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
FROM
student 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
FROM
score 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
FROM
student 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
FROM
student s
INNER JOIN
(SELECT
s_id,
AVG(s_score) AS avg_s
FROM
score
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
FROM
student 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
FROM
student 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
SELECT
s.`s_id`,
s.`s_name`,
NULL AS avg_s
FROM
student 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
FROM
student 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
FROM
score) ;
– 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
FROM
student 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`
FROM
student st
WHERE s_id IN
(SELECT
--查询学过张三老师开设课程的学生的编号id
s_id
FROM
score
WHERE c_id IN
(SELECT
--查询张三老师开设课程的编号
c_id
FROM
course
WHERE t_id IN
(SELECT
--查询张三老师的教师编号id
t_id
FROM
teacher
WHERE t_name = '张三'))) ;
#方法二：使用连接查询+子查询
SELECT
st.*
FROM
student st
INNER JOIN score sc
ON st.`s_id` = sc.`s_id`
WHERE sc.c_id IN
(SELECT
c_id
FROM
course
WHERE t_id IN
(SELECT
t_id
FROM
teacher
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`
FROM
student 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id
s_id
FROM
score
WHERE c_id IN
(SELECT
--查询张三老师开设课程的编号
c_id
FROM
course
WHERE t_id IN
(SELECT
--查询张三老师的教师编号id
t_id
FROM
teacher
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
FROM
student 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.*
FROM
student 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id
s_id
FROM
score
WHERE `c_id` = '02') ;
#方法二：使用子查询
SELECT
*
FROM
student st
WHERE st.`s_id` IN
(SELECT
s_id
FROM
score
WHERE c_id = '01')
AND st.`s_id` NOT IN
(SELECT
s_id
FROM
score
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
*
FROM
student
WHERE s_id IN
(SELECT
--提取出符合条件的学生编号id
c.`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
FROM
student st
LEFT JOIN score sc
ON st.`s_id` = sc.s_id
GROUP BY st.`s_id`
HAVING num <>
(SELECT
COUNT(DISTINCT c_id)
FROM
score)) c);
#方法二：使用子查询
SELECT
*
FROM
student
WHERE s_id NOT IN
(SELECT
--查询学完全部课程的学生的编号
s_id
FROM
score
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
*
FROM
student st
WHERE st.`s_id` IN
(SELECT
s1.s_id
FROM
score 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
*
FROM
student 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_id
HAVING 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
FROM
score 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
FROM
student 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_id
GROUP 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
FROM
score ;
– 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
FROM
score
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
FROM
course 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
FROM
score) 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
FROM
score
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
FROM
score
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
FROM
score) 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
FROM
student st
JOIN
(SELECT
s_id,
COUNT(s_score) AS num
FROM
score
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
FROM
student1 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
FROM
score
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
FROM
student st
INNER JOIN
(SELECT
s_id,
AVG(s_score) AS avg_s
FROM
score
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
FROM
student st
INNER JOIN score sc
ON st.s_id = sc.s_id
AND s_score < 60
AND c_id =
(SELECT
c_id
FROM
course
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_id
LEFT 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
FROM
student 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
FROM
student 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
FROM
student 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
FROM
score 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
FROM
student 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
FROM
score s1
INNER JOIN score s2
ON s1.s_score = s2.s_score AND s1.c_id <> s2.c_id
GROUP BY
s1.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
FROM
score 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
FROM
score 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
FROM
score
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.*
FROM
student 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;