Hive SQL50道练习题

关键时间，第一时间送达！

来源：大数据技术与架构

接着上一篇 Hive使用必知必会系列，本次分享一个Hive SQL 50道练习题。

建表

create table student(s_id string,s_name string,s_birth string,s_sex string) row format delimited fields terminated by '\t';

create table course(c_id string,c_name string,t_id string) row format delimited fields terminated by '\t';

create table teacher(t_id string,t_name string) row format delimited fields terminated by '\t';

create table score(s_id string,c_id string,s_score int) row format delimited fields terminated by '\t';

生成数据

vi /export/data/hivedatas/student.csv

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

vi /export/data/hivedatas/course.csv

01 语文 02
02 数学 01
03 英语 03

vi /export/data/hivedatas/teacher.csv

01  张三
02  李四
03  王五

vi /export/data/hivedatas/score.csv

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

导数据到hive

load data local inpath '/export/data/hivedatas/student.csv' into table student;

load data local inpath '/export/data/hivedatas/course.csv' into table course;

load data local inpath '/export/data/hivedatas/teacher.csv' into table teacher;

load data local inpath '/export/data/hivedatas/score.csv' into table score;

–注:–hive查询语法

SELECT [ALL | DISTINCT] select_expr, select_expr, ...
FROM table_reference
[WHERE where_condition]
[GROUP BY col_list [HAVING condition]]
[CLUSTER BY col_list
| [DISTRIBUTE BY col_list] [SORT BY| ORDER BY col_list]
]
[LIMIT number]

– 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 student.*,a.s_score as 01_score,b.s_score as 02_score
from student
  join score a on student.s_id=a.s_id and a.c_id='01'
  left join score b on student.s_id=b.s_id and b.c_id='02'
where  a.s_score>b.s_score;

–答案2

select student.*,a.s_score as 01_score,b.s_score as 02_score
from student
join score a on a.c_id='01'
join score b on b.c_id='02'
where a.s_id=student.s_id and b.s_id=student.s_id and a.s_score>b.s_score;

– 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 student.*,a.s_score as 01_score,b.s_score as 02_score
from student
join score a on student.s_id=a.s_id and a.c_id='01'
left join score b on student.s_id=b.s_id and b.c_id='02'
where a.s_score<b.s_score;

–答案2

select student.*,a.s_score as 01_score,b.s_score as 02_score
from student
join score a on a.c_id='01'
join score b on b.c_id='02'
where a.s_id=student.s_id and b.s_id=student.s_id and a.s_score<b.s_score;

– 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  student.s_id,student.s_name,tmp.平均成绩 from student
  join (
    select score.s_id,round(avg(score.s_score),1)as 平均成绩
        from score group by s_id)as tmp
  on tmp.平均成绩>=60
where student.s_id = tmp.s_id

–答案2

select student.s_id,student.s_name,round(avg (score.s_score),1) as 平均成绩 from student
join score on student.s_id = score.s_id
group by student.s_id,student.s_name
having avg (score.s_score) >= 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  student.s_id,student.s_name,tmp.avgScore from student
join (
select score.s_id,round(avg(score.s_score),1)as avgScore from score group by s_id)as tmp
on tmp.avgScore < 60
where student.s_id=tmp.s_id
union all
select  s2.s_id,s2.s_name,0 as avgScore from student s2
where s2.s_id not in
    (select distinct sc2.s_id from score sc2);

–答案2

select score.s_id,student.s_name,round(avg (score.s_score),1) as avgScore from student
inner join score on student.s_id=score.s_id
group by score.s_id,student.s_name
having avg (score.s_score) < 60
union all
select s2.s_id,s2.s_name,0 as avgScore from student s2
where s2.s_id not in
(select distinct sc2.s_id from score sc2);

– 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 student.s_id,student.s_name,(count(score.c_id) )as total_count,sum(score.s_score)as total_score
from student
left join score on student.s_id=score.s_id
group by student.s_id,student.s_name ;

– 6、查询'李'姓老师的数量:

select t_name,count(1) from teacher where t_name like '李%' group by t_name;

– 7、查询学过'张三'老师授课的同学的信息:

select student.* from student
join score on student.s_id =score.s_id
join  course on course.c_id=score.c_id
join  teacher on course.t_id=teacher.t_id and t_name='张三';

– 8、查询没学过'张三'老师授课的同学的信息:

select student.* from student
left join (select s_id from score
join course on course.c_id=score.c_id
join teacher on course.t_id=teacher.t_id and t_name='张三')tmp
on student.s_id =tmp.s_id
where tmp.s_id is null;

– 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 * from student
join (select s_id from score where c_id =1 )tmp1
    on student.s_id=tmp1.s_id
join (select s_id from score where c_id =2 )tmp2
    on student.s_id=tmp2.s_id;

– 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 student.* from student
join (select s_id from score where c_id =1 )tmp1
on student.s_id=tmp1.s_id
left join (select s_id from score where c_id =2 )tmp2
on student.s_id =tmp2.s_id
where tmp2.s_id 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

   select count(1) from course;

–再查询所需结果

select student.* from student
left join(
select s_id
from score
group by s_id
having count(c_id)=3)tmp
on student.s_id=tmp.s_id
where tmp.s_id is null;

–方法二(一步到位):

select student.* from student
join (select count(c_id)num1 from course)tmp1
left join(
      select s_id,count(c_id)num2
        from score group by s_id)tmp2
on student.s_id=tmp2.s_id and tmp1.num1=tmp2.num2
where tmp2.s_id 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 student.* from student
join (select c_id from score where score.s_id=01)tmp1
join (select s_id,c_id from score)tmp2
on tmp1.c_id =tmp2.c_id and student.s_id =tmp2.s_id
where student.s_id not in('01')
group by student.s_id,s_name,s_birth,s_sex;

– 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:
–备注:hive不支持group_concat方法,可用 concat_ws(’|’, collect_set(str)) 实现

select student.*,tmp1.course_id from student
join (select s_id ,concat_ws('|', collect_set(c_id)) course_id from score
      group by s_id having s_id not in (1))tmp1
  on student.s_id = tmp1.s_id
join (select concat_ws('|', collect_set(c_id)) course_id2
            from score  where s_id=1)tmp2
      on tmp1.course_id = tmp2.course_id2;

– 14、查询没学过'张三'老师讲授的任一门课程的学生姓名:

select student.* from student
left join (select s_id from score
join (select c_id from course join teacher on course.t_id=teacher.t_id and t_name='张三')tmp2
on score.c_id=tmp2.c_id )tmp
on student.s_id = tmp.s_id
where tmp.s_id is null;

– 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 student.s_id,student.s_name,tmp.avg_score from student
inner join (select s_id from score
      where s_score<60
        group by score.s_id having count(s_id)>1)tmp2
on student.s_id = tmp2.s_id
left join (
    select s_id,round(AVG (score.s_score)) avg_score
      from score group by s_id)tmp
      on tmp.s_id=student.s_id;

– 16、检索'01'课程分数小于60，按分数降序排列的学生信息:

select student.*,s_score from student,score
where student.s_id=score.s_id and s_score<60 and c_id='01'
order by s_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.s_id,tmp1.s_score as chinese,tmp2.s_score as math,tmp3.s_score as english,
    round(avg (a.s_score),2) as avgScore
from score a
left join (select s_id,s_score  from score s1 where  c_id='01')tmp1 on  tmp1.s_id=a.s_id
left join (select s_id,s_score  from score s2 where  c_id='02')tmp2 on  tmp2.s_id=a.s_id
left join (select s_id,s_score  from score s3 where  c_id='03')tmp3 on  tmp3.s_id=a.s_id
group by a.s_id,tmp1.s_score,tmp2.s_score,tmp3.s_score order by avgScore 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:
–及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90

select course.c_id,course.c_name,tmp.maxScore,tmp.minScore,tmp.avgScore,tmp.passRate,tmp.moderate,tmp.goodRate,tmp.excellentRates from course
join(select c_id,max(s_score) as maxScore,min(s_score)as minScore,
round(avg(s_score),2) avgScore,
round(sum(case when s_score>=60 then 1 else 0 end)/count(c_id),2)passRate,
round(sum(case when s_score>=60 and s_score<70 then 1 else 0 end)/count(c_id),2) moderate,
round(sum(case when s_score>=70 and s_score<80 then 1 else 0 end)/count(c_id),2) goodRate,
round(sum(case when s_score>=80 and s_score<90 then 1 else 0 end)/count(c_id),2) excellentRates
from score group by c_id)tmp on tmp.c_id=course.c_id;

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

select s1.*,row_number()over(order by s1.s_score desc) Ranking
    from score s1 where s1.c_id='01'order by noRanking asc
union all select s2.*,row_number()over(order by s2.s_score desc) Ranking
    from score s2 where s2.c_id='02'order by noRanking asc
union all select s3.*,row_number()over(order by s3.s_score desc) Ranking
    from score s3 where s3.c_id='03'order by noRanking asc;

– 20、查询学生的总成绩并进行排名:

select score.s_id,s_name,sum(s_score) sumscore,row_number()over(order by sum(s_score) desc) Ranking
from score ,student
where score.s_id=student.s_id
group by score.s_id,s_name order by sumscore desc;

– 21、查询不同老师所教不同课程平均分从高到低显示:
– 方法1

select course.c_id,course.t_id,t_name,round(avg(s_score),2)as avgscore from course    join teacher on teacher.t_id=course.t_id    join score on course.c_id=score.c_id    group by course.c_id,course.t_id,t_name order by avgscore desc;

– 方法2

select course.c_id,course.t_id,t_name,round(avg(s_score),2)as avgscore from course,teacher,score where teacher.t_id=course.t_id and course.c_id=score.c_id    group by course.c_id,course.t_id,t_name order by avgscore desc;

– 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 tmp1.* from    (select * from score where c_id='01' order by s_score desc limit 3)tmp1    order by s_score asc limit 2union all select tmp2.* from    (select * from score where c_id='02' order by s_score desc limit 3)tmp2    order by s_score asc limit 2union all select tmp3.* from    (select * from score where c_id='03' order by s_score desc limit 3)tmp3    order by s_score asc limit 2;

– 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],[0-60]及所占百分比

select c.c_id,c.c_name,tmp1.s0_60, tmp1.percentum,tmp2.s60_70, tmp2.percentum,tmp3.s70_85, tmp3.percentum,tmp4.s85_100, tmp4.percentumfrom course cjoin(select c_id,sum(case when s_score<60 then 1 else 0 end )as s0_60, round(100*sum(case when s_score<60 then 1 else 0 end )/count(c_id),2)as percentum from score group by c_id)tmp1 on tmp1.c_id =c.c_idleft join(select c_id,sum(case when s_score<70 and s_score>=60 then 1 else 0 end )as s60_70, round(100*sum(case when s_score<70 and s_score>=60 then 1 else 0 end )/count(c_id),2)as percentum from score group by c_id)tmp2 on tmp2.c_id =c.c_idleft join(select c_id,sum(case when s_score<85 and s_score>=70 then 1 else 0 end )as s70_85, round(100*sum(case when s_score<85 and s_score>=70 then 1 else 0 end )/count(c_id),2)as percentum from score group by c_id)tmp3 on tmp3.c_id =c.c_idleft join(select c_id,sum(case when s_score>=85 then 1 else 0 end )as s85_100, round(100*sum(case when s_score>=85 then 1 else 0 end )/count(c_id),2)as percentum     from score group by c_id)tmp4 on tmp4.c_id =c.c_id;

– 24、查询学生平均成绩及其名次:

select tmp.*,row_number()over(order by tmp.avgScore desc) Ranking from  (select student.s_id,          student.s_name,          round(avg(score.s_score),2) as avgScore  from student join score  on student.s_id=score.s_id  group by student.s_id,student.s_name)tmporder by avgScore desc;

– 25、查询各科成绩前三名的记录

–课程id为01的前三名

select score.c_id,course.c_name,student.s_name,s_score from scorejoin student on student.s_id=score.s_idjoin course on score.c_id='01' and course.c_id=score.c_idorder by s_score desc limit 3;  

–课程id为02的前三名

select score.c_id,course.c_name,student.s_name,s_score from scorejoin student on student.s_id=score.s_idjoin course on  score.c_id='02' and course.c_id=score.c_idorder by s_score desc limit 3; 

–课程id为03的前三名

select score.c_id,course.c_name,student.s_name,s_score from scorejoin student on student.s_id=score.s_idjoin course on score.c_id='03' and course.c_id=score.c_id order by s_score desc limit 3;

– 26、查询每门课程被选修的学生数:

select c.c_id,c.c_name,tmp.number from course c    join (select c_id,count(1) as number from score        where score.s_score<60 group by score.c_id)tmp    on tmp.c_id=c.c_id;

– 27、查询出只有两门课程的全部学生的学号和姓名:

select st.s_id,st.s_name from student st join (select s_id from score group by s_id having count(c_id) =2)tmp    on st.s_id=tmp.s_id;

– 28、查询男生、女生人数:

select tmp1.man,tmp2.women from    (select count(1) as man from student where s_sex='男')tmp1,    (select count(1) as women from student where s_sex='女')tmp2;

– 29、查询名字中含有'风'字的学生信息:

select * from student where s_name like '%风%';

– 30、查询同名同性学生名单，并统计同名人数:

select s1.s_id,s1.s_name,s1.s_sex,count(*) as sameNamefrom student s1,student s2where s1.s_name=s2.s_name and s1.s_id<>s2.s_id and s1.s_sex=s2.s_sexgroup by s1.s_id,s1.s_name,s1.s_sex;

– 31、查询1990年出生的学生名单:

select * from student where s_birth like '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 score.c_id,c_name,round(avg(s_score),2) as avgScore from score  join course on score.c_id=course.c_id    group by score.c_id,c_name order by avgScore desc,score.c_id asc;

– 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 score.s_id,s_name,round(avg(s_score),2)as avgScore from score join student on student.s_id=score.s_id    group by score.s_id,s_name having avg(s_score) >= 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 s_name,s_score as mathScore from student    join (select s_id,s_score            from score,course            where score.c_id=course.c_id and c_name='数学')tmp    on tmp.s_score < 60 and student.s_id=tmp.s_id;

– 35、查询所有学生的课程及分数情况:

select a.s_name, SUM(case c.c_name when '语文' then b.s_score else 0 end ) as chainese, SUM(case c.c_name when '数学' then b.s_score else 0 end ) as math, SUM(case c.c_name when '英语' then b.s_score else 0 end ) as english, SUM(b.s_score) as sumScore from student a join score b on a.s_id=b.s_id join course c on b.c_id=c.c_id    group by s_name,a.s_id;

– 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 student.s_id,s_name,c_name,s_score from student  join (select sc.* from score sc        left join(select s_id from score where s_score < 70 group by s_id)tmp        on sc.s_id=tmp.s_id where tmp.s_id is null)tmp2    on student.s_id=tmp2.s_id  join course on tmp2.c_id=course.c_idorder by s_id;

**-- 查询全部及格的信息**select sc.* from score sc  left join(select s_id from score where s_score < 60 group by s_id)tmp    on sc.s_id=tmp.s_idwhere  tmp.s_id is  null;**-- 或(效率低)**select sc.* from score scwhere sc.s_id not in (select s_id from score where s_score < 60 group by s_id);

– 37、查询课程不及格的学生:

select s_name,c_name as courseName,tmp.s_scorefrom studentjoin (select s_id,s_score,c_name from score,course where score.c_id=course.c_id and s_score < 60)tmpon student.s_id=tmp.s_id;

–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 student.s_id,s_name,s_score as score_01from studentjoin score on student.s_id=score.s_idwhere c_id='01' and s_score >= 80;

– 39、求每门课程的学生人数:

select course.c_id,course.c_name,count(1)as selectNumfrom coursejoin score on course.c_id=score.c_idgroup by course.c_id,course.c_name;

– 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 student.*,tmp3.c_name,tmp3.maxScorefrom (select s_id,c_name,max(s_score)as maxScore from score      join (select course.c_id,c_name from course join                  (select t_id,t_name from teacher where t_name='张三')tmp            on course.t_id=tmp.t_id)tmp2      on score.c_id=tmp2.c_id group by score.s_id,c_name      order by maxScore desc limit 1)tmp3join studenton student.s_id=tmp3.s_id;

– 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.s_id,a.c_id,a.s_score from score a,score b    where a.c_id <> b.c_id and a.s_score=b.s_score;

– 42、查询每门课程成绩最好的前三名:

select tmp1.* from  (select *,row_number()over(order by s_score desc) ranking      from score  where c_id ='01')tmp1where tmp1.ranking <= 3union allselect tmp2.* from  (select *,row_number()over(order by s_score desc) ranking      from score where c_id ='02')tmp2where tmp2.ranking <= 3union allselect tmp3.* from  (select *,row_number()over(order by s_score desc) ranking      from score where c_id ='03')tmp3where tmp3.ranking <= 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 distinct course.c_id,tmp.num from course join (select c_id,count(1) as num from score group by c_id)tmp    where tmp.num>=5 order by tmp.num desc ,course.c_id asc;

– 44、检索至少选修两门课程的学生学号:

select s_id,count(c_id) as totalCoursefrom scoregroup by s_idhaving count(c_id) >= 2;

– 45、查询选修了全部课程的学生信息:

select student.* from student, (select s_id,count(c_id) as totalCourse from score group by s_id)tmpwhere student.s_id=tmp.s_id and totalCourse=3;

–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
方法一

 select s_name,s_birth,      (year(CURRENT_DATE)-year(s_birth)-        (case when month(CURRENT_DATE) < month(s_birth) then 1          when month(CURRENT_DATE) = month(s_birth) and day(CURRENT_DATE) < day(s_birth) then 1          else 0 end)        ) as age    from student;

方法二:

select s_name,s_birth, floor((datediff(current_date,s_birth) - floor((year(current_date) - year(s_birth))/4))/365) as agefrom student;

– 47、查询本周过生日的学生:
–方法1

select * from student where weekofyear(CURRENT_DATE)+1 =weekofyear(s_birth);

–方法2

select s_name,s_sex,s_birth from student where substring(s_birth,6,2)='10'    and substring(s_birth,9,2)=14;

– 48、查询下周过生日的学生:
–方法1

select * from student where weekofyear(CURRENT_DATE)+1 =weekofyear(s_birth);

–方法2

select s_name,s_sex,s_birth from student where substring(s_birth,6,2)='10' and substring(s_birth,9,2)>=15    and substring(s_birth,9,2)<=21;

– 49、查询本月过生日的学生:
–方法1

select * from student where MONTH(CURRENT_DATE) =MONTH(s_birth);

–方法2

select s_name,s_sex,s_birth from student where substring(s_birth,6,2)='10';

– 50、查询12月份过生日的学生:

select s_name,s_sex,s_birth from student where substring(s_birth,6,2)='12';