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45道SQL练习题！
数据库准备
· 在服务器上创建好数据库后需要配置访问权限，否则会报错Access deniedfor user 'root'@'%' to database 'xxx'
· 解决方法：grant all on xxx.* to 'root'@'%'identified by 'password' with grant option;
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
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');
2.课程表
Course(CId,Cname,TId)
–CId 课程编号,Cname 课程名称,TId 教师编号
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');
3.教师表
Teacher(TId,Tname)
–TId 教师编号,Tname 教师姓名
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');
4.成绩表
SC(SId,CId,score)
–SId 学生编号,CId 课程编号,score 分数
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
select Student.*,01Score,02Score
from Student,
(select tb1.SId,01Score,02Score
from
(select SId,score as 01Score
from sc
where CId='01')as tb1,
(select SId,score as 02Score
from sc
where CId='02')as tb2
where tb1.SId=tb2.SId and 01Score>02Score)as tb3
where tb3.SId=Student.SId
1.1 查询同时存在” 01 “课程和” 02 “课程的情况
SELECT Student.*
FROM Student JOIN
(SELECT tb1.SId
FROM
(SELECT SId
FROM sc
WHERE CId='01') as tb1 join
(SELECT SId
FROM sc
WHERE CId='01') as tb2 on tb1.SId=tb2.SId) AS tb3
ON tb3.SId=Student.SId
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 )
可能不存在？？？“可能”可以作为条件吗？
1.3 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'
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 Student.Sname,tb1.*
from Student join
(select avg(score) as avg_score,SId
from sc
group by SId
HAVING avg(score)>60)as tb1
on Student.SId=tb1.SId
3.查询在 SC 表存在成绩的学生信息
select distinct Student.*
from Student JOIN
sc on Student.SId=sc.SId
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 )
SELECTStudent.Sname,Student.SId,ifnull(tb1.class_num,0),tb1.class_score
from Student left join
(select SId,count(1) as class_num,sum(score)as class_score
from sc
group by SId)as tb1
on Student.SId=tb1.SId
4.1查有成绩的学生信息
select Student.*
from Student
where Student.SId in
(select SId
from sc)
5.查询「李」姓老师的数量
select count(1)
from Teacher
where Tname like'李%'
6.查询学过「张三」老师授课的同学的信息
· 多表联合查询时，如果条件简单（没有分组之内的），可以考虑不用子查询
· 错误例子
select Student.*
from Student join
(select SId
from sc
where CId=
(select CId
from Course
where TId=
(select TId
from Teacher
where Tname like '张三')))as tb1
on Student.SId=tb1.SId
· 正确例子
select student.* from student,teacher,course,sc
where
student.sid= sc.sid
and course.cid=sc.cid
and course.tid = teacher.tid
and tname = '张三';
7.查询没有学全所有课程的同学的信息
· 没仔细思考需要查出的记录由哪几部分组成，没有选全包括选了部分，这个可以通过sc表得出，还包括没选课的，这个就需要计算了。
· 错误例子
select sid
from sc
group by sid
HAVING count(1)<
(select count(1)as class_num
from Course)
· 正确例子
select * from student
where student.sid not in (
select sc.sid from sc
group by sc.sid
having count(sc.cid)= (selectcount(cid) from course)
);
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
· 祥哥说的：少用子查询，效率低。
· 错误例子
select * from student
where student.sid in (
select sc.sid from sc
where sc.cid in(
select sc.cid from sc
where sc.sid = '01'
)
);
· 正确例子
select distinct student.*
from sc,student
where cid in(
select cid
from sc
where sid='01')
and sc.sid=student.sid
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 sid
from sc
where cid in(
SELECT cid
from sc
where sid='01')
and sid <> '01'
group by sid
having count(1)=(
select count(1)
from sc
where sid='01')
10.查询没学过”张三”老师讲授的任一门课程的学生姓名
· 每张表尽量只查询一次
select Student.*
from Student
where SId not in(
select sc.sid
from sc,teacher,course
where tname = '张三'
and teacher.tid=course.tid
and course.cid=sc.cid)
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 student.sid,avg(score),sname
from sc,student
where score<60
and student.SId=sc.sid
group by sid
having count(1)>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 student.*,score
from sc,student
where cid='01'
and student.sid=sc.sid
and score<60
order by score desc
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
· 这里同时使用了多个join
· 效果还不错
select student.sid,student.sname,tb1.cname,tb1.score,tb2.cname,tb2.score,tb3.cname,tb3.score,tb4.avg_score
from student
left join
(select sc.sid,cname,score
from sc,course
where sc.cid=course.cid
and course.cname='语文')as tb1
on student.sid=tb1.sid
left join
(select sc.sid,cname,score
from sc,course
where sc.cid=course.cid
and course.cname='数学')as tb2
on student.sid=tb2.sid
left join
(select sc.sid,cname,score
from sc,course
where sc.cid=course.cid
and course.cname='英语')as tb3
on student.sid=tb3.sid
left JOIN
(select sid,avg(score) as avg_score
from sc
group by sid) as tb4
on student.sid=tb4.sid
order by tb4.avg_score desc
14.查询各科成绩最高分、最低分和平均分：
select cname,max(score),min(score),avg(score)
from sc,course
where sc.cid=course.cid
group by sc.cid
14.1.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
· 利用sum，case when求及格率，中等率，优良率，优秀率
select cname,max(score),min(score),avg(score),
sum(case when sc.score>=60 then 1 else 0 end)/count(1) as 及格率,
sum(case when sc.score>=70 and sc.score<80then 1 else 0 end)/count(1) as 中等率,
sum(case when sc.score>=80 and sc.score<90then 1 else 0 end)/count(1) as 优良率,
sum(case when sc.score>=90 then 1 else 0 end)/count(1) as 优秀率
from sc,course
where sc.cid=course.cid
group by sc.cid
14.2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
· 多条件排序
select cid,count(1)
from sc
group by cid
order by COUNT(1) desc,cid
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
· 利用自交，达到排名的效果
select tb1.cid,tb1.sid,tb1.score,count(tb2.cid) asrank
from sc as tb1 left join sc as tb2 on tb1.score<tb2.score and tb1.cid=tb2.cid
group by tb1.cid,tb1.sid
order by cid,rank asc;
15.1 按各科成绩进行排序，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
· 使用变量以及if
set @r=0;
set @old=-1;
select Sname,tb1.total,tb1.rank
from(
select tb.*,if(@old<>tb.total,@r:=@r+1,@r)as rank,@old:=tb.total
from (
select sc.sid,sum(sc.score) astotal
from sc
group by sc.sid
order by total desc)as tb)as tb1,student
where student.SId=tb1.sid
ORDER BY tb1.rank
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
· 利用sum,case when求各分数段的人数以及百分比
select cname,tb.*
from course,(
select sc.cid,
sum(case when score<60 then 1 else 0 end)as '0-60',
sum(case when score<60 then 1 else 0 end)/count(1)*100 as '0-60百分比',
sum(case when score<70 and score>=60then 1 else 0 end)as '60-70',
sum(case when score<70 and score>=60then 1 else 0 end)/count(1)*100 as '60-70百分比',
sum(case when score<85 and score>=70then 1 else 0 end)as '70-85',
sum(case when score<85 and score>=70then 1 else 0 end)/count(1)*100 as '70-85百分比',
sum(case when score<=100 and score>=85then 1 else 0 end)as '85-100',
sum(case when score<=100 and score>=85then 1 else 0 end)/count(1)*100 as '85-100百分比'
from sc
group by sc.cid)as tb
where tb.cid=course.cid
18.查询各科成绩前三名的记录
· 利用where,count找前三名
select sc.*
from sc
where (
select count(1)
from sc as tb
where sc.cid=tb.cid and sc.score<tb.score
)<3
order by sc.cid,sc.score desc
· 利用自交找前三名
select a.sid,a.cid,a.score from sc a
left join sc b on a.cid = b.cid anda.score<b.score
group by a.cid, a.sid
having count(b.cid)<3
order by a.cid;
19.查询每门课程被选修的学生数
select cid, count(sid)
from sc
group by cid;
20.查询出只选修两门课程的学生学号和姓名
select sname, count(cid)
from sc ,student
where sc.sid=student.sid
group by sc.sid
having count(cid)=2;
21.查询男生、女生人数
· count(表达式)
select count(Ssex='男')as '男',count(Ssex='女') as '女'
from student
22.查询名字中含有「风」字的学生信息
select *
from student
where sname like '%风%'
23.查询同名同性学生名单，并统计同名人数
· 同名同姓：group by
select student.*,tb.count
from student,(
select sname,count(1)as count
from student
group by sname
having count(1)>1)as tb
where student.sname=tb.sname
24.查询 1990 年出生的学生名单
· 日期比较，可以使用字符串
select *
from student
where Sage between '1990-0-0 00:00:00' and '1991-0-0 00:00:00'
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 cid,avg(score)
from sc
group by cid
order by avg(score) desc,cid
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 sname,student.sid,avg(score)
from sc,student
where sc.sid=student.sid
group by sc.sid
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 sname,score
from student,sc,course
where cname='数学'
and student.sid=sc.sid
and course.cid=sc.cid
and 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）
· 当遇到“所有”这样的字眼时，应该考虑使用join
select sname,cname,score
from student left join sc on student.sid=sc.sid left join course on course.cid=sc.cid
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 sname,cname,score
from course left join sc on course.cid=sc.cid left join student on student.sid=sc.sid
where score>70
30.查询不及格的课程
select sname,cname,score
from course left join sc on course.cid=sc.cid left join student on student.sid=sc.sid
where 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 sc.sid,sname,score
from sc,student
where sc.sid=student.sid
and sc.cid = '01'
and score >80
32.求每门课程的学生人数
select cname,count(sid)
from sc,course
where sc.cid=course.cid
group by sc.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
· 最高：limit
select sname,score
fromteacher,sc,student,course
where tname='张三'
and teacher.tid=course.tid
and course.cid=sc.cid
and student.sid=sc.sid
order by 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
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 sc.*
from sc,(
select score,sid
from sc
group by score,sid
having count(score)>1 ) as tb
where sc.score=tb.score and sc.sid=tb.sid
36.查询每门功成绩最好的前两名
· 同18题
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 sc.cid,count(1)
from course left join sc on course.cid=sc.cid
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 sid
from sc
group by sid
having count(1)>=2
39.查询选修了全部课程的学生信息
select sid
from sc
group by sid
having count(1)=(select count(1) from course)
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
· 日期计算应使用函数TIMESTAMPDIFF
select *,TIMESTAMPDIFF(year,Sage,current_date)as age
from student
42.查询本周过生日的学生
· WEEKOFYEAR
select *
from student
where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE());
43.查询下周过生日的学生
select *
from student
where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE())+1;
44.查询本月过生日的学生
· MONTH
select *
from student
where MONTH(student.Sage)=MONTH(CURDATE());
45.查询下月过生日的学生
select *
from student
where MONTH(student.Sage)=MONTH(CURDATE())+1;
如果大家有其他的方法，或者想要知道哪方面的办公技巧，下方评论哦~
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