MySQL的五十道练习题
时间:2019-08-14 13:01 来源:linux.it.net.cn 作者:IT
MySQL的五十道练习题
目录
数据表介绍
题目
创建表，语句可以使用创建数据库及表
查询语句，不保证一定对
数据表介绍
–1.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
–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 教师姓名
–4.成绩表
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);
练习题目
1 查询" 01 “课程比” 02 “课程成绩高的学生的信息及课程分数
1.1 查询同时存在” 01 “课程和” 02 “课程的情况
1.2 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 )
1.3 查询不存在” 01 “课程但存在” 02 “课程的情况
2 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
3 查询在 SC 表存在成绩的学生信息
4 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 )
4.1 查有成绩的学生信息
5 查询「李」姓老师的数量
6 查询学过「张三」老师授课的同学的信息
7 查询没有学全所有课程的同学的信息
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"老师讲授的任一门课程的学生姓名
11 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
12 检索” 01 "课程分数小于 60，按分数降序排列的学生信息
13 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
14 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
15 按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺
15.1 各科成绩进行排序，并显示排名， Score 重复时合并名次
16 查询学生的总成绩，并进行排名，总分重复时保留名次空缺
16.1 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
17 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
18 查询各科成绩前三名的记录
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姓名
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，并统计同名人数
24 查询 1990 年出生的学生名单
25 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
26 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
27 查询课程名称为「数学」，且分数低于 60 的学生姓名和分数
28 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）
29 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
30 查询不及格的课程
31 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
32 求每门课程的学生人数
33 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
34 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
35 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
36 查询每门功成绩最好的前两名
37 统计每门课程的学生选修人数（超过 5 人的课程才统计）。
38 检索至少选修两门课程的学生学号
39 查询选修了全部课程的学生信息
40 查询各学生的年龄，只按年份来算
41 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一
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
创建操作
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 课程成绩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 查询同时存在 01 课程和 02 课程的情况
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 课程的情况(不存在时显示为 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 课程的情况
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 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 表存在成绩的学生信息
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 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 查有成绩的学生信息
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
#5 查询「李」姓老师的数量
select *,count(1) from teacher t where t.Tname LIKE ‘李%’
#6 查询学过「张三」老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 查询没有学全所有课程的同学的信息
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"老师讲授的任一门课程的学生姓名
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，姓名及其平均成绩
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，按分数降序排列的学生信息
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
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
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 按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺
#15.1 各科成绩进行排序，并显示排名， Score 重复时合并名次
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 查询学生的总成绩，并进行排名，总分重复时保留名次空缺
#16.1 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
set @prev = null
set @ranknum = 0
#17 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
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 查询各科成绩前三名的记录
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姓名
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，并统计同名人数
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 年出生的学生名单
Select * from student s where year(s.Sage)=1990
#25 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 sc.cid, avg(sc.score) from sc group by sc.cid ORDER BY avg(sc.score) desc, sc.cid asc
#26 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 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 的学生姓名和分数
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）
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绩在 70 分以上的姓名、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 且课程成绩在 80 分以上的学生的学号和姓名
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
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
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 查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩
select * from sc sc1 inner join sc sc2 on sc1.score = sc2.score where sc1.sid<sc2.sid
太麻烦
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 查询每门功成绩最好的前两名
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 by sc1.cid
where 是在两个表join完成后，再附上where条件
而 and 则是在表连接前过滤A表或B表里面哪些记录符合连接条件，同时会兼顾是left join还是right join。即
假如是左连接的话，如果左边表的某条记录不符合连接条件，那么它不进行连接，但是仍然留在结果集中（此时右边部分的连接结果为NULL）。on条件是在生成临时表时使用的条件，它不管on中的条件是否为真，都会返回左边表中的记录。
建议尽量用where来过滤条件
#37 统计每门课程的学生选修人数（超过 5 人的课程才统计）。
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 查询选修了全部课程的学生信息
select * from student s where (select count(1) from sc where sc.sid = s.sid) = (select count(1) from course)
#40 查询各学生的年龄，只按年份来算
select s.sid, s.sname, 2019-year(s.sage) sage from student s
#41 按照出生日期来算学生年龄，当前月日 < 出生年月的月日则，年龄减一
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
(责任编辑：IT)
| MySQL的五十道练习题 目录 数据表介绍 题目 创建表，语句可以使用创建数据库及表 查询语句，不保证一定对 数据表介绍 –1.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 –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 教师姓名 –4.成绩表 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); 练习题目 1 查询" 01 “课程比” 02 “课程成绩高的学生的信息及课程分数 1.1 查询同时存在” 01 “课程和” 02 “课程的情况 1.2 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 ) 1.3 查询不存在” 01 “课程但存在” 02 “课程的情况 2 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 3 查询在 SC 表存在成绩的学生信息 4 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 ) 4.1 查有成绩的学生信息 5 查询「李」姓老师的数量 6 查询学过「张三」老师授课的同学的信息 7 查询没有学全所有课程的同学的信息 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"老师讲授的任一门课程的学生姓名 11 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 12 检索” 01 "课程分数小于 60，按分数降序排列的学生信息 13 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 14 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 15 按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺 15.1 各科成绩进行排序，并显示排名， Score 重复时合并名次 16 查询学生的总成绩，并进行排名，总分重复时保留名次空缺 16.1 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 17 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 18 查询各科成绩前三名的记录 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姓名 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，并统计同名人数 24 查询 1990 年出生的学生名单 25 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 26 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 27 查询课程名称为「数学」，且分数低于 60 的学生姓名和分数 28 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） 29 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 30 查询不及格的课程 31 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 32 求每门课程的学生人数 33 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 34 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 35 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 36 查询每门功成绩最好的前两名 37 统计每门课程的学生选修人数（超过 5 人的课程才统计）。 38 检索至少选修两门课程的学生学号 39 查询选修了全部课程的学生信息 40 查询各学生的年龄，只按年份来算 41 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一 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 创建操作 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 课程成绩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 查询同时存在 01 课程和 02 课程的情况 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 课程的情况(不存在时显示为 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 课程的情况 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 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 表存在成绩的学生信息 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 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 查有成绩的学生信息 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 #5 查询「李」姓老师的数量 select *,count(1) from teacher t where t.Tname LIKE ‘李%’ #6 查询学过「张三」老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 查询没有学全所有课程的同学的信息 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"老师讲授的任一门课程的学生姓名 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，姓名及其平均成绩 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，按分数降序排列的学生信息 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 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 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 按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺 #15.1 各科成绩进行排序，并显示排名， Score 重复时合并名次 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 查询学生的总成绩，并进行排名，总分重复时保留名次空缺 #16.1 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 set @prev = null set @ranknum = 0 #17 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 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 查询各科成绩前三名的记录 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姓名 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，并统计同名人数 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 年出生的学生名单 Select * from student s where year(s.Sage)=1990 #25 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 sc.cid, avg(sc.score) from sc group by sc.cid ORDER BY avg(sc.score) desc, sc.cid asc #26 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 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 的学生姓名和分数 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） 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绩在 70 分以上的姓名、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 且课程成绩在 80 分以上的学生的学号和姓名 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 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 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 查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩 select * from sc sc1 inner join sc sc2 on sc1.score = sc2.score where sc1.sid<sc2.sid 太麻烦 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 查询每门功成绩最好的前两名 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 by sc1.cid where 是在两个表join完成后，再附上where条件 而 and 则是在表连接前过滤A表或B表里面哪些记录符合连接条件，同时会兼顾是left join还是right join。即 假如是左连接的话，如果左边表的某条记录不符合连接条件，那么它不进行连接，但是仍然留在结果集中（此时右边部分的连接结果为NULL）。on条件是在生成临时表时使用的条件，它不管on中的条件是否为真，都会返回左边表中的记录。 建议尽量用where来过滤条件 #37 统计每门课程的学生选修人数（超过 5 人的课程才统计）。 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 查询选修了全部课程的学生信息 select * from student s where (select count(1) from sc where sc.sid = s.sid) = (select count(1) from course) #40 查询各学生的年龄，只按年份来算 select s.sid, s.sname, 2019-year(s.sage) sage from student s #41 按照出生日期来算学生年龄，当前月日 < 出生年月的月日则，年龄减一 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 (责任编辑：IT) |