深耕MySQL - 50道SQL练习题_mysql题目训练，2024年最新醍醐灌顶_mysql训练

mysql> select sum(s_score) from score group by c_id having c_id=‘02’;

9、查询所有，课程成绩小于60分的学生的学号、姓名

select distinct a.s_id,a.s_name
from student a
left join score b on a.s_id=b.s_id
where b.s_score<60 or a.s_id not in (select s_id from score);

(1) 查询课程成绩小于60分的学生的学号

(2) 将步骤1的查询结果和student表联表查询出姓名，同时考虑没有考试成绩的学生

10、查询没有学全所有课的学生的学号、姓名

select a.s_id,a.s_name
from student a
left join score b on a.s_id=b.s_id
group by a.s_id,a.s_name
having count(b.c_id)<(select count(c_id) from course);

11、查询至少有一门课与学号为“01”的学生所学课程相同的学生的学号和姓名 （难）

select distinct a.s_id,a.s_name
from student a
right join score b on a.s_id=b.s_id
where b.c_id in (select c_id from score where s_id=‘01’) and a.s_id !=‘01’;

(1) 查询至少有一门课与学号为“01”的学生所学课程相同的学生的学号

(2) 根据步骤1的结果与student表联表查询出姓名，同时排除学号1的学生

12、查询和“01”号同学所学课程完全相同的其他同学的学号（难）

select s_id
from score where s_id!=‘01’
group by s_id
having count(s_id) = (select count(*) from score where s_id=‘01’);

13、查询没学过"张三"老师讲授的任一门课程的学生姓名

select s_id,s_name
from student
where s_id not in (select a.s_id
from score a
join course b on a.c_id=b.c_id
join teacher c on b.t_id=c.t_id
where c.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 a.s_id,a.s_name,avg(s_score) avg_score
from student a
join score b on a.s_id=b.s_id
where b.s_score<60
group by s_id
having count(b.s_id)>=2;

(1) 求出每个学生成绩不及格的课程数量及平均成绩

(2) 利用步骤1的结果与student表联表查询出学生姓名，其中课程成绩不及格数量要大于等于2门

16、检索"01"课程分数小于60，按分数降序排列的学生信息

select a.s_id,a.s_name
from student a
join score b on a.s_id=b.s_id
where b.s_score<60 and b.c_id=‘01’
order by b.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(难)

select a.s_id,a.s_score,b.avg_score
from score a
left join (select s_id, avg(s_score) avg_score
from score
group by s_id) b
on a.s_id =b.s_id
order by avg_score desc;

注意：上述虽然表达出了题目的意思，但是最好以横排形式展示，可以使用select嵌套子查询

方法2 ：

select s_id,s_name,
(select s_score from score where score.s_id=student.s_id and c_id=‘01’) 01_score,
(select s_score from score where score.s_id=student.s_id and c_id=‘02’) 02_score,
(select s_score from score where score.s_id=student.s_id and c_id=‘03’) 03_score,
(select avg(s_score) from score where score.s_id=student.s_id) avg_score
from student
order by avg_score desc;

使用select子查询嵌套查询时，需要将有歧义的字段区分出来比如： score.s_id=student.s_id

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

(1) 查询各科成绩最高分、最低分和平均分

select a.c_id,a.c_name, avg(b.s_score) avg_score,min(b.s_score) min_score,max(b.s_score) max_score
from course a j
oin score b on a.c_id=b.c_id
group by b.c_id;

(2) 每一个科目的及格率，中等率，优良率，优秀率需要使用select后嵌套子查询获得

select
b.c_id,b.c_name, avg(a.s_score) avg_score,min(a.s_score) min_score,max(a.s_score) max_score,
– 注意：这里相当于将 score a和course b联表查询后再与score这张表联表查询
(select count() from score where c_id=a.c_id and s_score>=60)/count() ‘及格率’,
(select count() from score where c_id=a.c_id and s_score between 70 and 80)/count() ‘中登率’,
(select count() from score where c_id=a.c_id and s_score between 80 and 90)/count() ‘优良率’,
(select count() from score where c_id=a.c_id and s_score>90) /count() ‘优秀率’
from score a
join course b
on a.c_id=b.c_id
group by a.c_id;

19、查询学生的总成绩并进行排名

select a.s_id,a.s_name,(case when sum(b.s_score) is null then 0 else sum(b.s_score) end) sum_score
from student a
left join score b on a.s_id=b.s_id
group by b.s_id
order by sum_score desc;

20、查询不同老师所教不同课程平均分,从高到低显示

select c_id,avg(s_score) avg_score
from score
group by c_id
order by avg_score desc;

21、查询学生平均成绩及其名次

select s_id,avg(s_score) avg_score,
row_number () over( order by avg(s_score) desc) row_num
from score
group by s_id;

22、按各科成绩进行排序，并显示排名(难)

序号函数：row_number() / rank() / dense_rank()

partition子句：窗口按照那些字段进行分组，窗口函数在不同的分组上分别执行。下面的例子就按照 c_id进行了分组。在每个 c_id上，按照order by的顺序分别生成从1开始的顺序编号。

order by子句：按照哪些字段进行排序，窗口函数将按照排序后的记录顺序进行编号。可以和partition子句配合使用，也可以单独使用。如果没有partition子句，则会按照所有score排序来生成序号。

– partition by c_id 按照c_id进行分组
– order by s_score desc 按照s_score倒叙排序
select c_id,s_id,row_number() over(partition by c_id order by s_score desc) num_row
from score;

23、查询每门功课成绩最好的前两名学生姓名

select a.s_id,a.s_name,b.num_row
from student a
– 将select查询作为一张子表，与student表联表查询
join (select
s_id,
c_id,
row_number() over(partition by c_id order by s_score desc) num_row
from score) b on a.s_id=b.s_id
where b.num_row<=2;

24、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 a.s_id,a.s_name,b.c_id,b.s_score,b.num_row
from student a
join (select
s_id,
c_id,
s_score,
row_number() over(partition by c_id order by s_score desc) num_row from score) b
on a.s_id=b.s_id
where b.num_row between 2 and 3;

25、查询各科成绩前三名的记录（不考虑成绩并列情况）

select a.s_id,a.s_name,b.c_id,b.s_score,b.num_row
from student a
join (select
s_id,
c_id,
s_score,row_number() over(partition by c_id order by s_score desc) num_row
from score) b
on a.s_id=b.s_id
where b.num_row<=3;

26、使用分段[100-85],[85-70],[70-60],[<60]来统计各科成绩，分别统计各分数段人数：课程ID和课程名称

select a.c_id,a.c_name,
(select count() from score where score.c_id=a.c_id and s_score<60) ‘小于60’,
(select count() from score where score.c_id=a.c_id and s_score between 60 and 70) ‘60-70’,
(select count() from score where score.c_id=a.c_id and s_score between 70 and 85) ‘70-85’,
(select count() from score where score.c_id=a.c_id and s_score between 85 and 100) ‘85-100’
from course a
join score b on a.c_id=b.c_id
group by c_id;

27、查询每门课程被选修的学生数

mysql> select c_id,count(c_id) count from score group by c_id;

28、查询出只有两门课程的全部学生的学号和姓名

select a.s_id,a.s_name,count(b.s_id) count
from student a
join score b on a.s_id=b.s_id
group by s_id
having count=2;

29、查询男生、女生人数

mysql> select count(s_sex) count_sex from student group by s_sex;

30、查询名字中含有"风"字的学生信息

mysql> select * from student where s_name like ‘%周%’;

31、查询1990年出生的学生名单

mysql> select s_id,s_name from student where year(s_birth)=1990;

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 a.s_id,a.s_name,avg(b.s_score) avg_score
from student a
join score b on a.s_id=b.s_id
group by b.s_id
having avg_score>=85;

33、查询每门课程的平均成绩，结果按平均成绩升序排序，平均成绩相同时，按课程号降序排列

select c_id, avg(s_score) avg_score from score group by c_id order by avg_score,c_id desc;

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 a.c_id,b.c_name,c.s_id,c.s_name,a.s_score
from score a
join course b on a.c_id=b.c_id
join student c on a.s_id=c.s_id
where a.s_score<60 and b.c_name=‘数学’;

35、查询所有学生的课程及分数情况

select b.s_id,b.s_name,c.c_id,c.c_name,a.s_score
from score a
join student b on a.s_id=b.s_id
join course c on a.c_id=c.c_id ;

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 c.s_name,b.c_name,a.s_score
from score a
join course b on a.c_id=b.c_id
join student c on a.s_id=c.s_id
where a.s_score>70;

37、查询不及格的课程并按课程号从大到小排列

mysql> select distinct c_id from score where s_score<60 order by c_id desc;

38、查询课程编号为03且课程成绩在80分以上的学生的学号和姓名

select a.s_id,a.s_name
from student a
join score b on a.s_id=b.s_id
where b.s_score>80 and b.c_id=03;

39、求每门课程的学生人数

mysql> select c_id, count(*) count from score group by 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姓名及其成绩

select a.s_score,b.s_id,b.s_name
from score a
join student b on a.s_id=b.s_id
join course c on a.c_id=c.c_id
join teacher d on c.t_id=d.t_id
where d.t_name=‘张三’
order by a.s_score desc
limit 1;

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 （难）

方法1：利用group by将s_id，c_id，s_score都重复的去除掉

select a.s_id,a.c_id,a.s_score
from score a
join score b on a.s_score=b.s_score
where a.c_id!=b.c_id and a.s_id=b.s_id
group by a.s_id,a.c_id,a.s_score;

方法2： 利用distinct将s_id，c_id，s_score都重复的去除掉

select distinct a.s_id,a.c_id,a.s_score
from score a
join score b on a.s_score=b.s_score
where a.c_id!=b.c_id and a.s_id=b.s_id ;

42、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

mysql> select c_id, count(*) count from score group by c_id order by c_id asc;

43、检索至少选修两门课程的学生学号

select s_id,count(*) count from score group by s_id having count>=2;

44、查询选修了全部课程的学生信息

方法1 ：

select a.s_id,a.s_name,a.s_birth,a.s_sex
from student a
join score b on a.s_id=b.s_id
group by a.s_id,a.s_name,a.s_birth,a.s_sex
having count(b.s_id) =(select count(*) from course);

方法2 ：

select * from student a
– 查询score表中s_id=student表中s_id的行数，即统计每个学生的选修课程数量
where (select count() from score where s_id = a.s_id)=(select count() from course)

45、查询各学生的年龄

– 时间差函数：timestampdiff
select TIMESTAMPDIFF(SECOND,“2020-02-27”,NOW()) – 31680773
select TIMESTAMPDIFF(MINUTE,“2020-02-27”,NOW()) – 528010
select TIMESTAMPDIFF(HOUR,“2020-02-27”,NOW()) – 8800
select TIMESTAMPDIFF(DAY,“2020-02-27”,NOW()) – 366
select TIMESTAMPDIFF(WEEK,“2020-02-27”,NOW()) – 52
select TIMESTAMPDIFF(MONTH,“2020-02-27”,NOW()) – 12
select TIMESTAMPDIFF(QUARTER,“2020-02-27”,NOW()) – 4

自我介绍一下，小编13年上海交大毕业，曾经在小公司待过，也去过华为、OPPO等大厂，18年进入阿里一直到现在。

深知大多数Go语言工程师，想要提升技能，往往是自己摸索成长或者是报班学习，但对于培训机构动则几千的学费，着实压力不小。自己不成体系的自学效果低效又漫长，而且极易碰到天花板技术停滞不前！

因此收集整理了一份《2024年Go语言全套学习资料》，初衷也很简单，就是希望能够帮助到想自学提升又不知道该从何学起的朋友，同时减轻大家的负担。

既有适合小白学习的零基础资料，也有适合3年以上经验的小伙伴深入学习提升的进阶课程，基本涵盖了95%以上Golang知识点，真正体系化！

由于文件比较大，这里只是将部分目录大纲截图出来，每个节点里面都包含大厂面经、学习笔记、源码讲义、实战项目、讲解视频，并且后续会持续更新

如果你觉得这些内容对你有帮助，可以添加V获取：vip1024b （备注Go）

一个人可以走的很快，但一群人才能走的更远。不论你是正从事IT行业的老鸟或是对IT行业感兴趣的新人，都欢迎扫码加入我们的的圈子（技术交流、学习资源、职场吐槽、大厂内推、面试辅导），让我们一起学习成长！

自己不成体系的自学效果低效又漫长，而且极易碰到天花板技术停滞不前！**

因此收集整理了一份《2024年Go语言全套学习资料》，初衷也很简单，就是希望能够帮助到想自学提升又不知道该从何学起的朋友，同时减轻大家的负担。
[外链图片转存中…(img-zm4g1Byx-1713064190207)]
[外链图片转存中…(img-S57usDOY-1713064190208)]
[外链图片转存中…(img-pORypJZZ-1713064190208)]
[外链图片转存中…(img-6NWAe1tq-1713064190209)]
[外链图片转存中…(img-sDZf1rL1-1713064190209)]

既有适合小白学习的零基础资料，也有适合3年以上经验的小伙伴深入学习提升的进阶课程，基本涵盖了95%以上Golang知识点，真正体系化！

由于文件比较大，这里只是将部分目录大纲截图出来，每个节点里面都包含大厂面经、学习笔记、源码讲义、实战项目、讲解视频，并且后续会持续更新

如果你觉得这些内容对你有帮助，可以添加V获取：vip1024b （备注Go）
[外链图片转存中…(img-bEhhkew4-1713064190210)]

一个人可以走的很快，但一群人才能走的更远。不论你是正从事IT行业的老鸟或是对IT行业感兴趣的新人，都欢迎扫码加入我们的的圈子（技术交流、学习资源、职场吐槽、大厂内推、面试辅导），让我们一起学习成长！