MySQL的45道操练题
目录A8站源码交易平台
数据表介绍
标题
创立表，句子能够运用创立数据库及表
查询句子，不确保必定对
数据表介绍
–1.学生表
Student(SId,Sname,Sage,Ssex)
–SId 学生编号,Sname 学生名字,Sage 出世年月,Ssex 学生性别
–2.课程表
Course(CId,Cname,TId)
–CId 课程编号,Cname 课程名称,TId 教师编号
–3.教师表
Teacher(TId,Tname)
–TId 教师编号,Tname 教师名字
–4.成果表
SC(SId,CId,score)
–SId 学生编号,CId 课程编号,score 分数
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’ , ‘孙风’ , ‘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’);
成果表 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);
操练标题
1 查询" 01 “课程比” 02 “课程成果高的学生的信息及课程分数
1.1 查询一起存在” 01 “课程和” 02 “课程的状况
1.2 查询存在” 01 “课程但可能不存在” 02 “课程的状况(不存在时显现为 null )
1.3 查询不存在” 01 “课程但存在” 02 “课程的状况
2 查询均匀成果大于等于 60 分的同学的学生编号和学生名字和均匀成果
3 查询在 SC 表存在成果的学生信息
4 查询一切同学的学生编号、学生名字、选课总数、一切课程的总成果(没成果的显现为 null )
4.1 查有成果的学生信息
5 查询「李」姓教师的数量
6 查询学过「张三」教师授课的同学的信息
7 查询没有学全一切课程的同学的信息
8 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
9 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
10 查询没学过"张三"教师教学的任一门课程的学生名字
11 查询两门及其以上不及格课程的同学的学号，名字及其均匀成果
12 检索” 01 "课程分数小于 60，按分数降序摆放的学生信息
13 按均匀成果从高到低显现一切学生的一切课程的成果以及均匀成果
14 查询各科成果最高分、最低分和均匀分：
以如下方式显现：课程 ID，课程 name，最高分，最低分，均匀分，及格率，中等率，优异率，优异率
及格为>=60，中等为：70-80，优异为：80-90，优异为：>=90
要求输出课程号和选修人数，查询成果按人数降序摆放，若人数相同，按课程号升序摆放
15 按各科成果进行排序，并显现排名， Score 重复时保存名次空缺
15.1 各科成果进行排序，并显现排名， Score 重复时兼并名次
16 查询学生的总成果，并进行排名，总分重复时保存名次空缺
16.1 查询学生的总成果，并进行排名，总分重复时不保存名次空缺
17 计算各科成果各分数段人数：课程编号，课程名称，[100-85]，[85-70]，[70-60]，[60-0] 及所占百分比
18 查询各科成果前三名的记载
19 查询每门课程被选修的学生数
20 查询出只选修两门课程的学生学号和名字
21 查询男生、女生人数
22 查询名字中含有「风」字的学生信息
23 查询同名同性学生名单，并计算同名人数
24 查询 1990 年出世的学生名单
25 查询每门课程的均匀成果，成果按均匀成果降序摆放，均匀成果相一起，按课程编号升序摆放 A8站源码交易平台
26 查询均匀成果大于等于 85 的一切学生的学号、名字和均匀成果
27 查询课程名称为「数学」，且分数低于 60 的学生名字和分数
28 查询一切学生的课程及分数状况（存在学生没成果，没选课的状况）
29 查询任何一门课程成果在 70 分以上的名字、课程名称和分数
30 查询不及格的课程
31 查询课程编号为 01 且课程成果在 80 分以上的学生的学号和名字
32 求每门课程的学生人数
33 成果不重复，查询选修「张三」教师所授课程的学生中，成果最高的学生信息及其成果
34 成果有重复的状况下，查询选修「张三」教师所授课程的学生中，成果最高的学生信息及其成果
35 查询不同课程成果相同的学生的学生编号、课程编号、学生成果
36 查询每门功成果最好的前两名
37 计算每门课程的学生选修人数（超越 5 人的课程才计算）。
38 检索至少选修两门课程的学生学号
39 查询选修了悉数课程的学生信息
40 查询各学生的年纪，只按年份来算
41 依照出世日期来算，当时月日 < 出世年月的月日则，年纪减一
42 查询本周过生日的学生
43 查询下周过生日的学生
44 查询本月过生日的学生
45 查询下月过生日的学生
创立操作
show databases
use sql_exercise
create table student(
Sid int primary key,
Sname varchar(10) not null,
Sage datetime,
Ssex enum(‘男’,‘女’)
)
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 , ‘孙风’ , ‘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’);
create table course(
Cid int primary key,
Cname varchar(10) not null,
Tid int
)
insert into Course values(01 , ‘语文’ , 02);
insert into Course values(02 , ‘数学’ , 01);
insert into Course values(03 , ‘英语’ , 03);
create table Teacher(
Tid int primary key,
Tname varchar(10) not null
)
insert into Teacher values(‘01’ , ‘张三’);
insert into Teacher values(‘02’ , ‘李四’);
insert into Teacher values(‘03’ , ‘王五’);
alter table course add foreign key (Tid) references Teacher(Tid);
create table SC(
Sid int,
Cid int,
score decimal(18,1),
foreign key(Sid) references student(Sid),
foreign key(Cid) references course(Cid)
)
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);
查询操作
#1 查询 01 课程比 02 课程成果高的学生的信息及课程分数
select s.sname, s.sage, s.ssex, sc.cid, sc.score from student s right join sc on s.sid = sc.sid where ((select score from sc sc1 where sc1.cid = 1 and sc1.sid = s.sid ) >(select score from sc sc2 where sc2.cid = 2 and sc2.sid = s.sid ))
#1.1 查询一起存在 01 课程和 02 课程的状况
select s.sname, s.sage, s.ssex, sc.cid, sc.score from student s right join sc on s.sid = sc.sid where (((select sc1.cid from sc sc1 where s.sid = sc1.sid and sc1.cid = 1) is not null) and ((select sc2.cid from sc sc2 where s.sid = sc2.sid and sc2.cid = 2) is not null))
#1.2 查询存在 01 课程但可能不存在 02 课程的状况(不存在时显现为 null )
select s.sname, s.sage, s.ssex, sc.cid, sc.score from student s right join sc on s.sid = sc.sid where (((select sc1.cid from sc sc1 where s.sid = sc1.sid and sc1.cid = 1) is not null) and ((select sc2.cid from sc sc2 where s.sid = sc2.sid and sc2.cid = 2) is null))
#1.3 查询不存在 01 课程但存在 02 课程的状况
select s.sname, s.sage, s.ssex, sc.cid, sc.score from student s right join sc on s.sid = sc.sid where (((select sc1.cid from sc sc1 where s.sid = sc1.sid and sc1.cid = 1) is null) and ((select sc2.cid from sc sc2 where s.sid = sc2.sid and sc2.cid = 2) is not null))
#2 查询均匀成果大于等于 60 分的同学的学生编号和学生名字和均匀成果
select s.sname , avg(score) from student s right join sc on s.sid = sc.sid GROUP BY s.sname having avg(score) > 60
#3 查询在 SC 表存在成果的学生信息
select s.sname , s.sage, s.ssex from student s right join sc on s.sid = sc.sid GROUP BY s.sname
select s.sid, s.sname , s.sage, s.ssex from student s where ((select sc.cid from sc where sc.sid = s.sid ORDER BY sc.cid limit 1) is null)
#4 查询一切同学的学生编号、学生名字、选课总数、一切课程的总成果(没成果的显现为 null )
select s.sid, s.sname, count(1), sum(score) from (student s left join sc on s.sid = sc.sid) GROUP BY s.sid ORDER BY sum(score) desc
#4.1 查有成果的学生信息
select * from (student s left join sc on s.sid = sc.sid) left join course c on sc.cid = c.cid
select * from student s left join sc on s.sid = sc.sid
union
select * from student s right join sc on s.sid = sc.sid
A8站源码交易平台
#5 查询「李」姓教师的数量
select *,count(1) from teacher t where t.Tname LIKE ‘李%’
#6 查询学过「张三」教师授课的同学的信息
select * from student s where exists (select sc.cid from (sc left join course c on sc.cid = c.cid) left join teacher t on c.tid =t.tid where sc.sid = s.sid and t.Tname =‘张三’ )
#7 查询没有学全一切课程的同学的信息
select * from student s where exists (select c1.cid from course c1 where c1.cid not in (select sc.cid from sc where sc.sid = s.sid))
#8 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 * from student s left join sc on s.sid = sc.sid where (sc.cid in (SELECT sc1.cid from sc sc1 where sc1.sid =01))
#9 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
select * from student s where (select count(1) from sc where sc.sid = s.sid and sc.cid in (SELECT sc1.cid from sc sc1 where sc1.sid =01))=3
select * from student s left join sc on s.sid = sc.sid left join course c on sc.cid = c.cid left join teacher t on c.tid = t.tid ORDER BY s.sid
#10 查询没学过"张三"教师教学的任一门课程的学生名字
select * from student s where not exists (select * from (sc left join course c on sc.Cid = c.cid) left join teacher t on c.tid = t.tid where (sc.sid = s.sid and t.tname = ‘张三’))
#11 查询两门及其以上不及格课程的同学的学号，名字及其均匀成果
select *from student s where (select count(1) from sc where sc.sid = s.sid) > 1
#12 检索 01 课程分数小于 60，按分数降序摆放的学生信息
select * from student s left join sc on s.sid = sc.sid where sc.cid = 01 and (select sc.score from sc where sc.sid = s.sid and sc.cid =01) <60 ORDER BY sc.score
#13 按均匀成果从高到低显现一切学生的一切课程的成果以及均匀成果
select s.sid, s.sname, sum(sc.score), AVG(sc.score) from (student s left join sc on s.sid = sc.sid) left join course c on sc.cid = c.cid where sc.cid
GROUP BY s.sid ORDER BY AVG(sc.score) desc
#14 查询各科成果最高分、最低分和均匀分：
#以如下方式显现：课程 ID，课程 name，最高分，最低分，均匀分，及格率，中等率，优异率，优异率
#及格为>=60，中等为：70-80，优异为：80-90，优异为：>=90
#要求输出课程号和选修人数，查询成果按人数降序摆放，若人数相同，按课程号升序摆放
select c.cid, c.cname, max(sc.score),min(sc.score),avg(sc.score) from (course c left join sc on c.cid = sc.cid) left join student s on sc.sid = s.sid
select c.cid, c.cname,sc.score,count(1) ‘选修人数’,count(if(sc.score>60,1,null))/count(1),count(if(sc.score>70 and sc.score <80,1,null))/count(1),count(if(sc.score>80 and sc.score <90,1,null))/count(1),count(if(sc.score>90,1,null))/count(1) from (course c left join sc on c.cid = sc.cid) left join student s on sc.sid = s.sid group by c.cid ORDER by count(1) desc, c.cid asc
#15 按各科成果进行排序，并显现排名， Score 重复时保存名次空缺
#15.1 各科成果进行排序，并显现排名， Score 重复时兼并名次
set @prev = null
set @ranknum = 0
select @ranknum := @ranknum +1 不重复名次, s.sname, c.cname,sc.score from (student s left join sc on s.sid = sc.sid ) left join course c on sc.cid = c.cid where c.cid = 01 order by @prev := sc.score desc
#16 查询学生的总成果，并进行排名，总分重复时保存名次空缺
#16.1 查询学生的总成果，并进行排名，总分重复时不保存名次空缺
set @prev = null
set @ranknum = 0
#17 计算各科成果各分数段人数：课程编号，课程名称，[100-85]，[85-70]，[70-60]，[60-0] 及所占百分比
select count(1),count(if(sc.score>85,1,null))/count(1),count(if(sc.score>70 and sc.score <85,1,null))/count(1),count(if(sc.score>60 and sc.score <70,1,null))/count(1), count(if(sc.score< 60,1,null))/count(1), c.cid, c.cname
from sc right join course c on sc.cid = c.cid GROUP BY c.cid
#18 查询各科成果前三名的记载
select s.sid, s.sname, sc.score,c.cname from student s right join sc on s.sid = sc.sid right join course c on sc.cid = c.cid where c.cid =01 ORDER BY sc.score desc limit 3
#19 查询每门课程被选修的学生数
select * from student s where ((select count(1) from sc where sc.sid = s.sid and sc.cid in (select c.cid from course c)) = (select count(1) from course c))
#20 查询出只选修两门课程的学生学号和名字
select * from student s where ((select count(1) from sc where sc.sid = s.sid and sc.cid in (select c.cid from course c)) = 2)
#21 查询男生、女生人数
select s.ssex, count(1) from student s GROUP BY s.ssex
#22 查询名字中含有「风」字的学生信息
select * from student s where s.sname like ‘%风%’
#23 查询同名同姓学生名单，并计算同名人数
select * from student s_ where exists (select s1_.sname from student s1_ where s1_.sid <> s_.sid and s1_.sname = s_.sname)
select count(1),s.sname from student s where exists (select s1.sname from student s1 where s1.sid <> s.sid and s1.sname = s.sname)
#24 查询 1990 年出世的学生名单
Select * from student s where year(s.Sage)=1990
#25 查询每门课程的均匀成果，成果按均匀成果降序摆放，均匀成果相一起，按课程编号升序摆放
select sc.cid, avg(sc.score) from sc group by sc.cid ORDER BY avg(sc.score) desc, sc.cid asc
#26 查询均匀成果大于等于 85 的一切学生的学号、名字和均匀成果
select s.Sid, s.sname, avg(sc.score) from student s left join sc on s.sid = sc.sid GROUP BY s.sid having avg(sc.score) > 85
#27 查询课程名称为「数学」，且分数低于 60 的学生名字和分数
select s.Sid, s.sname,sc.score from student s left join sc on s.sid = sc.sid where (((select c.cname from course c where c.cid= sc.cid) = ‘数学’) and sc.score < 60)
#28 查询一切学生的课程及分数状况（存在学生没成果，没选课的状况）
select s.sid, s.sname, c.cid, c.cname, sc.score from student s left join sc on s.sid = sc.sid left join course c on sc.cid = c.cid left join teacher t on c.tid = t.tid ORDER BY s.sid
#29 查询任何一门课程成果在 70 分以上的名字、课程名称和分数
select s.sname, c.cname, sc.score from student s left join sc on s.sid = sc.sid left join course c on sc.cid = c.cid where sc.score > 70 and c.cid = 01
#30 查询不及格的课程
select s.sid, s.sname, c.cname, sc.score from student s left join sc on s.sid = sc.sid left join course c on sc.Cid = c.cid where sc.score < 60
#31 查询课程编号为 01 且课程成果在 80 分以上的学生的学号和名字
select s.sid, s.sname, sc.cid,sc.score from student s left join sc on s.sid = sc.sid where sc.score >= 80 and sc.cid = 01
#32 求每门课程的学生人数
select count(1), c.cname from sc left join course c on sc.cid = c.cid group by c.cid
#33 成果不重复，查询选修「张三」教师所授课程的学生中，成果最高的学生信息及其成果
select * from student s left join sc on s.sid = sc.sid where sc.cid in (select c.cid from course c where c.tid=(select t.tid from teacher t where t.tname = ‘张三’)) order by sc.score desc limit 1
#34 成果有重复的状况下，查询选修「张三」教师所授课程的学生中，成果最高的学生信息及其成果
select * from student s left join sc on s.sid = sc.sid left join course c on sc.cid = c.cid where c.tid = (select t.tid from teacher t where t.tname = ‘张三’) and sc.score = (select max(sc.score) from sc left join course c on sc.cid = c.cid where c.tid = (select t.tid from teacher t where t.tname = ‘张三’))
#35 查询不同课程成果相同的学生的学生编号、课程编号、学生成果
select * from sc sc1 inner join sc sc2 on sc1.score = sc2.score where sc1.sid<sc2.sid
太费事
select s.sid, s.sname, sc.cid, sc.score from student s left join sc on s.sid = sc.sid where sc.score in (select sc1.score from student s1 left join sc sc1 on s1.sid = sc1.sid where sc1.cid = sc.cid and s1.sid <> s.sid) order by sc.cid
36 查询每门功成果最好的前两名
Select a.sid,c.cid,sc1.score from sc sc1 left join sc sc2 on (sc1.cid=sc2.cid and sc1.score<sc2.score) group="" by="" sc1.sid,sc1.cid="" having="" count(sc2.cid)<2="" order="" sc1.cid
where 是在两个表join完成后，再附上where条件
而 and 则是在表衔接前过滤A表或B表里边哪些记载契合衔接条件，一起会统筹是left join仍是right join。即
假如是左衔接的话，假如左面表的某条记载不契合衔接条件，那么它不进行衔接，可是依然留在成果会集（此刻右边部分的衔接成果为NULL）。on条件是在生成暂时表时运用的条件，它不论on中的条件是否为真，都会回来左面表中的记载。
主张尽量用where来过滤条件
#37 计算每门课程的学生选修人数（超越 5 人的课程才计算）。
select count(1), sc.cid from sc group by sc.cid having count(1) > 5
#38 检索至少选修两门课程的学生学号
select s.sid, s.sname from student s where (select count(1) from sc where sc.sid = s.sid)>2
#39 查询选修了悉数课程的学生信息
select * from student s where (select count(1) from sc where sc.sid = s.sid) = (select count(1) from course)
#40 查询各学生的年纪，只按年份来算
select s.sid, s.sname, 2019-year(s.sage) sage from student s
#41 依照出世日期来算学生年纪，当时月日 < 出世年月的月日则，年纪减一
select *,now(),if(date_add(now(), interval (convert(-year(now()),signed) +convert(year(s.sage),signed)) year)> s.sage, year(now())-year(s.sage),year(now())-year(s.sage)-1) from student s
#42 查询本周过生日的学生
select * from student s where week(date_add(s.sage, interval (convert(year(now()),signed) - convert(year(s.sage),signed)) year)) = week(now())
select s.sage, week(date_add(s.sage, interval (convert(year(now()),signed) - convert(year(s.sage),signed)) year)) , now(), week(now()) from student s
#43 查询下周过生日的学生
select * from student s where week(date_add(s.sage, interval (convert(year(now()),signed) - convert(year(s.sage),signed)) year)) = (week(now())+1)
44 查询本月过生日的学生
select s.sage from student s where month(s.sage) = month(now())
45 查询下月过生日的学生
select s.sage from student s where month(s.sage) = month(now())+2
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