用到的表的介绍
student: sid（学生ID）,sname（学生姓名）,sage,ssex
course: cid（课程ID）,cname,tid（教师ID）
teacher: tid,tname
sc：sid,cid,score
只是附上题目和代码，没有运行结果
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
(select t1.sid,class1,class2
from
(select sid,score as class1 from sc where cid='01' )as t1
join
(select sid,score as class2 from sc where cid='02') as t2
on t1.sid=t2.sid and t1.class1>t2.class2
)r
left join student
on r.sid=student.sid
;
1.1 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
(select * from sc where cid='01')t1
left join
(select * from sc where cid='02')t2
on t1.sid=t2.sid
;
1.2 查询同时存在01和02课程的情况
select *
from
(select * from sc where cid='01'
)t1
join (select * from sc where cid='02'
)t2
on t1.sid =t2.sid
;
1.3 查询选择了02课程但没有01课程的情况
select * from sc where cid='02' and sid not in (
select sid from sc where cid='01')
;
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 t1.sid,t1.meanscore,student.sname
from
(select sid,avg(score)as meanscore
from sc
group by sid
having avg(score)>=60
)t1
left join student
on t1.sid=student.sid
;
3.查询在 SC 表存在成绩的学生信息
select distinct student.*
from sc
join student
on sc.sid=student.sid
;
4.查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和
select r.*,student.sname
from
(select sid, count(distinct cid) as qcourse,sum(score)as sumscore
from sc
group by sid
)r
left join student
on r.sid=student.sid
;
5.查询「李」姓老师的数量
select count(*) from teacher where tname like ('李%')
;
6.查询学过「张三」老师授课的同学的信息
select student.*
from
(select * from teacher where tname= '张三'
)t1
left join course
on t1.tid=course.tid
left join sc
on course.cid=sc.cid
left join student
on sc.sid=student.sid
;
@7.查询没有学全所有课程的同学的信息
select student.*,qcourse
from
(select sid,count(distinct cid) as qcourse
from sc
group by sid
having count(distinct cid)<(select count(distinct cid) from course)
)
left join student
on sc.sid=student.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 student.*
from sc
join student
on sc.sid=student.sid
where sc.cid in (select cid from sc where sid ='01'
)
and sc.sid !='01'
group by sc.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
10.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 not in(select sid from sc
left join course
on sc.cid=course.cid
left join teacher
on course.tid=teacher.tid
where teacher.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 t1.*,student.sname
from
(select sid,avg(score)as meanscore
from sc
where sid in (select sid,count(distinct cid)
from sc
where score<60
group by sid
having count(distinct cid)>=2)
)t1
left join student
on t1.sid=student.sid
;
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.*,t1.score
from
(select *
from sc
where cid='01'
and score<60
)t1
left join student
on t1.sid=student.sid
order by t1.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
select sc.*,t1.meanscore
from sc
left join
(select sid,avg(score) as meanscore
from sc
group by cid
)t1
on sc.sid=t1.sid
order by t1.meanscore
;
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 cid as '课程 ID',
count(sid) as '课程人数',
max(score) as '最高分',
min(score) as '最低分',
avg(score) as '平均分',
cast(cast sum(及格) as double)/(cast count(sid) as double) as '及格率',
SUM(中等) / COUNT(sid) AS '中等率',
SUM(优良) / COUNT(sid) AS '优良率',
SUM(优秀) / COUNT(sid) AS '优秀率'
from
(select *,
case when score >=60 then 1 else 0 end as '及格',
case when score>=70 and score<80 then 1 else 0 end as '中等 ',
case when score >=80 and score<90 then 1 else 0 end as '优良',
case when score >=90 then 1 else 0 end as '优秀'
from sc
)a
group by cid
order by count(sid)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
select a.*,count(a.score) as '排名'
from sc a
left join sc b
on a.sid=b.sid
and a.score < b.score
group by a.cid,a.sid,a.score
order by a.cid
;
--用sc中的score和自己进行对比，来计算“比当前分数高的分数有几个”。
@15.1 按各科成绩进行行排序，并显示排名， Score 重复时合并名次
select a.*,count(b.score)+1 as '排名'
from sc a
left join sc b
on a.sid=b.sid
and a.score < b.score
group by a.cid,a.sid,a.score
order by a.cid
;
@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
select a.*,@rank:=@rank+1 as rank
from
(select sid,
sum(score)
from sc
group by sid
order by sum(score)desc
)a,
(select @rank:=0
)b
;
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 a.*,course.cname
from
(select cid,
sum(case when score>=85 and score <=100 then 1 else 0 end )/count(distinct sid) as '[100-85]',
sum(case when score>=70 and score <85 then 1 else 0 end )/count(distinct sid) as '[85-75]',
sum(case when score>=60 and score <70 then 1 else 0 end )/count(distinct sid) as '[70-60]',
sum(case when score <60 then 1 else 0 end )/count(distinct sid) as '[60-0]'
from sc
group by cid
)a
left join course
on a.cid =course.cid
;
18.查询各科成绩前三名的记录
select a.*,count(b.score)+1 as ranking
from sc as a
left join sc as b
on a.cid=b.cid and a.score<b.score
group by a.cid,a.sid
having ranking <=3
order by a.cid,ranking
--having where不能使用别名,group by order by中可以使用别名
--
select *
from
(select *,row_number()over(partition by cid order by score desc)as ranking
from sc)
where ranking>=3
;
19.查询每门课程被选修的学生数
select cid,count(distinct sid) as num
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 sc.sid,student.sname
from sc
left join student
on sc.sid=student.sid
group by sc.cid
having count(cid)=2
;
21. 查询男生、女生人数
select count(case when ssex='男' then sid else null end)as '男生人数',
count(case when ssex='女' then sid else null end)as '女生人数'
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
select sname,count(ssex)
from student
group by sname
having count(ssex)>1
;
24.查询 1990 年出生的学生名单
select *
from student
where year(sage) like'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 cid,avg(score)as meanscore
from sc
group by cid
order by meanscore 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 a.sid,student.sname,a.meanscore
from
(select sid,avg(score) as meanscore
from sc
group by sid
having avg(score)>=85
)a
left join student
on a.sid=student.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 student.sname,b.socre
from
(select cid
from course
where cname ='数学'
)a
left join
(select *
from sc
where score<60
)b
on a.cid=b.cid
left join student
on b.sid=student.sid
;
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 sname,sc.cid,sc.score
from sc
join student
on sc.sid=student.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 sname,sc.cid,sc.score
from sc
join student
on sc.sid=student.sid
where sc.score >70
;
30.查询不及格的课程
select distinct cid
from sc
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
from
(select *
from sc
where cid='01'and score>80
)a
left join student
on a.sid=student.sid
;
32.求每门课程的学生人数
select cid, count(sid)as cnt
from sc
group by 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 student.*,sc.score
from
(select tid
from teacher
where tname='张三'
)a
left join course
on a.tid=course.tid
left join sc
on course.cid=sc.cid
left join student
on sc.sid=student.sid
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
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 distinct a.*
from sc as a
join sc as b
on a.score=b.score
and a.cid!=b.cid
;
36. 查询每门成绩最好的前两名
select *
from
(select *,
row_number()over(partition by cid order by score desc)as ranking
)
where ranking <=2
;
--方法2自交
select *
from sc
where (select count(*)
from sc as a
where sc.cid=a.cid
and sc.score<a.score
)<2
order by cid ,sc.score desc
;
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 cid,count(sid)
from sc
group by sc
having count(sid)>5
;
38.检索至少选修两门课程的学生学号
select cid,count(sid)
from sc
group by sc
having count(sid)>2
;
39.查询选修了全部课程的学生信息
select student.*
from
(select sid
from sc
where cid =(select count(distinct cid) from course
)
)a
left join student
on a.sid=student.sid
;
40.查询各学生的年龄，只按年份来算
select sname,year(now())-year(age) as '年级'
from student
;
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
select sname,
case when (date_format(now(),'%m-%d')-date_format(sage,'%m-%d')<0
then year(now())-year(sage)+1
else year(now())-year(sage)
end as '年龄'
from student
;
--date_format转化为时间戳，利用时间戳求差值
42.查询本周过生日的学生
select *
from student
where weekofyear(date_format(now(),'%y%m%d'))
=
weekofyear(date_format(sage,'%y%m%d')-date_format(sage,'%y')*10000+date_format(now(),'%y')*10000)
;
--出生日期的年份去掉，得到月份和日期，再拼接上当前的年份
--weekofyear函数是计算出当前日期所在周数
43. 查询下周过生日的学生
select *
from student
where weekofyear(date_format(now(),'%y%m%d'))+1
=
weekofyear(date_format(sage,'%y%m%d')-date_format(sage,'%y')*10000+date_format(now(),'%y')*10000)
;
44.查询本月过生日的学生
select sname
from student
where month(sage)=month(now())
;
45.查询下月过生日的学生
select sname
from student
where month(sage)=month(now())+1
;
来源：https://www.cnblogs.com/daxixixixi/p/12259464.html