05-数据库_数据库高级_SQL进阶练习(部分习题)

目录

五,SQL进阶练习

1,数据表介绍

2,数据表建立SQL

3,练习题目(部分题目)

1. 查询" 01 "课程⽐" 02 "课程成绩⾼的学⽣的信息及课程分数

2. 查询同时存在" 01 "课程和" 02 "课程的情况

3. 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 )

4. 查询不存在" 01 "课程但存在" 02 "课程的情况

5.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

6.查询在 SC 表存在成绩的学⽣信息

7.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 )

8.查询「李」姓⽼师的数量

9.查询学过「张三」⽼师授课的同学的信息

10.查询没有学全所有课程的同学的信息

11.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

12.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

13.查询没学过"张三"⽼师讲授的任⼀⻔课程的学⽣姓名

14.查询两⻔及其以上不及格课程的同学的学号,姓名及其平均成绩

15.检索" 01 "课程分数小于 60,按分数降序排列的学生信息

16.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

17.查询各科成绩最⾼分、最低分和平均分:

18.按各科平均成绩进行排序,并显示排名, Score 重复时保留名次空缺

19.按各科平均成绩进行排序,并显示排名, Score 重复时不保留名次空缺


五,SQL进阶练习

1,数据表介绍

学⽣表

Student(SId,Sname,Sage,Ssex)

SId 学⽣编号,Sname 学⽣姓名,Sage 出⽣年⽉,Ssex 学⽣性别

课程表

Course(CId,Cname,TId)

CId 课程编号,Cname 课程名称,TId 教师编号

教师表

Teacher(TId,Tname)

TId 教师编号,Tname 教师姓名

成绩表

SC(SId,CId,score)

SId 学⽣编号,CId 课程编号,score 分数

2,数据表建立SQL

-- 数据SQL
-- 学⽣表 Student
create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex
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' , '孙⻛' , '1990-12-20' , '男');
insert into Student values('04' , '李云' , '1990-12-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '⼥');
insert into Student values('06' , '吴兰' , '1992-01-01' , '⼥');
insert into Student values('07' , '郑⽵' , '1989-01-01' , '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' , '2012-06-06' , '⼥');
insert into Student values('12' , '赵六' , '2013-06-13' , '⼥');
insert into Student values('13' , '孙七' , '2014-06-01' , '⼥');
-- 科⽬表 Course
create table Course(CId varchar(10),Cname nvarchar(10),TId 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');
-- 教师表 Teacher
create table Teacher(TId varchar(10),Tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
-- 成绩表 SC
create table SC(SId varchar(10),CId varchar(10),score decimal(18,1));
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);

3,练习题目(部分题目)

1. 查询" 01 "课程⽐" 02 "课程成绩⾼的学⽣的信息及课程分数

我的:

select s1.sid,s1.score 
from sc as s1,sc as s2 
where s1.sid=s2.sid and s1.cid='01' and s2.cid='02' and s1.score>s2.score;

老师讲解

先分别查询01和02课程的学员id和分数

select sid,score from sc where cid = '01';
select sid,score from sc where cid = '02';

发现返回的结果中sid并不完全一致,因此可以对两个结果做join联结,条件是 sid要相等

select s1.sid,s1.score from
(select sid,score from sc where cid = '01') as s1
join
(select sid,score from sc where cid = '02') as s2
on s1.sid = s2.sid
where s1.score > s2.score;

通过以上的sql,得到了符合条件的学员id和分数,再联结学生表,获取学员信息

select stu.sid,stu.sname,s.score
from student as stu
join
(
   select s1.sid,s1.score from
    (select sid,score from sc where cid = '01') as s1
    join
    (select sid,score from sc where cid = '02') as s2
    on s1.sid = s2.sid
    where s1.score > s2.score
) as s
on stu.sid = s.sid;


2. 查询同时存在" 01 "课程和" 02 "课程的情况

select s1.* from
(select sid,score from sc where cid = '01') as s1
join
(select sid,score from sc where cid = '02') as s2
on s1.sid = s2.sid;


3. 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 s1.*,s2.* from
(select sid,score from sc where cid = '01') as s1
left join
(select sid,score from sc where cid = '02') as s2
on s1.sid = s2.sid;


4. 查询不存在" 01 "课程但存在" 02 "课程的情况

select * from sc
where sid not in (select sid from sc where cid = '01')
and cid = '02'


5.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 sc.sid,sname,round(avg(score),2) as avg_score
from sc,student
where sc.sid = student.sid
group by sc.sid,sname having avg_score >= 60;

注意:

1,使用group by时,select中出现的列必须同样出现在group by中(除了聚集函数);

2,使用group by 和 having在检索结果中继续过滤

3,round函数保留小数点位数,不使用的话会默认保留当前列中最多的小数点个数


6.查询在 SC 表存在成绩的学⽣信息

两个表联结并对结果去重

select distinct stu.* from student as stu join sc on stu.sid = sc.sid;


7.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 stu.sid,stu.sname,count(sc.cid) as total_num,sum(sc.score) as sum_score
from student as stu
left join sc on stu.sid = sc.sid
group by stu.sid,stu.sname;


8.查询「李」姓⽼师的数量

select count(TId) as num
from teacher
where Tname like '李%';


9.查询学过「张三」⽼师授课的同学的信息

在成绩表中找到课程id和学生id。根据课程id找到教师id,根据教师id找到教师的名字。根据学生id找到学生信息。

-- 嵌套查询方式
select * from student
where SId in (
	select SId from sc
	where CId = (
		select CId from course
		where TId = (select TId from teacher where TName='张三')
	)
);

-- join联结方式
select * from student as stu
join (
	select sc.SId from sc
	join(
		select course.CId from course
		join teacher
		on teacher.Tname='张三' and course.TId=teacher.TId
	) as course_teacher
	on sc.CId=course_teacher.CId
) as sc_course_teacher
on stu.SId=sc_course_teacher.SId;


10.查询没有学全所有课程的同学的信息

select stu.SId,stu.Sname,stu.Sage,stu.Ssex from student as stu
join (
	select SId 
	from sc
	group by SId
	having count(sc.CId)<>(select count(CId) from course)
) as stu_list
on stu.SId=stu_list.SId
group by stu.SId,stu.Sname,stu.Sage,stu.Ssex;


11.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 stu.*
from student as stu
join sc on sc.sid = stu.sid
where sc.cid in (select cid from sc where sid = '01');


12.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

s1的课程id和s2的课程id相同,且上的课程数目和'01'号同学一致即可

select s2.sid,student.sname
from sc as s1
join sc as s2
on s1.cid = s2.cid and s1.sid = '01' and s2.sid != '01'
join student on s2.sid = student.sid
group by s2.sid,student.sname
having count(s2.cid) = (select count(*) from sc where sid = '01');


13.查询没学过"张三"⽼师讲授的任⼀⻔课程的学⽣姓名

select student.* from student where student.SId not in(
	select SId from sc where CId in(
		select CId from course where TId=(
			select TId from teacher where Tname='张三'
		)
	)
)


14.查询两⻔及其以上不及格课程的同学的学号,姓名及其平均成绩

select stu.sid,stu.sname,round(avg(sc.score),2) as avg_score
from student as stu
join sc on stu.sid = sc.sid
where sc.score < 60
group by stu.sid,stu.sname having count(sc.cid) >= 2;


15.检索" 01 "课程分数小于 60,按分数降序排列的学生信息

select sc.sid,stu.sname,sc.score
from sc join student as stu
on sc.sid = stu.sid
where sc.cid = '01' and sc.score < 60
order by sc.score desc;


16.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

方案一:

select sc.*,s2.avg_score
from sc
join (select sid,avg(score) as avg_score from sc group by sid) as s2
on sc.sid = s2.sid
order by s2.avg_score desc,sc.sid;

方案二:

select
stu.sname,
a.score as '语文',
b.score as '数学',
c.score as '英语',
avg(d.score) as '平均成绩'
from student as stu
left join sc as a on stu.sid = a.sid and a.cid = '01'
left join sc as b on stu.sid = b.sid and b.cid = '02'
left join sc as c on stu.sid = c.sid and c.cid = '03'
left join sc as d on stu.sid = d.sid
group by stu.sname,语文,数学,英语
order by 平均成绩 desc;


17.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

不考虑各种率的简单查询:

select sc.cid, c.cname,
max(sc.score) as '最高分',
min(sc.score) as '最低分',
round(avg(sc.score),2) as '平均分',
count(sc.cid) as '选修人数'
from sc join course as c on sc.cid = c.cid
group by sc.cid,c.cname
order by '选修人数' desc,sc.cid;

使用CASE WHEN

CASE WHEN sc.score >= 60 THEN 1 ELSE 0 END

相当于编程中 if

if  sc.score >= 60:

    return 1

else:

    return 0

select
sc.cid,
c.cname,
max(sc.score) as '最高分',
min(sc.score) as '最低分',
round(avg(sc.score),2) as '平均分',
count(sc.cid) as '选修人数',
sum(CASE WHEN sc.score >= 60 THEN 1 ELSE 0 END) / count(sc.cid) as '及格率',
sum(CASE WHEN sc.score >= 70 and sc.score < 80 THEN 1 ELSE 0 END) / count(sc.cid) as '中等率',
sum(CASE WHEN sc.score >= 80 and sc.score < 90 THEN 1 ELSE 0 END) / count(sc.cid) as '优良率',
sum(CASE WHEN sc.score >= 90 THEN 1 ELSE 0 END) / count(sc.cid) as '优秀率'
from sc join course as c on sc.cid = c.cid
group by sc.cid,c.cname
order by '选修人数' desc,sc.cid;


18.按各科平均成绩进行排序,并显示排名, Score 重复时保留名次空缺

-- 按照各学科进行分组,计算平均成绩
select cid,round(avg(score),2) as avg_sc from sc group by cid;


-- 按照各学科的平均成绩,做自联结,进行比较
select s1.* ,s2.*
from
(select cid,round(avg(score),2) as avg_sc from sc group by cid) as s1
join
(select cid,round(avg(score),2) as avg_sc from sc group by cid) as s2;


-- 检索出表s1中成绩大于等于表s2中的记录
select s1.* ,s2.*
from
(select cid,round(avg(score),2) as avg_sc from sc group by cid) as s1
join
(select cid,round(avg(score),2) as avg_sc from sc group by cid) as s2
on s1.avg_sc >= s2.avg_sc;


 对于平均分最高的科目,另一张相同的表在联结的时候分数大于等于它的只有1个(自身)。同理,对于平均分位列第二的科目,另一张相同的表中分数大于等于它的只有2个(第一名和它自己)。

-- 按照s2进行分组,统计s1的平均分出现的次数,
select s2.cid,s2.avg_sc,count(distinct s1.avg_sc) as rank
from
(select cid,round(avg(score),2) as avg_sc from sc group by cid) as s1
join
(select cid,round(avg(score),2) as avg_sc from sc group by cid) as s2
on s1.avg_sc >= s2.avg_sc
group by s2.cid,s2.avg_sc
order by rank


19.按各科平均成绩进行排序,并显示排名, Score 重复时不保留名次空缺

 这里声明一个变量,排序后每条记录加一,这样就能展示出名次了

-- @i 是sql中定义变量的意思
select b.cid,b.avg_sc,@i := @i+1 as rank
from (select @i := 0) as a,
(select cid,round(avg(score),2) as avg_sc from sc group by cid order by avg_sc desc) as b;


更新数据

-- 当修改完数据之后
update sc set score = 104.0 where sid = 01 and cid = 01;
insert into sc values('07','04',62.0);

 

与18题对比


以下题目视频课中未给出讲解,有缘再补系列^(* ̄(oo) ̄)^

20.查询学⽣的总成绩,并进⾏排名,总分重复时保留名次空缺

21.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

22.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

23.查询各科成绩前三名的记录

24.查询每⻔课程被选修的学⽣数

25.查询出只选修两⻔课程的学⽣学号和姓名

26.查询男⽣、⼥⽣⼈数

27.查询名字中含有「⻛」字的学⽣信息

28.查询同名同性学⽣名单,并统计同名⼈数

29.查询 1990 年出⽣的学⽣名单

30.查询每⻔课程的平均成绩,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

31.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

32.查询课程名称为「数学」,且分数低于 60 的学⽣姓名和分数

33.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)

34.查询任何⼀⻔课程成绩在 70 分以上的姓名、课程名称和分数

35.查询不及格的课程

36.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

37.求每⻔课程的学⽣⼈数

38.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

39.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

40.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

41.查询每⻔课程成绩最好的前两名

42.统计每⻔课程的学⽣选修⼈数(超过 5 ⼈的课程才统计)

43.检索⾄少选修两⻔课程的学⽣学号

44.查询选修了全部课程的学⽣信息

45.查询各学⽣的年龄,只按年份来算

46.按照出⽣⽇期来算,当前⽉⽇ < 出⽣年⽉的⽉⽇则,年龄减⼀

TIMESTAMPDIFF() 从⽇期时间表达式中减去间隔

https://dev.mysql.com/doc/refman/5.7/en/date-and-time-functions.html

47.查询本周过⽣⽇的学⽣

返回⽇期从范围内的数字⽇历星期1到53

48.查询下周过⽣⽇的学⽣

49.查询本⽉过⽣⽇的学⽣

50.查询下⽉过⽣⽇的学⽣

章节汇总在这里(づ ̄3 ̄)づ╭❤~@&再见萤火虫&【05-数据库】


对学习Java感兴趣的同学欢迎加入QQ学习交流群:1126298731

有问题欢迎提问,大家一起在学习Java的路上打怪升级!(o゜▽゜)o☆[BINGO!]

猜你喜欢

转载自blog.csdn.net/qq_41528502/article/details/112799473