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);
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】
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