超经典sql练习题，在teradata上实现

Posted 2020-12-28 papio

题目来源：https://blog.csdn.net/flycat296/article/details/63681089

teradata实现：

drop table student;
create table student(
s_id varchar(10),
sname varchar(20),
sage date,
sex varchar(20)
);

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‘ , ‘1990-05-20‘ , ‘男‘);
insert into Student values(‘04‘ , ‘李云‘ , ‘1990-08-06‘ , ‘男‘);
insert into Student values(‘05‘ , ‘周梅‘ , ‘1991-12-01‘ , ‘女‘);
insert into Student values(‘06‘ , ‘吴兰‘ , ‘1992-03-01‘ , ‘女‘);
insert into Student values(‘07‘ , ‘郑竹‘ , ‘1989-07-01‘ , ‘女‘);
insert into Student values(‘08‘ , ‘王菊‘ , ‘1990-01-20‘ , ‘女‘);
select * from student;

create table course(
c_id varchar(10),
cname varchar(20),
t_id 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‘);
select * from course;
create table teacher(
t_id varchar(10),
tname varchar(20)
);

insert into Teacher values(‘01‘ , ‘张三‘);
insert into Teacher values(‘02‘ , ‘李四‘);
insert into Teacher values(‘03‘ , ‘王五‘);
select * from teacher;
create table sc(
s_id varchar(10),
c_id 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);
select * from sc order by s_id , c_id;

/*1*/
select  t1.s_id,t1.c_id,t1.score,t2.c_id,t2.score 
from sc t1 inner join sc t2
on t1.s_id=t2.s_id 
and t1.c_id=‘01‘
and t2.c_id=‘02‘
and t1.score>t2.score;

select a.s_id,a.c_id,b.s_id,b.c_id 
from (select *from sc where c_id=‘01‘) a
left join (select * from sc where c_id=‘02‘) b
on a.s_id=b.s_id
where a.score >b.score;

/*1.1*/
select t1.s_id,t1.c_id,t2.s_id,t2.c_id
from sc t1 inner join sc t2
on t1.s_id=t2.s_id 
and t1.c_id=‘01‘
and t2.c_id=‘02‘;

select *from 
(select *from sc where c_id=‘01‘) a 
left join (select * from sc where c_id=‘02‘) b
on a.s_id=b.s_id 
where b.s_id is not null;

/*1.2*/
select *from 
(select *from sc where c_id=‘01‘) a 
left join (select * from sc where c_id=‘02‘) b
on a.s_id=b.s_id 
;

/*1.3*/
select * from 
sc where c_id=‘02‘
and s_id not in
(select s_id from sc where c_id=‘01‘);

/*2*/
select a.s_id,b.sname,avg(a.score)
from sc a left join student b
on a.s_id=b.s_id 
group by a.s_id,b.sname
having avg(a.score)>=60;

/*3*/
select * from student where s_id in (select s_id from sc group by s_id)

/*4*/
select a.s_id,a.sname,count(b.s_id),sum(b.score)
from student a left join sc b
on a.s_id=b.s_id
group by a.s_id , a.sname;

/*4.1*/

select a.s_id,b.s_id, a.countclass,a.totlescore
from
(select s_id,count(s_id) countclass,sum(score) totlescore from sc group by s_id) a
left join 
student b
on  a.s_id=b.s_id

 /*5*/
 select count(*) from teacher where tname like ‘李%‘;
 /*6*/
select *
from student a left join sc b
on a.s_id=b.s_id
left join course  c
on b.c_id= c.c_id
left join teacher  d
on c.t_id =d.t_id
where d.tname=‘张三‘;
/*7*/
select a.s_id,a.sname,a.sage,a.sex  from student a left join sc b 
on a.s_id =b.s_id 
having count(b.s_id)<3
group by a.s_id,a.sname,a.sage,a.sex 
;
/*8*/
select * from student where s_id 
in (select distinct s_id from sc where c_id in (select c_id from sc where s_id=‘01‘)  )

/*9*/
select * from student where 
s_id in(select s_id from sc where c_id in 
(select c_id from sc where s_id=‘01‘ ) and s_id<>‘01‘
group by s_id having count(s_id)>=3) ;

/*10*/
select * from student where s_id not in 
(select s_id from sc where c_id in 
(select c_id from course where t_id in 
(select t_id from teacher where tname=‘张三‘)));

/*11*/
select a.s_id,a.sname, b.avg_score
from student a right join (select s_id ,avg(score) avg_score from sc where score<60 group by s_id having count(score)>=2) b
on a.s_id =b.s_id; 

/*12*/
select a.s_id,a.sname,a.sage,a.sex, b.score 
from student  a right join sc b 
on a.s_id=b.s_id 
where b.c_id=‘01‘
and b.score<60
order by b.score desc;

/*13*难点：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_id, max(case c_id when ‘01‘ then score else 0 end ) a,
max(case c_id when ‘02‘ then score else 0 end ) b,
max(case c_id when ‘03‘ then score else 0 end) c,
avg(score)
from sc
group by s_id 
order by 5 desc;

/*14*/
select a.c_id,a.cname,b.highest,b.lowest,b.avgscore,c.jigelv,d.middle,e.excellent,f.great,g.people_number
from course a left join
(select c_id,max(score) highest, min(score) lowest ,avg(score) avgscore
from sc group by c_id) b
on a.c_id=b.c_id
left join 
(select c_id,(sum(case  when score>=60 then 1 else 0 end)*1.00/count(*)*100) jigelv  from sc group by c_id) c
on a.c_id=c.c_id
left join 
(select c_id ,(sum(case when score>=70 and score<80 then 1 else 0 end)*1.00/count(*)*100) middle from sc group by c_id) d
on a.c_id=d.c_id
left join
(select c_id,(sum(case when score>=80 and score<90 then 1 else 0 end )*1.00/count(*)*100) excellent from sc group by c_id)e
on a.c_id=e.c_id
left join
(select c_id,(sum(case when score>=90 then 1 else 0 end )*1.00/count(*)*100) great from sc group by c_id ) f
on a.c_id=f.c_id
left join 
(select c_id,count(*) people_number from sc group by c_id) g
on a.c_id=g.c_id
order by g.people_number,a.c_id;
/*15,15.1    row_number()over() rank()over() dense_rank()*/
select s_id, c_id,score,row_number()over(partition by c_id order by score desc ) rank1
from sc;


select s_id,c_id,score,rank()over(partition by c_id order by score desc) rank1
from sc;

select s_id,c_id,score,dense_rank() over(partition by c_id order by score desc) rank1
from sc;

/*16*/
select s_id,sum(score),rank()over(order by sum(score) desc) from sc group by s_id;

/*16.1*/
select s_id,sum(score),dense_rank()over(order by sum(score) desc) from sc group by s_id;

/*17*/
select c_id, sum(case when score<=60 then 1 else 0 end ) a1, 
(sum(case when score<=60 then 1 else 0 end)*1.00/count(*) ) a2
from sc group by c_id
order by c_id;

/*18*/
select * from 
(select c_id,s_id,score,rank()over( partition by c_id order by score desc ) rank1
from sc ) b where b.rank1<=3
/*方法二：难点*/
select a.c_id,a.s_id ,a.score,b.score from sc a
left join sc b 
on a.c_id=b.c_id and a.score<b.score
group by a.s_id,a.c_id,a.score
having count(b.s_id)<3
order by a.c_id,a.score desc;

select * from sc a where 
(select count(*) from sc where c_id=a.c_id and score>a.score)<3
order by a.c_id, a.score desc
/*19*/
select c_id ,count(*) from sc group by c_id;
/*20*/
select s_id from sc group by s_id having count(s_id)=2
/*21*/
select sex,count(sex) from student group by sex
/*22*/
select * from student where sname like‘%风%‘;
/*23*/

select a.s_id,b.countnumber from student a
left join (select sname,sex,count(*) countnumber from student group by sname,sex)b
on a.sname=b.sname
and a.sex=b.sex
where b.countnumber>1;
/*24 to_char()的使用*/
select s_id ,sname from student where to_char(sage,‘yyyy‘)=1990
/*25*/
select c_id,avg(score) avgscore from sc group by c_id order by avgscore desc,c_id;

/*26*/
select a.s_id,a.sname,avg(b.score) avgscore
from student a left join sc b
on a.s_id=b.s_id 
group by a.s_id ,a.sname
having avg(b.score)>85;
/*27*/
select a.sname,b.score

from 
student a left join sc b
on a.s_id=b.s_id
left join course c
on b.c_id=c.c_id 
where c.cname=‘数学‘
and b.score<60;

/*28*/
select a.s_id,a.sname,b.c_id,b.score,c.cname
from student a left join sc b
on a.s_id=b.s_id
left join course c
on b.c_id=c.c_id
order by a.s_id,b.c_id;
/*29%%*/
select a.sname,c.cname,b.score
from (select s_id,c_id,score from sc where score>70) b 
left join
course c on b.c_id=c.c_id
left join student a
on b.s_id=a.s_id;

/*30*/
select c_id,count(*)
from sc where  score<60
group by c_id
;
/*31*/
select a.c_id,count(*) from 
course a left join sc b
on a.c_id=b.c_id
group by a.c_id

/*32*/
select c_id,count(*) 
from 
sc group by c_id;

/*33 %%成绩不重复*/
select top 1* from sc 
where c_id in (select c_id from course where t_id in (select t_id from teacher where tname=‘张三‘))
order by score desc;

/*34 %%成绩重复，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 a.s_id,a.sname,b.score,b.c_id from 
(select c_id,max(score) maxscore
from sc  
group by c_id
 ) e
 left join 
 sc b 
 on e.maxscore=b.score and e.c_id=b.c_id
 left join course c
 on b.c_id=c.c_id
 left join teacher d
 on c.t_id=d.t_id
 left join student a
 on a.s_id=b.s_id 
 where d.tname=‘张三‘;
 
 /*dense_rank()*/
 select e.s_id,e.c_id,e.score
 from (select s_id ,c_id, score,dense_rank()over(partition by c_id order by s_id)rank1
 from sc) e
 left join course c
 on e.c_id=c.c_id
 left join teacher d
 on c.t_id=d.t_id 
 where d.tname=‘张三‘ and e.rank1=1;
 
 select   top 1*
 from (select s_id ,c_id, score,dense_rank()over(partition by c_id order by s_id)rank1
 from sc) e
 left join course c
 on e.c_id=c.c_id
 left join teacher d
 on c.t_id=d.t_id 
 where d.tname=‘张三‘ ;
 
 /*不同学生课程相同，分数相同*/
 select a.s_id,a.c_id,a.score,b.s_id,b.c_id,b.score from  sc a left join sc b 
 on a.score=b.score and a.s_id>b.s_id and a.c_id=b.c_id
 where b.score is not null;

 /*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 c.s_id,max(c.c_id) c_id,max(c.score) score from sc c
left join (select s_id ,avg(score)a from sc group by s_id)b
on c.s_id=b.s_id
where c.score=b.a
group by c.s_id
having count(0)=(select count(0) from sc where s_id=c.s_id)

 /*存在三行，如何归并成一行*/
 select a.s_id,a.c_id,b.s_id,b.c_id,a.score,b.score from
 (select s_id,c_id, score,rank()over(partition by s_id order by score) rank1 from sc) a
 inner join 
 (select s_id,c_id, score,rank()over(partition by s_id order by score) rank1 from sc)b
 on a.rank1=b.rank1
 and a.s_id=b.s_id
 and a.c_id<b.c_id;
 /*36*/
 select * from 
 (select s_id,c_id,score,rank()over(partition by c_id order by score) rank1
 from sc )a  where  a.rank1<3;
 
 /*37*/
select c_id,count(*)from sc
group by c_id
having count(*)>5;
 
 /*38*/
 select s_id from sc 
 group by s_id
 having count(c_id)>=2

 
/*39*/
select s_id from sc
group by s_id 
having count(c_id)=3;

/*40*/
select s_id,sname,extract(year from date)-extract(year from sage) age from student;