搜集了50个经典SQL语句，以便加强对Hive的理解，包含了基本操作，UDF函数，以及很多常用统计函数，与mySQL写法有一定差别，用来做HQL练习
1.数据准备
home目录下新建data文件夹存放准备数据
mkdir data
创建student数据文本
vi student.txt
添加如下数据：
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
创建course数据文本
vi course.txt
添加如下数据：
01 语文 02
02 数学 01
03 英语 03
创建teacher数据文本
vi teacher.txt
添加如下数据：
01 张三
02 李四
03 王五
创建score数据文本
vi score.txt
添加如下数据：
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
进入hive，创建测试表
建表语句：
create table if not exists student(
s_id string,
s_name string,
s_birth string,
s_sex string
)
row format delimited
fields terminated by '/t'
stored as textfile
create table if not exists course(
c_id string,
c_name string,
t_id string
)
row format delimited
fields terminated by '/t'
stored as textfile
create table if not exists teacher(
t_id string,
t_name string
)
row format delimited
fields terminated by '/t'
stored as textfile
create table if not exists score(
s_id string,
c_id string,
s_score int
)
row format delimited
fields terminated by '/t'
stored as textfile
加载本地数据到hive表中
备注：这里有个坑，因为集群hive部署角色分配是每台节点上都有hive代理，所以我随意任何一台机器执行hive命令都是能够访问Hive Cli 的，所以当我选择在worknode1节点创建txt的时候再访问node执行load指令的时候执行失败，报错信息是找不到路径，所以这里给出两种解决方案：
一、把文件放到Hive server节点的目录才能被hive访问到，执行
scp -r data/ root@toolnode1:/home/
把文件复制到toolnode1节点再执行load指令成功执行
二、把文件上传到hdfs目录，再把路径替换成为hdfs路径
依次执行加载命令：
load data local inpath '/home/data/student.txt' into table student;
load data local inpath '/home/data/course.txt' into table course;
load data local inpath '/home/data/score.txt' into table score;
load data local inpath '/home/data/teacher.txt' into table teacher;
题目
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 a.*,b.s_score as 01_score,c.s_score as 02_score
from student a
join score b on a.s_id = b.s_id and b.c_id = '01'
left join score c on a.s_id = c.s_id and c.c_id = '02'
where b.s_score > c.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 a.*,b.s_score as 01_score,c.s_score as 02_score
from student a
join score b on a.s_id = b.s_id and b.c_id = '01'
left join score c on a.s_id = c.s_id and c.c_id = '02'
where b.s_score < c.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 * from (
select a.s_id,a.s_name,round(avg(b.s_score),2) as avg_score
from student a join score b on a.s_id = b.s_id
group by a.s_id,a.s_name
) t where t.avg_score>=60
这里round函数保留两位小数没有生效，round参数改为0就能取整了，找不到原因
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 * from (
select a.s_id,a.s_name,round(avg(b.s_score),2) as avg_score
from student a join score b on a.s_id = b.s_id
group by a.s_id,a.s_name
) t where t.avg_score<60
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.s_id,a.s_name,count(b.c_id) num_subject,sum(b.s_score) total_score
from student a
join score b on a.s_id = b.s_id
group by a.s_id,a.s_name
6、查询"李"姓老师的数量:
select count(*)
from teacher where t_name like '李%'
7、查询学过"张三"老师授课的同学的信息:
select distinct a.*
from student a
join score b on a.s_id = b.s_id
left join course c on b.c_id = c.c_id
left join teacher d on c.t_id = d.t_id
where d.t_name = '张三'
8、查询没学过"张三"老师授课的同学的信息:
select * from student a
where a.s_id not in(select distinct a.s_id
from student a
join score b on a.s_id = b.s_id
left join course c on b.c_id = c.c_id
left join teacher d on c.t_id = d.t_id
where d.t_name = '张三')
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 a
JOIN score b ON a.s_id=b.s_id AND b.c_id='01'
JOIN score c ON a.s_id=c.s_id AND c.c_id='02'
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 student a
JOIN score b ON a.s_id=b.s_id AND b.c_id='01'
WHERE a.s_id NOT IN(SELECT s_id FROM score WHERE c_id='02');
11、查询没有学全所有课程的同学的信息:
select distinct
a.*
from student a
join score b
left join score c on a.s_id=c.s_id and b.c_id=c.c_id
where c.s_score 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 distinct
a.*
from student a
join score b on b.s_id='01'
join score c on b.c_id=c.c_id and a.s_id=c.s_id
where a.s_id<>'01'
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 p.s_id,p.s_name,p.s_birth,p.s_sex from(
SELECT a.s_id as s_id,a.s_name as s_name,a.s_birth as s_birth,a.s_sex as s_sex,count(b.c_id) as num_course
FROM student a
JOIN score b ON a.s_id=b.s_id
WHERE b.s_id!='01' AND b.c_id IN (SELECT c_id FROM score WHERE s_id='01')
GROUP BY a.s_id,a.s_name,a.s_birth,a.s_sex
) p
right join (SELECT COUNT(c_id) as 01_num_course FROM score WHERE s_id='01') q on p.num_course=q.01_num_course
SELECT s.s_name
FROM student s
JOIN score sc ON s.s_id=sc.s_id join
(SELECT collect_set(c_id) AS sub,count(c_id) AS num
FROM score
WHERE s_id='01' ) sc2
WHERE array_contains(sc2.sub,sc.c_id) and s.s_id!='01'
GROUP BY s.s_id,
s.s_name,
s.s_birth,
s.s_sex,
sc2.num
HAVING count(sc.c_id)=sc2.num
14、查询没学过"张三"老师讲授的任一门课程的学生姓名:
select * from student a where a.s_id not in(select w.s_id from(
SELECT distinct
a.*
FROM student a
JOIN score b ON a.s_id=b.s_id
right join (SELECT c_id FROM course a join (SELECT t_id FROM teacher WHERE t_name='张三') b on a.t_id=b.t_id) t on b.c_id=t.c_id
) as w
)
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.s_id,a.s_name,round(avg(b.s_score),2)
from student a
join score b on a.s_id =b.s_id
where a.s_id in (select w.s_id from (select c.s_id,count(d.s_score) num_subject
from student c
join score d on c.s_id = d.s_id and d.s_score<60
group by c.s_id
having num_subject>=2) as w)
group by a.s_id,a.s_name
同样round不生效的问题依然存在
16、检索"01"课程分数小于60，按分数降序排列的学生信息:
select *
from student a
join score b on a.s_id = b.s_id and b.c_id = '01' and b.s_score<60
order by b.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,a.s_name,b.c_id,c.c_name,b.s_score,w.avg_score
from student a
join score b on a.s_id = b.s_id
left join course c on b.c_id = c.c_id
left join (select a.s_id,round(avg(b.s_score),2) avg_score
from student a
join score b on a.s_id = b.s_id
group by a.s_id) as w
on a.s_id = w.s_id
order by w.avg_score 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.c_id,
b.c_name,
max(a.s_score) max_score,
min(a.s_score) min_score,
avg(a.s_score) avg_score,
count(if(a.s_score>=60,a.s_score,null))/count(a.s_score)*100 as jg,
count(if(a.s_score>=70 and a.s_score<80,a.s_score,null))/count(a.s_score)*100 as zd,
count(if(a.s_score>=80 and a.s_score<90,a.s_score,null))/count(a.s_score)*100 as yl,
count(if(a.s_score>=90,a.s_score,null))/count(a.s_score)*100 as yx
FROM score a
JOIN course b ON a.c_id = b.c_id
GROUP BY a.c_id,
b.c_name
19、按各科成绩进行排序，并显示排名:– row_number() over()分组排序功能
备注：重点掌握，统计很好用的函数
select
*,
row_number() over(distribute by c_id sort by s_score desc) as rm
from score
20、查询学生的总成绩并进行排名:
select s_id,
sum(s_score) as total_score,
row_number() over(sort by sum(s_score) desc) as rm
from score
group by s_id
21、查询不同老师所教不同课程平均分从高到低显示:
select c.t_id,c.t_name,b.c_id,b.c_name,round(avg(a.s_score),2) avg_score
from score a
join course b on a.c_id = b.c_id
left join teacher c on b.t_id = c.t_id
group by c.t_id,c.t_name,b.c_id,b.c_name
order by avg_score 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 *
from (select *,
row_number() over(distribute by c_id sort by s_score) rm
from score) as a
join student b on a.s_id = b.s_id
where a.rm =2 or a.rm =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],[0-60]及所占百分比
select
a.c_id,
b.c_name,
count(if(a.s_score>85 and a.s_score<=100,a.s_score,null)) as 85_100,
round(count(if(a.s_score>85 and a.s_score<=100,a.s_score,null))/count(a.s_score)*100,2) as percentage,
count(if(a.s_score>70 and a.s_score<=85,a.s_score,null)) as 70_85,
round(count(if(a.s_score>70 and a.s_score<=85,a.s_score,null))/count(a.s_score)*100,2) as percentage,
count(if(a.s_score>60 and a.s_score<=70,a.s_score,null)) as 60_70,
round(count(if(a.s_score>60 and a.s_score<=70,a.s_score,null))/count(a.s_score)*100,2) as percentage,
count(if(a.s_score>0 and a.s_score<=60,a.s_score,null)) as 0_60,
round(count(if(a.s_score>0 and a.s_score<=60,a.s_score,null))/count(a.s_score)*100,2) as percentage
from score a
join course b on a.c_id=b.c_id
group by a.c_id,
b.c_name
24、查询学生平均成绩及其名次:
select
a.*,
row_number() over(order by a.avg_score desc)
from (select a.s_id,a.s_name,round(avg(b.s_score),2) avg_score
from student a
join score b on a.s_id = b.s_id
group by a.s_id,a.s_name) a
25、查询各科成绩前三名的记录三个语句
select a.* from (
select
s_id,
c_id,
s_score,
row_number() over(distribute by c_id sort by s_score desc) as rm
from score
) a
where a.rm<=3
