Hive SQL50道练习题

王知无-import_bigdata

发布于 2020-04-07 16:33:04

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发布于 2020-04-07 16:33:04

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

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 2
union 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 2
union 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.percentum
from course c
join(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_id
left 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_id
left 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_id
left 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)tmp
order 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 score
join student on student.s_id=score.s_id
join course on  score.c_id='01' and course.c_id=score.c_id
order 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 score
join student on student.s_id=score.s_id
join course on  score.c_id='02' and course.c_id=score.c_id
order 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 score
join student on student.s_id=score.s_id
join 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 sameName
from student s1,student s2
where s1.s_name=s2.s_name and s1.s_id<>s2.s_id and s1.s_sex=s2.s_sex
group 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_id
order 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_id
where  tmp.s_id is  null;
**-- 或(效率低)**
select sc.* from score sc
where 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_score
from student
join (select s_id,s_score,c_name
      from score,course
      where score.c_id=course.c_id and s_score < 60)tmp
on 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_01
from student
join score on student.s_id=score.s_id
where 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 selectNum
from course
join score on course.c_id=score.c_id
group 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.maxScore
from (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)tmp3
join student
on 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')tmp1
where tmp1.ranking <= 3
union all
select tmp2.* from
  (select *,row_number()over(order by s_score desc) ranking
      from score where c_id ='02')tmp2
where tmp2.ranking <= 3
union all
select tmp3.* from
  (select *,row_number()over(order by s_score desc) ranking
      from score where c_id ='03')tmp3
where 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 totalCourse
from score
group by s_id
having 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)tmp
where 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 age
from 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';

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