score WHERE s_id=a.s_id AND c_id='02') AS c2_score,
(SELECT s_score FROM score WHERE s_id=a.s_id AND c_id='03') AS c3_score,
AVG(s_score) AS avg_s 
FROM score a GROUP BY s_id ORDER BY avg_s DESC;
#方法二：多表连接，但表的个数很多是效率较低
SELECT
sc.`s_id`, s1.s_score,s2.s_score,s3.s_score,AVG(sc.s_score) AS avg_s
FROM score sc
LEFT JOIN score s1 ON sc.`s_id`=s1.s_id AND s1.c_id='01'
LEFT JOIN score s2 ON sc.`s_id`=s2.s_id AND s2.c_id='02'
LEFT JOIN score s3 ON sc.`s_id`=s3.s_id AND s3.c_id='03'
GROUP BY sc.`s_id`
ORDER BY avg_s 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

SELECT sc.c_id,c_name,
MAX(s_score) AS max_s,
MIN(s_score) AS min_s,
AVG(s_score) AS avg_s,
SUM(CASE WHEN s_score>60 THEN 1 ELSE 0 END)/COUNT(s_score) AS 及格率,
SUM(CASE WHEN s_score>=70 AND s_score<80 THEN 1 ELSE 0 END)/COUNT(s_score) AS 中等率,
SUM(CASE WHEN s_score>=80 AND s_score<90 THEN 1 ELSE 0 END)/COUNT(s_score) AS 优良率,
SUM(CASE WHEN s_score>=90 THEN 1 ELSE 0 END)/COUNT(s_score) AS 优秀率
FROM score sc
INNER JOIN course c ON sc.c_id=c.c_idGROUP BY c_id;

– 19、按各科成绩进行排序，并显示排名

#使用窗口函数，row_number(),MySQL是8.0版本以上才支持窗口函数
SELECT c_id,s_score,row_number () over (PARTITION BY c_id 
ORDER BY s_score DESC
) AS rank1 
FROMscore ;

– 20、查询学生的总成绩并进行排名

SELECT a.s_id,a.sum_s,row_number () over (ORDER BY sum_s DESC) AS rank1 
FROM(SELECT s_id,SUM(s_score) AS sum_s FROMscore GROUP BY s_id) a ;

– 21、查询不同老师所教不同课程平均分从高到低显示

SELECT t.t_id,t.t_name,c.c_id,c.c_name,sc.avg_s 
FROMcourse c 
INNER JOIN teacher t ON c.t_id = t.t_id
LEFT JOIN(
SELECT c_id,AVG(s_score) AS avg_s FROM score GROUP BY c_id
) sc
ON c.c_id=sc.c_id
ORDER BY sc.avg_s DESC;

– 22、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

SELECT st.*,c_id,a.s_score,rank1 
FROM
student st
JOIN(SELECT s_id,c_id,s_score,row_number () over (PARTITION BY c_id ORDER BY s_score) AS rank1 FROMscore) a ON st.s_id=a.s_id
WHERE a.rank1 IN (2, 3);

– 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],[0-60]及所占百分比

SELECT c_id,
SUM(CASE WHEN s_score >=85 THEN 1 ELSE 0 END) AS "100-85_num",
SUM(CASE WHEN s_score >=85 THEN 1 ELSE 0 END)/COUNT(s_score) AS "100-85_num",
SUM(CASE WHEN s_score>=70 AND s_score<85 THEN 1 ELSE 0 END) AS "85-70_num",
SUM(CASE WHEN s_score>=70 AND s_score<85 THEN 1 ELSE 0 END)/COUNT(s_score) AS "85-70_num",
SUM(CASE WHEN s_score>=60 AND s_score<70 THEN 1 ELSE 0 END) AS "70-60_num",
SUM(CASE WHEN s_score>=60 AND s_score<70 THEN 1 ELSE 0 END)/COUNT(s_score) AS "70-60_num",
SUM(CASE WHEN s_score<=60 THEN 1 ELSE 0 END) AS "0-60_num",
SUM(CASE WHEN s_score<=60 THEN 1 ELSE 0 END)/COUNT(s_score) AS "0-60_num"
FROM score
GROUP BY c_id;

– 24、查询学生平均成绩及其名次

#和第20题相似，只需把聚合函数sum改成avg即可
SELECT a.s_id,a.avg_s,row_number () over (ORDER BY avg_s DESC) AS rank1 
FROM(SELECT s_id,AVG(s_score) AS avg_s FROMscore GROUP BY s_id) a ;

• 24.1查询学生平均成绩及其名次，添加名次rank,（相同分数的相同名次，并列排名）

#注意rank()/row_number()/dense_rank()之间的差别
SELECT a.s_id,a.avg_s,rank () over (ORDER BY avg_s DESC) AS rank1 
FROM(SELECT s_id,AVG(s_score) AS avg_s FROMscore GROUP BY s_id) a ;

– 25、查询各科成绩前三名的记录

SELECT a.c_id,a.s_score 
FROM(SELECT c_id,s_score,row_number () over (PARTITION BY c_id ORDER BY s_score DESC) AS rank1 FROMscore) a 
WHERE rank1 <= 3 ;

– 26、查询每门课程被选修的学生数

SELECT c_id,COUNT(s_score) AS num FROM score GROUP BY c_id;

– 27、查询出只有两门课程的全部学生的学号和姓名

SELECT st.s_id,st.s_name 
FROMstudent st JOIN (SELECT s_id,COUNT(s_score) AS num FROMscore GROUP BY s_id HAVING num = 2) a ON st.s_id = a.s_id ;

28/29题

-- 28、查询男生、女生人数
SELECT s_sex,COUNT(*) FROM student GROUP BY s_sex;
-- 29、查询名字中含有"风"字的学生信息
SELECT * FROM student WHERE s_name LIKE '%风%';

– 30、查询同名同性学生名单，并统计同名人数

SELECT st.`s_name`,st.`s_sex`,COUNT(*) AS num 
FROMstudent1 st 
GROUP BY st.`s_name`,st.`s_sex`
HAVING num>1 ;

31-35

-- 31、查询1990年出生的学生名单
SELECT * FROM student WHERE s_birth LIKE '1990%';-- s_bith的数据类型是varchar
-- 32、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_id,AVG(s_score) AS avg_s 
FROMscore 
GROUP BY c_id 
ORDER BY avg_s DESC,c_id ;
-- 33、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 st.s_id,st.s_name,sc.avg_s 
FROMstudent st INNER JOIN (SELECT s_id,AVG(s_score) AS avg_s FROMscore GROUP BY s_id HAVING avg_s > 85) sc ON st.s_id = sc.s_id ;
-- 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 st.s_name,sc.s_score 
FROMstudent st INNER JOIN score sc ON st.s_id = sc.s_id AND s_score < 60 AND c_id = (SELECT c_id FROMcourse WHERE c_name = '数学') ;-- 35、查询所有学生的课程及分数情况；SELECT st.s_id,st.s_name,
SUM(CASE c.c_name WHEN '语文' THEN sc.s_score ELSE 0 END) AS 语文,
SUM(CASE c.c_name WHEN '数学' THEN sc.s_score ELSE 0 END) AS 数学,
SUM(CASE c.c_name WHEN '英语' THEN sc.s_score ELSE 0 END) AS 英语FROM student st  LEFT JOIN score sc ON st.s_id=sc.s_idLEFT JOIN course c ON sc.c_id=c.c_id
GROUP BY st.s_id,st.s_name;

题目：36-50

-- 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 st.s_name,c.c_name,sc.s_score 
FROMstudent st LEFT JOIN score sc ON st.s_id = sc.s_id LEFT JOIN course c ON sc.c_id = c.c_id 
WHERE sc.s_score >70 ;-- 37、查询不及格的学生id,姓名，及其课程名称，分数
SELECT st.s_id,st.s_name,c.c_name,sc.s_score 
FROMstudent st LEFT JOIN score sc ON st.s_id = sc.s_id LEFT JOIN course c ON sc.c_id = c.c_id 
WHERE sc.s_score < 60 ;-- 38、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 st.s_id,st.s_name,sc.c_id,sc.s_score 
FROMstudent st LEFT JOIN score sc ON st.s_id = sc.s_id 
WHERE sc.c_id = '01' AND sc.s_score >= 80;-- 39、求每门课程的学生人数
SELECT c.c_name,COUNT(1) AS 学生人数 
FROMscore s INNER JOIN course c ON s.c_id = c.c_id 
GROUP BY c.c_id ;-- 40、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 st.*,sc.c_id,sc.s_score 
FROMstudent st LEFT JOIN score sc ON st.s_id = sc.s_id 
WHERE c_id IN (SELECT c_id FROM course WHERE t_id IN (SELECT t_id FROM teacher WHERE t_name = '张三')) 
GROUP BY sc.c_id 
ORDER BY sc.s_score DESC;-- 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 s1.s_id,s1.c_id,s1.s_score 
FROMscore s1 INNER JOIN score s2 
ON s1.s_score = s2.s_score AND s1.c_id <> s2.c_id
GROUP BYs1.s_id,s1.c_id,s1.s_score ;-- 42、查询每门功课成绩最好的前两名
SELECT a.c_id,a.s_score,a.rank1 
FROM(SELECT s.c_id,s.s_score,row_number () over (PARTITION BY c_id ORDER BY s_score) AS rank1 FROMscore s) a 
WHERE a.rank1 < 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）。
-- 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 sc.c_id,COUNT(sc.s_score) AS num 
FROMscore sc 
GROUP BY sc.c_id 
HAVING num > 5 
ORDER BY num DESC,c_id ;-- 44、检索至少选修两门课程的学生学号
SELECT s_id 
FROMscore 
GROUP BY s_id HAVING COUNT(1) >= 2 ;-- 45、查询选修了全部课程的学生信息
SELECT st.* 
FROMstudent st INNER JOIN score sc ON st.s_id = sc.s_id 
GROUP BY st.s_id 
HAVING COUNT(sc.s_score) = (SELECT DISTINCT COUNT(c_id) FROM course) ;-- 46、查询各学生的年龄
SELECT st.*,YEAR(NOW())-YEAR(st.s_birth)AS age FROM student st;-- 47、查询本周过生日的学生
SELECT * FROM student WHERE WEEK(CURDATE())=WEEK(s_birth);-- 48、查询下周过生日的学生
SELECT * FROM student WHERE WEEK(CURDATE())=WEEK(s_birth)-1;-- 49、查询本月过生日的学生
SELECT * FROM student WHERE MONTH(CURDATE())=MONTH(s_birth);-- 50、查询下月过生日的学生
SELECT * FROM student WHERE MONTH(CURDATE())=MONTH(s_birth)-1;

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