HQL练习_02，经典sql50题

题目

Student(Sid,Sname,Sage,Ssex)学生表

Sid：学号

Sname：学生姓名

Sbirth：学生生日

Ssex：学生性别

01 赵雷 1990-01-01 男

02 钱电 1990-12-21 男

03 孙风 1990-05-20 男

04 李云 1990-08-06 男

05 周梅 1991-12-01 女

06 吴兰 1992-03-01 女

07 郑竹 1989-07-01 女

08 王菊 1990-01-20 女

Course(Cid,Cname,Tid)课程表

Cid：课程编号

Cname：课程名称

Tid：教师编号

01 语文 02

02 数学 01

03 英语 03

04 hadoop 01

 

Teacher(Tid,Tname)教师表

Tid：教师编号：

Tname：教师名字

01 张三

02 李四

03 王五

SC(Sid,Cid,score)成绩表

Sid：学号Cid：课程编号score：成绩

01    01    80
01    02    90
01    03    99
02    01    70
02    02    60
02    03    80
03    01    80
03    02    80
03    03    80
04    01    50
04    02    30
04    03    20
05    01    76
05    02    87
06    01    31
06    03    34
07    02    89

07    03    98
01    04    50
07    04    60

建表：

建表：
CREATE TABLE IF NOT EXISTS `myhive2.student2` (
sid int,
sname string,
sbirth string,
ssex string
) 
row format delimited 
fields terminated by '\t'
;

load data local inpath "/zgm/student2.txt" into table myhive2.student2;


CREATE TABLE IF NOT EXISTS myhive2.course2 (
cid int,
cname string,
tid int
) 
row format delimited 
fields terminated by '\t'
;

load data local inpath "/zgm/course2.txt" into table myhive2.course2;


CREATE TABLE IF NOT EXISTS myhive2.teacher2 (
tid int,
tname string
) 
row format delimited 
fields terminated by '\t'
;

load data local inpath "/zgm/teacher2.txt" into table myhive2.teacher2;



CREATE TABLE IF NOT EXISTS myhive2.sc2 (
sid int,
cid int,
score int
) 
row format delimited 
fields terminated by '\t'
;

load data local inpath "/zgm/sc2.txt" into table myhive2.sc2;

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 c.*,a.score as `01课程成绩`, b.score as `02课程成绩`
from student2 c
join sc2 a on c.sid=a.sid and a.cid=1
left join sc2 b on a.sid=b.sid and b.cid=2 
where a.score>b.score or b.sid is null;

2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数:

select c.*,b.score as `01课程成绩`, a.score as `02课程成绩`
from student2 c
join sc2 a on c.sid=a.sid and a.cid=2
left join sc2 b on a.sid=b.sid and b.cid=1 
where a.score>b.score or b.sid is null;

 

3、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 a.sid,a.sname,avg(b.score) as `平均成绩`
from student2 a
join sc2 b on a.sid=b.sid 
group by a.sid,a.sname
having `平均成绩`>=60;

4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩:
(包括有成绩的和无成绩的)

select a.sid,a.sname,avg(nvl(b.score,0)) as `平均成绩`   //或者nvl(avg(b,score),0) as `平均成绩`
from student2 a
left join sc2 b on a.sid=b.sid 
group by a.sid,a.sname
having `平均成绩`<=60;

结果：
a.sid   a.sname 平均成绩
4       李云    33.333333333333336
6       吴兰    32.5
8       王菊    0.0

-----------------------------------------------------
第二种不符合题目要求
select a.sid,a.sname,avg(b.score) as `平均成绩`
from student2 a
left join sc2 b on a.sid=b.sid 
group by a.sid,a.sname
having `平均成绩`<=60;

结果：
a.sid   a.sname 平均成绩
4       李云    33.333333333333336
6       吴兰    32.5

5、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:

select a.sid,a.sname,count(b.cid),sum(nvl(score,0))
from student2 a
left join sc2 b on a.sid=b.sid
group by a.sid,a.sname;

 

6、查询"李"姓老师的数量:

select count(1)
from 
teacher2 a
where a.tname like "李%";

7、查询学过"张三"老师授课的同学的信息:

//
select  d.*
from student2 d
where d.sid in(
select c.sid from
 sc2 c  
join course2 a on c.cid=a.cid
join teacher2 b on a.tid=b.tid
where b.tname='张三');

3个job
73.366 seconds

//
select  d.*
from student2 d
left join sc2 c on c.sid=d.sid 
join course2 a on c.cid=a.cid
join teacher2 b on a.tid=b.tid
where b.tname='张三' 
group by d.sid,d.sname,d.sbirth,d.ssex;

1个job，59.455 seconds

//
select  distinct d.*
from student2 d
left join sc2 c on c.sid=d.sid 
join course2 a on c.cid=a.cid
join teacher2 b on a.tid=b.tid
where b.tname='张三' 
;
1个job 54.246 seconds,

更正：不用left join 直接join就可以！！！

8、查询没学过"张三"老师授课的同学的信息:

 

 

select d.*
from student2 d
left join
(select c.sid
from sc2 c
join course2 a on c.cid=a.cid
join teacher2 b on a.tid=b.tid
where b.tname='张三' 
)t
on d.sid=t.sid
where t.sid is null;

3job  83.587 seconds, 

---------------------------------------------------------------------------------------
select d.sid,d.sname,d.sbirth,d.ssex
from student2 d
join teacher2 b on  b.tname='张三'   
join course2 a on b.tid=a.tid
left join sc2 c on d.sid=c.sid and a.cid=c.cid
group by d.sid,d.sname,d.sbirth,d.ssex
having sum(case when c.score is null then 0 else 1 end)=0;

1个job  55.531 seconds



join 后的情况：

d.sid   d.sname d.sbirth        d.ssex  b.tid   b.tname a.cname c.score
1       赵雷    1990-01-01      男      1       张三    数学    90
1       赵雷    1990-01-01      男      1       张三    hadoop  90
1       赵雷    1990-01-01      男      1       张三    hadoop  50
2       钱电    1990-12-21      男      1       张三    数学    60
2       钱电    1990-12-21      男      1       张三    hadoop  NULL
3       孙风    1990-05-20      男      1       张三    数学    80
3       孙风    1990-05-20      男      1       张三    hadoop  90
4       李云    1990-08-06      男      1       张三    数学    30
4       李云    1990-08-06      男      1       张三    hadoop  NULL
5       周梅    1991-12-01      女      1       张三    数学    87
5       周梅    1991-12-01      女      1       张三    hadoop  NULL
6       吴兰    1992-03-01      女      1       张三    数学    NULL
6       吴兰    1992-03-01      女      1       张三    hadoop  NULL
7       郑竹    1989-07-01      女      1       张三    数学    89
7       郑竹    1989-07-01      女      1       张三    hadoop  60
8       王菊    1990-01-20      女      1       张三    数学    NULL
8       王菊    1990-01-20      女      1       张三    hadoop  NULL

----------------------------------------------------------------------------------
select d.*
from student2 d
where d.sid not in 
(select c.sid
from sc2 c
join course2 a on c.cid=a.cid
join teacher2 b on a.tid=b.tid
where b.tname='张三' 
);
6个job

9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息:

select a.*
from student2 a 
 join sc2 b on a.sid=b.sid and b.cid=1
join sc2 c on a.sid=c.sid and c.cid=2;

1个job

10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息:

select a.*
from student2 a
join sc2 b on a.sid=b.sid and b.cid=1
left join sc2 c on a.sid=c.sid and c.cid=2
where c.sid is null;

 

11、查询没有学全所有课程的同学的信息:
–先查询出课程的总数量–再查询所需结果

select distinct a.*
from student2 a
left join course2 b
left join sc2 c on c.sid=a.sid and c.cid=b.cid
where c.sid is null ;

 

 

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 a.sid
from student2 a
left join sc2 b on b.sid=1
join sc2 c on a.sid=c.sid and b.cid=c.cid
where a.sid<>1
group by a.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

//以下都不行
select  a.sid,a.sname,a.sbirth,a.ssex
from student2 a
join sc2 b on a.sid=b.sid and a.sid <> 4
left join sc2 c on c.sid=4 and c.cid=b.cid
group by a.sid,a.sname,a.sbirth,a.ssex
having sum(case when c.sid is null then 1 else 0 end) <=0
;

select  a.sid,a.sname,b.cid,c.sid,c.cid
from student2 a
join sc2 b on a.sid=b.sid and a.sid <> 4
left join sc2 c on c.sid=4 and c.cid=b.cid;
group by a.sid,a.sname,a.sbirth,a.ssex
having sum(case when c.sid is null then 1 else 0 end) <=0
;


select a.sid,a.sname,b.sid,b.cid,c.sid,c.cid
from student2 a
join sc2 b on b.sid=4
right join sc2 c on c.sid=a.sid and c.cid=b.cid;
group by a.sid,a.sname,a.sbirth,a.ssex
having sum(case when c.sid is null then 1 else 0 end) <=0;  


select *
from student2 a
left join sc2 b on b.sid=5
full outer join sc2 c on a.sid=c.sid and b.cid=c.cid;
where a.sid<>1
group by a.sid; 

14、查询没学过"张三"老师讲授的任一门课程的学生姓名:

select a.sid,a.sname
from
student2 a
join teacher2 b on b.tname="李四"
join course2 c on c.tid=b.tid
left join sc2 d on d.sid=a.sid and d.cid=c.cid
group by a.sid,a.sname
having sum(if(d.score is not null,1,0))=0;

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 a.sid,a.sname,avg(b.score) 
from
student2 a 
left join sc2 b on a.sid=b.sid
where b.score<40
group by a.sid,a.sname
having count(*)>=2; 


//不过
这样求得平均成绩是小于60的哪几门的平均成绩

16、检索"01"课程分数小于60，按分数降序排列的学生信息:

select a.*,b.score
from student2 a
join sc2 b on a.sid=b.sid and b.cid=1
where b.score <60
order by b.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 a.*,avg(score) as av,
sum(if(b.cid=1,b.score,0)) as `01成绩`,
sum(if(b.cid=2,b.score,0)) as `02成绩`,
sum(if(b.cid=3,b.score,0)) as `03成绩`,
sum(if(b.cid=4,b.score,0)) as `04成绩`
from student2 a
left join sc2 b on a.sid=b.sid 
group by a.sid,a.sname,a.sbirth,a.ssex
order by av desc;
 

******18.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:

select 
a.cid,a.cname,
max(b.score) as `最高分`,
min(b.score) as `最低分`,
avg(b.score) as `平均分`,
round(sum(case when b.score >=60 then 1 else 0 end)/count(*),2) as `及格率`,
round(sum(case when b.score >60 and b.score <=60 then 1 else 0 end)/count(*),2) as `中等率`,
round(sum(case when b.score >80 and b.score <=90 then 1 else 0 end)/count(*),2) as `优良率`,
round(sum(case when b.score >90 then 1 else 0 end)/count(*),2) as `优秀率`
from course2 a
join sc2 b on a.cid=b.cid
group by a.cid,a.cname;

19、按各科成绩进行排序，并显示排名:– row_number() over()分组排序功能

select 
b.cid,a.sname,b.score,
row_number() over(partition by b.cid order by b.score desc) as `排名`
from student2 a 
join sc2 b on a.sid=b.sid
;

20、查询学生的总成绩并进行排名:
 

select 
a.sid,a.sname,sum(b.score) as ss
from student2 a 
join sc2 b on a.sid=b.sid
group by a.sid,a.sname
order by ss desc
;

21、查询不同老师所教不同课程平均分从高到低显示:

select
t.tname,t.cname,av,
row_number() over(partition by t.tname order by av desc) as rn 
from(
select 
a.tname,b.cname,avg(c.score) as av
from teacher2 a
join course2 b on a.tid=b.tid 
join sc2 c on  b.cid=c.cid
group by a.tname,b.cname
) t
;

这个好像写麻烦了

22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩:

select 
t.*
from
(
select 
a.*,b.score,
row_number() over(partition by b.cid order by b.score desc) as rn
from student2 a 
join sc2  b on a.sid=b.sid
) t
where t.rn=2 or t.rn=3;

23、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],

select
a.cname,
sum(case when b.score>85 and  b.score<=100  then 1 else 0 end) as `100-85`,
sum(case when b.score>70 and  b.score<=85  then 1 else 0 end) as `85-70`,
sum(case when b.score>60 and  b.score<=70  then 1 else 0 end) as `70-60`
from course2 a
join sc2 b on a.cid=b.cid
group by a.cname;

24、查询学生平均成绩及其名次:

select 
sname,av,
row_number() over(order by av desc) 
from 
(select 
a.sname,avg(b.score) as av
from student2 a 
join sc2 b on a.sid=b.sid
group by a.sname)t
;



优化：

select
a.sid,b.sname,avg(a.score) as av,
row_number() over(order by avg(a.score) desc) as rn
from sc2 a
join student2 b on a.sid=b.sid
group by a.sid,b.sname;


原来2个job，150s,现在两个job，106s

25、查询各科成绩前三名的记录三个语句
 

select 
t.*
from
(
select 
b.cid,a.sname,b.score,
row_number() over(partition by b.cid order by b.score desc) as rn
from student2 a 
join sc2  b on a.sid=b.sid
) t
where t.rn<4;

26、查询每门课程被选修的学生数:

select
a.cname,count(b.cid) as `选课人数`
from course2 a 
join sc2 b on a.cid=b.cid
group by a.cname;

27、查询出只有两门课程的全部学生的学号和姓名:

select
c.sname,count(1) as `选修门数`
from course2 a 
join sc2 b on a.cid=b.cid
join student2 c on b.sid=c.sid
group by c.sname
having `选修门数`=2;

 

28、查询男生、女生人数:
 

select 
a.ssex,count(1)
from student2 a
group by a.ssex;

29、查询名字中含有"风"字的学生信息:

select 
a.*
from student2 a
where a.sname like "%风%";

30、查询同名同性学生名单，并统计同名人数:

select 
a.sname,a.ssex
from student2 a
group by a.sname,a.ssex
having count(*)>1
;

 

31、查询1990年出生的学生名单:

select *
from student2 
where year(sbirth)=1990;

32、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 av
from sc2
group by cid
order by av desc,cid;

33、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,a.sname,avg(b.score) as av
from student2 a
join sc2 b on a.sid=b.sid
group by a.sid,a.sname
having av>=85
;

34、查询课程名称为"数学"，且分数低于60的学生姓名和分数:

select 
c.sname,b.score
from course2 a
join sc2 b on a.cid=b.cid and a.cname='数学' and b.score<60
join student2 c on b.sid=c.sid
; 

 

35、查询所有学生的课程及分数情况:

select 
b.sname,
sum(if(a.cid=1,a.score,0)) as `1号课程成绩`,
sum(if(a.cid=2,a.score,0)) as `2号课程成绩`,
sum(if(a.cid=3,a.score,0)) as `3号课程成绩`,
sum(if(a.cid=4,a.score,0)) as `4号课程成绩`
from sc2 a right join student2 b on a.sid=b.sid
group by b.sname;

36、查询任何一门课程成绩在70分以上的学生姓名、课程名称和分数:
 

select
a.sid,a.sname,c.cname,b.score
from 
(select
t1.sid ,t2.sname
from sc2 t1 join student2 t2 on t1.sid=t2.sid
group by t1.sid,t2.sname
having min(t1.score)>70) a
join sc2 b on a.sid=b.sid
join course2 c on b.cid=c.cid
;

37、查询课程不及格的学生:

select 
a.sname,b.score
from student2 a join
sc2 b on a.sid=b.sid
where b.score<60;

38、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 
a.sid,a.sname
from student2 a join
sc2 b on a.sid=b.sid
where b.score>80 and b.cid=1;

39、求每门课程的学生人数 
 

select
a.cid,count(a.sid) 
from sc2 a 
group by a.cid;

40、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.sname,b.score
from student2 a join sc2 b on a.sid=b.sid
join course2 c on b.cid=c.cid
join teacher2 d on c.tid=d.tid and d.tname="张三"
order by b.score desc 
limit 1;

41、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.sid,a.cid,a.score,b.sid,b.cid,a.score
from  sc2 a left join sc2 b 
on a.score=b.score
where a.cid<>b.cid and a.sid <>b.sid
;

42、查询每门课程成绩最好的前三名:

 

select
t.sname,t.rn
from(
select 
a.sname,
row_number() over(partition by b.cid order by b.score desc) as rn
from student2 a join sc2 b on a.sid=b.sid
)t
where t.rn<=3;

43、统计每门课程的学生选修人数（超过5人的课程才统计）:–

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 
a.cid,count(a.sid) as cs
from sc2 a 
group by a.cid
having cs>=5
order by cs desc ,cid asc;

44、检索至少选修两门课程的学生学号:

select 
a.sid
from student2 a join sc2 b on a.sid=b.sid
group by a.sid
having count(b.cid)>2;

45、查询选修了全部课程的学生信息:

 

select c.*
from student2 c
left semi join 
(
select 
a.sid
from sc2 a,(select count(*) as co from course2 ) b
group by a.sid,b.co
having count(*)=b.co
)t
on c.sid=t.sid
;

46、查询各学生的年龄(周岁):
– 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,
year(current_timestamp)-year(a.sbirth)-(case when date_format(current_timestamp,"mm-dd")

47、查询本周过生日的学生:
 

select
sname,sbirth
from student2 a 
where weekofyear(current_timestamp)=weekofyear(a.sbirth)
;

48、查询下周过生日的学生:

select
sname,sbirth
from student2 a 
where weekofyear(current_timestamp)+1=weekofyear(a.sbirth)
;

 

 49、查询本月过生日的学生:

select
sname,sbirth
from student2 a 
where month(current_timestamp)=month(a.sbirth)
;

 50、查询12月份过生日的学生:
 

select * from student2 where month(sbirth)=12;