Hive SQL50道练习题

hive学习之经典sql 50题 hive版
建表:

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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

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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

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01  语文  02
02  数学  01
03  英语  03

vi /export/data/hivedatas/teacher.csv

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01  张三
02  李四
03  王五

vi /export/data/hivedatas/score.csv

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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

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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

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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:

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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

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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:

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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

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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:

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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

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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)

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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

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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:

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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 studentjoin score on student.s_id =score.s_idjoin  course on course.c_id=score.c_idjoin  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 studentleft 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')tmpon  student.s_id =tmp.s_idwhere 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 studentjoin (select s_id from score where c_id =1 )tmp1    on student.s_id=tmp1.s_idjoin (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 studentjoin (select s_id from score where c_id =1 )tmp1    on student.s_id=tmp1.s_idleft join (select s_id from score where c_id =2 )tmp2    on student.s_id =tmp2.s_idwhere 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 studentleft join(      select s_id        from score          group by s_id            having count(c_id)=3)tmpon student.s_id=tmp.s_idwhere tmp.s_id is null;
–方法二(一步到位):
select student.* from studentjoin (select count(c_id)num1 from course)tmp1left join(      select s_id,count(c_id)num2        from score group by s_id)tmp2on student.s_id=tmp2.s_id and tmp1.num1=tmp2.num2where 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 studentjoin (select c_id from score where score.s_id=01)tmp1join (select s_id,c_id from score)tmp2    on tmp1.c_id =tmp2.c_id and student.s_id =tmp2.s_idwhere 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 studentjoin (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_idjoin (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 studentinner join (select s_id from score      where s_score<60        group by score.s_id having count(s_id)>1)tmp2on student.s_id = tmp2.s_idleft 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,scorewhere 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 avgScorefrom score aleft join (select s_id,s_score  from score s1 where  c_id='01')tmp1 on  tmp1.s_id=a.s_idleft join (select s_id,s_score  from score s2 where  c_id='02')tmp2 on  tmp2.s_id=a.s_idleft join (select s_id,s_score  from score s3 where  c_id='03')tmp3 on  tmp3.s_id=a.s_idgroup 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 coursejoin(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) excellentRatesfrom 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 ascunion all select s2.*,row_number()over(order by s2.s_score desc) Ranking    from score s2 where s2.c_id='02'order by noRanking ascunion 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;
后续部分参见:
https://blog.csdn.net/Thomson617/article/details/83280617
Hive下的SQL语法总结：

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(1).Hive不支持join的非等值连接,不支持or
分别举例如下及实现解决办法。
  不支持不等值连接
       错误:select * from a inner join b on a.id<>b.id
       替代方法:select * from a inner join b on a.id=b.id and a.id is null;
 不支持or
       错误:select * from a inner join b on a.id=b.id or a.name=b.name
       替代方法:select * from a inner join b on a.id=b.id
                union all
                select * from a inner join b on a.name=b.name
  两个sql union all的字段名必须一样或者列别名要一样。
        
(2).分号字符:不能智能识别concat(‘;’,key)，只会将‘；’当做SQL结束符号。
    •分号是SQL语句结束标记，在HiveQL中也是，但是在HiveQL中，对分号的识别没有那么智慧，例如：
        •select concat(key,concat(';',key)) from dual;
    •但HiveQL在解析语句时提示：
        FAILED: Parse Error: line 0:-1 mismatched input '<EOF>' expecting ) in function specification
    •解决的办法是，使用分号的八进制的ASCII码进行转义，那么上述语句应写成：
        •select concat(key,concat('\073',key)) from dual;

(3).不支持INSERT INTO 表 Values（）, UPDATE, DELETE等操作.这样的话，就不要很复杂的锁机制来读写数据。
    INSERT INTO syntax is only available starting in version 0.8。INSERT INTO就是在表或分区中追加数据。

(4).HiveQL中String类型的字段若是空(empty)字符串, 即长度为0, 那么对它进行IS NULL的判断结果是False，使用left join可以进行筛选行。

(5).不支持 ‘< dt <’这种格式的范围查找，可以用dt in(”,”)或者between替代。

(6).Hive不支持将数据插入现有的表或分区中，仅支持覆盖重写整个表，示例如下：
    INSERT OVERWRITE TABLE t1 SELECT * FROM t2;
    
(7).group by的字段,必须是select后面的字段，select后面的字段不能比group by的字段多.
    如果select后面有聚合函数,则该select语句中必须有group by语句
    而且group by后面不能使用别名
    
(8).hive的0.13版之前select , where 及 having 之后不能跟子查询语句(一般使用left join、right join 或者inner join替代)

(9).先join(及inner join) 然后left join或right join

(10).hive不支持group_concat方法,可用 concat_ws('|', collect_set(str)) 实现

(11).not in 和 <> 不起作用,可用left join tmp on tableName.id = tmp.id where tmp.id is null 替代实现

– 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

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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)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;
后续部分参见:
https://blog.csdn.net/Thomson617/article/details/83281254
Hive下的SQL经验总结：

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(1).不支持非等值连接，一般使用left join、right join 或者inner join替代。
    •SQL中对两表内联可以写成：
        select * from dual a,dual b where a.key = b.key;
    •Hive中应为:
        select * from dual a join dual b on a.key = b.key; 
    而不是传统的格式：
        SELECT t1.a1 as c1, t2.b1 as c2 FROM t1, t2 WHERE t1.a2 = t2.b2 
        
(2).分号字符:不能智能识别concat(‘;’,key)，只会将‘；’当做SQL结束符号。
    •分号是SQL语句结束标记，在HiveQL中也是，但是在HiveQL中，对分号的识别没有那么智慧，例如：
        •select concat(key,concat(';',key)) from dual;
    •但HiveQL在解析语句时提示：
        FAILED: Parse Error: line 0:-1 mismatched input '<EOF>' expecting ) in function specification
    •解决的办法是，使用分号的八进制的ASCII码进行转义，那么上述语句应写成：
        •select concat(key,concat('\073',key)) from dual;

(3).不支持INSERT INTO 表 Values（）, UPDATE, DELETE等操作.这样的话，就不要很复杂的锁机制来读写数据。
    INSERT INTO syntax is only available starting in version 0.8。INSERT INTO就是在表或分区中追加数据。

(4).HiveQL中String类型的字段若是空(empty)字符串, 即长度为0, 那么对它进行IS NULL的判断结果是False，使用left join可以进行筛选行。

(5).不支持 ‘< dt <’这种格式的范围查找，可以用dt in(”,”)或者between替代。

(6).Hive不支持将数据插入现有的表或分区中，仅支持覆盖重写整个表，示例如下：
    INSERT OVERWRITE TABLE t1 SELECT * FROM t2;
    
(7).group by的字段,必须是select后面的字段，select后面的字段不能比group by的字段多.
    如果select后面有聚合函数,则该select语句中必须有group by语句;
    而且group by后面不能使用别名;
    有聚合函数存在就必须有group by.
    
(8).select , where 及 having 之后不能跟子查询语句(一般使用left join、right join 或者inner join替代)

(9).先join(及inner join) 然后left join或right join

(10).hive不支持group_concat方法,可用 concat_ws('|', collect_set(str)) 实现

(11).not in 和 <> 不起作用,可用left join tmp on tableName.id = tmp.id where tmp.id is null 替代实现

(12).hive 中‘不等于’不管是用！ 或者<>符号实现，都会将空值即null过滤掉，此时要用
        where （white_level<>'3' or  white_level is null） 
    或者 where (white_level!='3' or white_level is null )  来保留null 的情况。

(13).union all 后面的表不加括号,不然执行报错;
    hive也不支持顶层的union all，使用子查询来解决;
    union all 之前不能有DISTRIBUTE BY | SORT BY| ORDER BY | LIMIT 等查询条件

– 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
select s_name,s_birth,(year(CURRENT_DATE)-year(s_birth)-    (case when month(CURRENT_DATE) > month(s_birth) then 0          when day(CURRENT_DATE) > day(s_birth) then          1 else 0 end)) 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)+1 =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';
所有代码亲测有效!
如果因为hive版本及测试环境造成无法运行的还请自行修正!
hive sql中的部分方法总结:

[AppleScript] 纯文本查看 复制代码

1.case when ... then ... else ... end

2.length(string)

3.cast(string as bigint)

4.rand()       返回一个0到1范围内的随机数

5.ceiling(double)    向上取整

6.substr(string A, int start, int len)

7.collect_set(col)函数只接受基本数据类型，它的主要作用是将某字段的值进行去重汇总，产生array类型字段

8.concat()函数
    1、功能：将多个字符串连接成一个字符串。
    2、语法：concat(str1, str2,...)
    返回结果为连接参数产生的字符串，如果有任何一个参数为null，则返回值为null。

    9.concat_ws()函数
    1、功能：和concat()一样，将多个字符串连接成一个字符串，但是可以一次性指定分隔符～（concat_ws就是concat with separator）
    2、语法：concat_ws(separator, str1, str2, ...)
    说明：第一个参数指定分隔符。需要注意的是分隔符不能为null，如果为null，则返回结果为null。

    10.nvl(expr1, expr2)：空值转换函数  nvl(x,y)    Returns y if x is null else return x

11.if(boolean testCondition, T valueTrue, T valueFalse)

12.row_number()over()分组排序功能,over()里头的分组以及排序的执行晚于 where group by  order by 的执行。

13.获取年、月、日、小时、分钟、秒、当年第几周
    select 
        year('2018-02-27 10:00:00')       as year
        ,month('2018-02-27 10:00:00')      as month
        ,day('2018-02-27 10:00:00')        as day
        ,hour('2018-02-27 10:00:00')       as hour
        ,minute('2018-02-27 10:00:00')     as minute
        ,second('2018-02-27 10:00:00')     as second
        ,weekofyear('2018-02-27 10:00:00') as weekofyear
  获取当前时间:
        1).current_timestamp
        2).unix_timestamp()
        3).from_unixtime(unix_timestamp())
        4).CURRENT_DATE

作者：IT界的壹个小学生
链接：https://www.jianshu.com/p/faee67f104f7
来源：简书