超經典sql練習題,在teradata上實現
阿新 • • 發佈:2018-08-05
count(0 like highlight student max 來源 1.2 生成 into
題目來源:https://blog.csdn.net/flycat296/article/details/63681089
teradata實現:
drop table student; create table student( s_id varchar(10), sname varchar(20), sage date, sex varchar(20) ); insert into Student values(‘01‘ , ‘趙雷‘ ,‘1990-01-01‘ , ‘男‘); insert into Student values(‘02‘ , ‘錢電‘ , ‘1990-12-21‘ , ‘男‘); insert into Student values(‘03‘ , ‘孫風‘ , ‘1990-05-20‘ , ‘男‘); insert into Student values(‘04‘ , ‘李雲‘ , ‘1990-08-06‘ , ‘男‘); insert into Student values(‘05‘ , ‘周梅‘ , ‘1991-12-01‘ , ‘女‘); insert into Student values(‘06‘ , ‘吳蘭‘ , ‘1992-03-01‘ , ‘女‘); insert into Student values(‘07‘ , ‘鄭竹‘ , ‘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‘ , ‘語文‘ , ‘02‘); insert into Course values(‘02‘ , ‘數學‘ , ‘01‘); insert into Course values(‘03‘ , ‘英語‘ , ‘03‘); select * from course; create table teacher( t_id varchar(10), tname varchar(20) ); insert into Teacher values(‘01‘ , ‘張三‘); 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=‘張三‘; /*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=‘張三‘))); /*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*難點:按平均成績從高到低顯示所有學生的所有課程的成績以及平均成績*/ 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:難點*/ 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‘%風%‘; /*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=‘數學‘ 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 %%成績不重復*/ 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=‘張三‘)) order by score desc; /*34 %%成績重復,查詢選修「張三」老師所授課程的學生中,成績最高的學生信息及其成績*/ 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=‘張三‘; /*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=‘張三‘ 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=‘張三‘ ; /*不同學生課程相同,分數相同*/ 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,查詢不同課程成績相同的學生的學生編號、課程編號、學生成績*/ 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,如何歸並成一行*/ 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;
超經典sql練習題,在teradata上實現