题目是网上经典的50道sql练习题,答案是自己写的(仅供参考)
字段名 | 含义 | 类型 |
---|---|---|
SId | 学生编号 | varchar(10) |
Sname | 学生姓名 | varchar(10) |
Sage | 出生年月 | datetime |
Ssex | 学生性别 | varchar(10) |
字段名 | 含义 | 类型 |
---|---|---|
CId | 课程编号 | varchar(10) |
Cname | 课程名称 | nvarchar(10) |
TId | 教师编号 | varchar(10) |
字段名 | 含义 | 类型 |
---|---|---|
TId | 教师编号 | varchar(10) |
Tname | 教师姓名 | varchar(10) |
字段名 | 含义 | 类型 |
---|---|---|
SId | 学生编号 | varchar(10) |
CId | 课程编号 | varchar(10) |
score | 分数 | decimal(18,1) |
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');
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');
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');
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);
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 查有成绩的学生信息
9 查询「李」姓老师的数量
10 查询学过「张三」老师授课的同学的信息
11 查询没有学全所有课程的同学的信息
12 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
13 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
14 查询没学过"张三"老师讲授的任一门课程的学生姓名
15 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
16 检索" 01 "课程分数小于 60,按分数降序排列的学生信息
17 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
18 查询各科成绩最高分、最低分和平均分:
19 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
20 按各科成绩进行排序,并显示排名, Score 重复时合并名次
(注意是左链接,安装我们的连接条件,如果是内连的话就会丢失第一名的信息)
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 查询每门课程被选修的学生数
25 查询出只选修两门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「风」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
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 查询任何一门课程成绩在 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 求每门课程的学生人数
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 查询每门功成绩最好的前两名
42 统计每门课程的学生选修人数(超过 5 人的课程才统计)
43 检索至少选修两门课程的学生学号
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
47 查询本周过生日的学生
48 查询下周过生日的学生
49 询本月过生日的学生
50 查询下月过生日的学生
Student(SId,Sname,Sage,Ssex) 学生表 //学生编号,学生姓名,学生年龄,学生性别
Course(CId,Cname,TId) 课程表 //课程编号,课程名称,教师编号
Teacher(TId,Tname) 教师表 // 教师编号,教师名字
SC(SId,CId,score) 成绩表 //学生编号,课程编号,成绩
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 *
from Student right join tmp
(select s1.SId,s1.CId,s1.score,s2.CId,s2.score
from SC s1 join SC s2
on s1.SId=s2.SId
where s1.CId='01' and s2.CId='02'
and s1.score>s2.score) as tmp
on Student.SId=tmp.SId
2 查询同时存在" 01 "课程和" 02 "课程的情况
select *
from SC s1 join SC s2
on s1.SId=s2.SId
where s1.CId='01' and s2.CId='02'
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 *
from SC s1 left join SC s2
on s1.SId=s2.SId
where s1.CId='01'
4 查询不存在" 01 "课程但存在" 02 "课程的情况
① select *
from SC s1 left join SC s2
on s1.SId=s2.SId
where s1.CId='02'
and s2.CId!='01'
and s2.CId!='02'
② select *
from SC S
where S.CId='02'
and S.SId not in (
select s1.CId
from SC s1
where s1.Cid='01'
)
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 Student.SId,Student.Sname,tmp.avg_score
from Student right join (
select avg(s1.score)as avg_score,s1.SId
from SC s1
group by s1.score
where avg_score>60) tmp
on Student.SId=tmp.SId
6 查询在 SC 表存在成绩的学生信息
1 select *
from student s1
where s1.SId in(
SELECT distinct s2.SId
from SC s2
where s2.score is not null
)
2 select distinct(s1.SId),s1.Sname,s1.Sage,s1.Ssex
from student s1 RIGHT JOIN SC s2
on s1.SId=s2.SId and s2.score is not null
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 * from
student left join
(SELECT SC.SId,count(*) as num,SUM(SC.score) as sum
from SC
GROUP BY SC.SId) as tmp
on student.SId=tmp.SId
8 查有成绩的学生信息
select * from
student left join
(SELECT SC.SId,count(*) as num,SUM(SC.score) as sum
from SC
GROUP BY SC.SId) as tmp
on student.SId=tmp.SId
where tmp.sum is not null
9 查询「李」姓老师的数量
select *
FROM teacher t
where t.Tname like '李%'
10 查询学过「张三」老师授课的同学的信息
1 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';
2 select * from student
where student.SId in (
select s1.SId
from SC s1
where s1.CId in (
select c.CId
from course c
where c.TId in(
select t.TId
FROM teacher t
where t.Tname = '张三')))
11 查询没有学全所有课程的同学的信息
SELECT * from student s right join (
select SC.SId,COUNT(SC.CId) num
from SC
GROUP BY SC.SId
HAVING num<(
SELECT COUNT(distinct Course.CId)
from Course)) as tmp
on tmp.SId=s.SId
12 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 right join(
select distinct s1.SId
from SC s1
where s1.CId in(
select CId
from SC
where SId='01'
) and s1.SId!='01') as tmp
on student.SId=tmp.SId
13 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
(select * from student where SId != '01') B inner join (select SId,group_concat(CId) gc from SC group by SId ORDER BY CId) A
on A.SId=B.SId where A.gc=(
select group_concat(CId)
from SC where SId='01'
ORDER BY CId
);
14 查询没学过"张三"老师讲授的任一门课程的学生姓名
select Sname from student where SId in(
select distinct SId
from SC
where CId not in(
select CId
from course
where TId in (
select TId
from teacher
where Tname='张三')))
15 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 s.SId,s.Sname,AVG(score) avg_score
from student s join SC
on s.SId = sc.SId
where s.SId in (
select SId
from sc
where score<60
group by SId
having count(score)>1
)
GROUP BY s.SId
16 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
where s.SId in(
select sid
from sc
where cid='01' and score<60
ORDER BY score desc)
17 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 sid,sum(score) tot_score, avg(score) avg_score
from sc
GROUP BY sid
ORDER BY avg_score desc;
18 查询各科成绩最高分、最低分和平均分:
select min(score),max(score),AVG(score)
from sc
GROUP BY cid
19 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 sc.CId,Cname,max(score),min(score),avg(score),
count(*) 选修人数,
sum(case when sc.score>=60 then 1 else 0 end)/count(score) as 及格率,
sum(case when score>=70 and score<80 then 1 else 0 end)/count(*) as 中等率,
sum(case when score>=80 and score<90 then 1 else 0 end)/count(*) as 优良率,
sum(case when score>=90 then 1 else 0 end)/count(*) as 优秀率
from sc join course
on sc.CId = course.CId
GROUP BY sc.CId
ORDER BY 选修人数 desc,sc.CId;
20 按各科成绩进行排序,并显示排名, Score 重复时合并名次
(注意是左链接,安装我们的连接条件,如果是内连的话就会丢失第一名的信息)
select s1.cid,s1.sid,s1.score,count(s2.score)+1 as rank
from sc s1 left join sc s2
on s1.score<s2.score and s1.cid=s2.cid
group by s1.cid,s1.sid
order by s1.cid,rank
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
set @rank:=0;
set @pre:=null;
select * ,
case
when @pre=tot then @rank
when @pre:=tot then @rank:=@rank + 1
end
as rank
from (
select sid,sum(score) as tot
from sc
GROUP BY sid
order by tot desc) tmp
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
select Cname,sc.CId,
sum(if(score<60,1,0))as '[0-60]',
sum(if(score<60,1,0))/count(*) as '%[0-60]',
sum(if(score>=60 and score <70,1,0)) as '[60-70]',
sum(if(score>=60 and score <70,1,0))/count(*) as '%[60-70]',
sum(if(score<85 and score>=70,1,0))as '[70-85]',
sum(if(score<85 and score>=70,1,0))/count(*)as '%[70-85]',
sum(if(score>=85,1,0)) as '[85-100]',
sum(if(score>=85,1,0))/count(*) as '%[85-100]'
from sc left join course
on sc.CId = course.CId
GROUP BY sc.CId
23 查询各科成绩前三名的记录
select s1.CId,s1.SId,s1.score,count(s1.score) as num
from sc s1 left join sc s2
on s1.cid=s2.cid and s1.score<s2.score
GROUP BY s1.cid,s1.sid
HAVING num<3
ORDER BY s1.cid,s1.score desc;
24 查询每门课程被选修的学生数
select cid,count(sid)
from sc
GROUP BY cid;
25 查询出只选修两门课程的学生学号和姓名
select sc.SId,Sname
from sc join student
on sc.SId=student.SId
GROUP BY sid
having count(cid)=2
26 查询男生、女生人数
select Ssex,count(Ssex) as num
from student
GROUP BY Ssex
27 查询名字中含有「风」字的学生信息
select *
from student
where Sname like '%风%'
28 查询同名同性学生名单,并统计同名人数
select *
from student
GROUP BY Sname,Ssex
HAVING count(*)>1
29 查询 1990 年出生的学生名单
1 select *
from student
where Sage like '1990%'
2 select *
from student
where YEAR(Sage)=1990;
30 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) as avg_score
from sc
GROUP BY cid
order by avg_score desc,cid;
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
select student.*,avg(score) avg_score
from sc join student
on sc.SId=student.SId
GROUP BY sc.sid
having avg_score>=85
32 查询课程名称为「数学」,且分数低于 60 的学生姓名和分数
SELECT student.Sname,sc.score
from student join sc join course
on student.SId=sc.SId and sc.CId=course.CId
where course.Cname='数学' and sc.score<60
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)
SELECT Sname,sc.CId,sc.score
from student left join sc
on student.SId=sc.SId
34 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,sc.score
from student join sc join course
on student.SId=sc.SId and sc.CId=course.CId
where sc.score>70
35 查询不及格的课程
select cid
from sc
where score<60
GROUP BY cid
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
select student.SId,student.Sname
from sc,student
where sc.CId='01' and score>=80
and sc.SId=student.SId
37 求每门课程的学生人数
select cid,count(sid) as num
from sc
GROUP BY cid
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
select student.*,max(score)
from teacher join course
on teacher.TId=course.TId
join sc on sc.CId=course.CId
join student on student.SId=sc.SId
where Tname='张三'
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
select student.*,score=max(score)
from teacher join course
on teacher.TId=course.TId
join sc on sc.CId=course.CId
join student on student.SId=sc.SId
where Tname='张三'
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
select *
from sc
GROUP BY cid,score
having count(sid)>1
41 查询每门功成绩最好的前两名
#这里我理解的排名允许并列,如果两个的分数相同,那么他们排名相同。所以需要加上distinct
select s1.CId,s1.SId,s1.score
from sc s1 left join sc s2
on s1.CId=s2.CId and s1.score<s2.score
GROUP BY s1.CId,s1.SId
having count(DISTINCT s2.score)<2
order by s1.CId, s1.score desc;
42 统计每门课程的学生选修人数(超过 5 人的课程才统计)
SELECT CId,SId,COUNT(SId) num
from sc
GROUP BY CId
HAVING count(SId)>5
43 检索至少选修两门课程的学生学号
SELECT SId
from sc
GROUP BY SId
HAVING count(CId)>=2
44 查询选修了全部课程的学生信息
SELECT student.*
from sc left join student
on sc.SId=student.SId
GROUP BY SId
HAVING count(CId)=(
SELECT count(*) from course
)
45 查询各学生的年龄,只按年份来算
SELECT Sname,year(CURRENT_DATE) - year(Sage) as age
from student
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
/*第一个参数可以控制返回的单位,
MICROSECOND 微秒
SECOND 秒
MINUTE 分钟
HOUR 小时
DAY 天
WEEK 周
MONTH 月份
YEAR 年份*/
SELECT Sname,TIMESTAMPDIFF(YEAR,Sage,CURRENT_DATE) as age
from student
47 查询本周过生日的学生
SELECT *
from student
where WEEKOFYEAR(CURRENT_DATE)=WEEKOFYEAR(Sage)
48 查询下周过生日的学生
/*
weekofyear的返回值是0-53,所以如果是-1的话,可能会出现负数,于是变成+1
又因为+1可能会超过53,所以应该对53取模
*/
SELECT *
from student
where WEEKOFYEAR(Sage)=(WEEKOFYEAR(CURRENT_DATE)+1)%53
49 询本月过生日的学生
SELECT *
from student
where MONTH(Sage)=MONTH(CURRENT_DATE)
50 查询下月过生日的学生
/*
和48一样的道理
*/
SELECT *
from student
where MONTH(Sage)=(MONTH(CURRENT_DATE)+1)%12