Hive sql语句必练50题-入门到精通(1)
spark-shell
–master spark://node1:7077
–executor-memory 1g
–total-executor-cores 2
–jars /export/server/hive/lib/mysql-connector-java-5.1.32.jar
–driver-class-path /export/server/hive/lib/mysql-connector-java-5.1.32.jar
./spark-sql --master local[2] --driver-class-path /export/server/hive/lib/mysql-connector-java-5.1.32.jar
2018年10月20日 12:22:19 Thomson617 阅读数:917
hive学习之经典sql 50题 hive版
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;
0: jdbc:hive2://node1:10000> select * from student;
±--------------±----------------±-----------------±---------------±-+
| student.s_id | student.s_name | student.s_birth | student.s_sex |
±--------------±----------------±-----------------±---------------±-+
| 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 |
±--------------±----------------±-----------------±---------------±-+
No rows affected (0.307 seconds)
0: jdbc:hive2://node1:10000> select * from course;
±-------------±---------------±-------------±-+
| course.c_id | course.c_name | course.t_id |
±-------------±---------------±-------------±-+
| 01 | 语文 | 02 |
| 02 | 数学 | 01 |
| 03 | 英语 | 03 |
±-------------±---------------±-------------±-+
0: jdbc:hive2://node1:10000> select * from teacher;
±--------------±----------------±-+
| teacher.t_id | teacher.t_name |
±--------------±----------------±-+
| 01 | 张三 |
| 02 | 李四 |
| 03 | 王五 |
±--------------±----------------±-+
0: jdbc:hive2://node1:10000> select * from score;
±------------±------------±---------------±-+
| score.s_id | score.c_id | score.s_score |
±------------±------------±---------------±-+
| 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 |
±------------±------------±---------------±-+
–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
±--------------±----------------±-----------------±---------------±----------±----------±-+
| student.s_id | student.s_name | student.s_birth | student.s_sex | 01_score | 02_score |
±--------------±----------------±-----------------±---------------±----------±----------±-+
| 02 | 钱电 | 1990-12-21 | 男 | 70 | 60 |
| 04 | 李云 | 1990-08-06 | 男 | 50 | 30 |
±--------------±----------------±-----------------±---------------±----------±----------±-+
2 rows selected (65.461 seconds)
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 select student.*,a.s_score as 01_score,b.s_score as 02_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 st.s_id,st.s_name,avg(sc.s_score) select student.s_id,student.s_name,round(avg (score.s_score),1) as avgScore from student ±--------------±----------------±----------±-+ – 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: select student.s_id,student.s_name,tmp.avgScore from student –答案2 select score.s_id,student.s_name,round(avg (score.s_score),1) as avgScore from student ±----------±------------±--------------±-+ 8 select st.s_id,st.s_name,count(sc.c_id),sum(sc.s_score) select student.s_id,student.s_name,(count(score.c_id)) as total_count,sum(score.s_score)as total_score select student.s_id,student.s_name,(count(score.c_id) )as total_count,sum(score.s_score)as total_score – 6、查询"李"姓老师的数量: – 7、查询学过"张三"老师授课的同学的信息: ±--------------±----------------±-----------------±---------------±-+ – 8、查询没学过"张三"老师授课的同学的信息: select student.* from student 1 select student.* from student – 11、查询没有学全所有课程的同学的信息: select count(1) from course; select student.* from student ±--------------±----------------±-----------------±---------------±-+ –方法二(一步到位): select student.* from student – 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 – 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: select student.*,tmp1.course_id from student – 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 – 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 – 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 – 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, – 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 course.c_id,course.c_name,tmp.maxScore,tmp.minScore,tmp.avgScore,tmp.passRate,tmp.moderate,tmp.goodRate,tmp.excellentRates from course – 19、按各科成绩进行排序,并显示排名: select s1.,row_number()over(order by s1.s_score desc) Ranking – 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 – 21、查询不同老师所教不同课程平均分从高到低显示: select course.c_id,course.t_id,t_name,round(avg(s_score),2)as avgscore from course – 方法2 select course.c_id,course.t_id,t_name,round(avg(s_score),2)as avgscore from course,teacher,score – 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 – 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 – 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 – 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 –课程id为02的前三名 select score.c_id,course.c_name,student.s_name,s_score –课程id为03的前三名 select score.c_id,course.c_name,student.s_name,s_score – 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 – 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 – 28、查询男生、女生人数: select tmp1.man,tmp2.women from – 29、查询名字中含有"风"字的学生信息: select * from student where s_name like ‘%风%’; select s1.s_id,s1.s_name,s1.s_sex,count(*) as sameName – 31、查询1990年出生的学生名单: select * from student where s_birth like ‘1990%’; select score.c_id,c_name,round(avg(s_score),2) as avgScore from score – 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 – 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 – 35、查询所有学生的课程及分数情况: select a.s_name, – 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 – 查询全部及格的信息 – 37、查询课程不及格的学生: select s_name,c_name as courseName,tmp.s_score –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 – 39、求每门课程的学生人数: select course.c_id,course.c_name,count(1)as selectNum – 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 – 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 – 42、查询每门课程成绩最好的前三名: select tmp1.* from – 43、统计每门课程的学生选修人数(超过5人的课程才统计): select distinct course.c_id,tmp.num from course – 44、检索至少选修两门课程的学生学号: select s_id,count(c_id) as totalCourse – 45、查询选修了全部课程的学生信息: select student.* –46、查询各学生的年龄(周岁): select s_name,s_birth,(year(CURRENT_DATE)-year(s_birth)- – 47、查询本周过生日的学生: select * from student where weekofyear(CURRENT_DATE)+1 =weekofyear(s_birth); select s_name,s_sex,s_birth from student – 48、查询下周过生日的学生: select * from student where weekofyear(CURRENT_DATE)+1 =weekofyear(s_birth); select s_name,s_sex,s_birth from student – 49、查询本月过生日的学生: select * from student where MONTH(CURRENT_DATE)+1 =MONTH(s_birth); select s_name,s_sex,s_birth from student where substring(s_birth,6,2)=‘10’; select s_name,s_sex,s_birth from student where substring(s_birth,6,2)=‘12’; hive sql中的部分方法总结: 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 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。 作者:Thomson617
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
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
from student st
join score sc
on st.s_id=sc.s_id
group by st.s_id,st.s_name
having avg(sc.s_score)>=60;
–答案2
join score on student.s_id = score.s_id
group by student.s_id,student.s_name;
having avg (score.s_score) >= 60;
| student.s_id | student.s_name | avgscore |
±--------------±----------------±----------±-+
| 01 | 赵雷 | 89.7 |
| 02 | 钱电 | 70.0 |
| 03 | 孙风 | 80.0 |
| 05 | 周梅 | 81.5 |
| 07 | 郑竹 | 93.5 |
±--------------±----------------±----------±-+
– (包括有成绩的和无成绩的)
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);
1
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);
| _u1.s_id | _u1.s_name | _u1.avgscore |
±----------±------------±--------------±-+
| 04 | 李云 | 33.3 |
| 06 | 吴兰 | 32.5 |
| 08 | 王菊 | 0.0 |
±----------±------------±--------------±-+
– 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:
from student st
left join score sc
on st.s_id=sc.s_id
group by st.s_id,sc.c_id;
from student
left join score
on student.s_id=score.s_id
group by student.s_id;
group by student.s_id,score.c_id;
from student
left join score
on student.s_id=score.s_id
group by student.s_id,student.s_name ;
select t_name,count(1) from teacher where t_name like ‘李%’ group by t_name;
1
±--------±-----±-+
| t_name | _c1 |
±--------±-----±-+
| 李四 | 1 |
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’;
| student.s_id | student.s_name | student.s_birth | student.s_sex |
±--------------±----------------±-----------------±---------------±-+
| 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 |
| 07 | 郑竹 | 1989-07-01 | 女 |
±--------------±----------------±-----------------±---------------±-+
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 student.* from student
inner join score
on student.c_id=score.c_id
where score.c_id in(select c_id from score where c_id);
±--------------±----------------±-----------------±---------------±-----------±-----------±-+
| student.s_id | student.s_name | student.s_birth | student.s_sex | tmp1.s_id | tmp2.s_id |
±--------------±----------------±-----------------±---------------±-----------±-----------±-+
| 01 | 赵雷 | 1990-01-01 | 男 | 01 | 01 |
| 02 | 钱电 | 1990-12-21 | 男 | 02 | 02 |
| 03 | 孙风 | 1990-05-20 | 男 | 03 | 03 |
| 04 | 李云 | 1990-08-06 | 男 | 04 | 04 |
| 05 | 周梅 | 1991-12-01 | 女 | 05 | 05 |
±--------------±----------------±-----------------±---------------±-----------±-----------±-+
4
5
– 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:
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;
–先查询出课程的总数量
1
–再查询所需结果
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;
| student.s_id | student.s_name | student.s_birth | student.s_sex |
±--------------±----------------±-----------------±---------------±-+
| 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 |
±--------------±----------------±-----------------±---------------±-+
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;
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;
–备注:hive不支持group_concat方法,可用 concat_ws(’|’, collect_set(str)) 实现
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;
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;
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;
where student.s_id=score.s_id and s_score<60 and c_id=‘01’
order by s_score desc;
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;
–及格为>=60,中等为:70-80,优良为:80-90,优秀为:>=90
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;
– row_number() over()分组排序功能(mysql没有该方法)
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;
from score ,student
where score.s_id=student.s_id
group by score.s_id,s_name order by sumscore desc;
– 方法1
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;
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;
(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;
from course c
join(select c_id,sum(case when s_score<60 then 1 else 0 end )as s0_60,
round(100sum(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(100sum(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(100sum(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(100sum(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;
(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;
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;
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;
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;
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;
join (select s_id from score group by s_id having count(c_id) =2)tmp
on st.s_id=tmp.s_id;
(select count(1) as man from student where s_sex=‘男’)tmp1,
(select count(1) as women from student where s_sex=‘女’)tmp2;
1
– 30、查询同名同性学生名单,并统计同名人数:
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;
1
– 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:
join course on score.c_id=course.c_id
group by score.c_id,c_name order by avgScore desc,score.c_id asc;
join student on student.s_id=score.s_id
group by score.s_id,s_name having avg(s_score) >= 85;
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;
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;
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;
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);
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;
from student
join score on student.s_id=score.s_id
where c_id=‘01’ and s_score >= 80;
from course
join score on course.c_id=score.c_id
group by course.c_id,course.c_name;
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;
where a.c_id <> b.c_id and a.s_score=b.s_score;
(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;
– 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
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;
from score
group by s_id
having count(c_id) >= 2;
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;
– 按照出生日期来算,当前月日 < 出生年月的月日则,年龄减一
(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;
–方法1
–方法2
where substring(s_birth,6,2)=‘10’
and substring(s_birth,9,2)=14;
–方法1
–方法2
where substring(s_birth,6,2)=‘10’
and substring(s_birth,9,2)>=15
and substring(s_birth,9,2)<=21;
–方法1
–方法2
– 50、查询12月份过生日的学生:
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!
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
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
来源:CSDN
原文:https://blog.csdn.net/Thomson617/article/details/83281254
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