HQL练习

Hive学习笔记总结

05. Hql练习

1. hql基础练习

题目和数据来源：http://www.w2b-c.com/article/150326(去掉-)

create和load

create table students(Sno int,Sname string,Sex string,Sage int,Sdept string)row format delimited fields terminated by ','stored as textfile;
create table course(Cno int,Cname string) row format delimited fields terminated by ',' stored as textfile;
create table sc(Sno int,Cno int,Grade int)row format delimited fields terminated by ',' stored as textfile;

load data local inpath '/home/hadoop/hivedata/students.txt' overwrite into table student;
load data local inpath '/home/hadoop/hivedata/sc.txt' overwrite into table sc;
load data local inpath '/home/hadoop/hivedata/course.txt' overwrite into table course;

1.查询全体学生的学号与姓名

hive> select Sno,Sname from students;

2.查询选修了课程的学生姓名

select distinct Sname from students, sc where students.Sno = sc.Sno;

或：

select distinct Sname from students inner join sc on students.Sno = sc.Sno;

3.查询学生的总人数

select count(*) from students;

4.计算1号课程的学生平均成绩

select avg(Grade) from sc where Cno = 1;

5.查询各科成绩平均分

select Cname,avg(Grade) from sc, course where sc.Cno = course.Cno group by sc.Cno;

//Grade要么出现在group关键词之后，要么使用聚合函数。

6.查询选修1号课程的学生最高分数

select max(Grade) from sc where Cno = 1;

7.求各个课程号及相应的选课人数

select Cno,count(*) from sc group by Cno;

8.查询选修了3门以上的课程的学生学号

select Sno from sc group by Sno having count(Cno) >3 ;

9.查询学生信息，结果按学号全局有序

select * from students order by Sno;

10.查询学生信息，结果区分性别按年龄有序

set mapred.reduce.tasks=2;
select * from students distribute by sex sort by sage;

11.查询每个学生及其选修课程的情况

select students.*,sc.* from students join sc on (students.Sno =sc.Sno);

12.查询学生的得分情况

13.查询选修2号课程且成绩在90分以上的所有学生。

select students.Sname from sc,students where sc.Cno = 2 and sc.Grade > 90 and sc.Sno = students.Sno;

或者：

select students.Sname,sc.Grade from students join sc on students.Sno=sc.Sno where  sc.Cno=2 and sc.Grade>90;

14.查询所有学生的信息，如果在成绩表中有成绩，则输出成绩表中的课程号

select students.Sname,sc.Cno from students join sc on students.Sno=sc.Sno;

15.重写以下子查询为LEFT SEMI JOIN

SELECT a.key, a.value FROM a WHERE a.key exist in (SELECT b.key FROM B);

查询目的：查找A中，key值在B中存在的数据。

可以被重写为：

select a.key,a.value from a left semi join b on a.key = b.key;

16.查询与“刘晨”在同一个系学习的学生

select s1.Sname from students s1 where sdept in (select sdept from students where sname = '刘晨');

或者：

select s1.Sname from students s1 left semi join students s2 on s1.Sdept=s2.Sdept and s2.Sname='刘晨';

注意比较：

select * from students s1 left join students s2 on s1.Sdept=s2.Sdept and s2.Sname='刘晨';
select * from students s1 right join students s2 on s1.Sdept=s2.Sdept and s2.Sname='刘晨';
select * from students s1 inner join students s2 on s1.Sdept=s2.Sdept and s2.Sname='刘晨';
select * from students s1 left semi join students s2 on s1.Sdept=s2.Sdept and s2.Sname='刘晨';

2. 执行顺序

标准顺序：

select--from--where--group by--having--order by

join操作中，on条件与where条件的区别

数据库在通过连接两张或多张表来返回记录时，都会生成一张中间的临时表，然后再将这张临时表返回给用户。

join发生在where字句之前，在使用left jion时，on和where条件的区别如下：

1、on条件是在生成临时表时使用的条件，它不管on中的条件是否为真，都会返回左边表中的记录。（右边置为Null了）

2、where条件是在临时表生成好后，再对临时表进行过滤的条件。这时已经没有left join的含义（必须返回左边表的记录）了，条件不为真的就全部过滤掉。

假设有两张表：

表1：tab1

id size
1  10
2  20
3  30

表2：tab2

size name
10   AAA
20   BBB
20   CCC

两条SQL:

1、select * from tab1 left join tab2 on tab1.size = tab2.size where tab2.name='AAA'
2、select * from tab1 left join tab2 on tab1.size = tab2.size and tab2.name='AAA'

第一条SQL的过程：

1、中间表

on条件:

tab1.size = tab2.size
tab1.id tab1.size tab2.size tab2.name
1 10 10 AAA
2 20 20 BBB
2 20 20 CCC
3 30 (null) (null)

2、再对中间表过滤

where 条件：

tab2.name='AAA'
tab1.id tab1.size tab2.size tab2.name
1 10 10 AAA

第二条SQL的过程：

1、中间表

on条件:

tab1.size = tab2.size and tab2.name='AAA'
(条件不为真也会返回左表中的记录) tab1.id tab1.size tab2.size tab2.name
1 10 10 AAA
2 20 (null) (null)
3 30 (null) (null)

其实以上结果的关键原因就是left join,right join,full join的特殊性，

不管on上的条件是否为真都会返回left或right表中的记录，full则具有left和right的特性的并集。

** 而inner join没这个特殊性，则条件放在on中和where中，返回的结果集是相同的。**

3. Hive实战--级联求和（累计报表）

需求：

有如下访客访问次数统计表 t_access_times

访客	月份	访问次数
A	2015-01	5
A	2015-01	15
B	2015-01	5
A	2015-01	8
B	2015-01	25
A	2015-01	5
A	2015-02	4
A	2015-02	6
B	2015-02	10
B	2015-02	5

需要输出报表：t_access_times_accumulate

月访问：当月的总次数；累计访问总计：截止到当月的月访问次数之和。

访客	月份	月访问总计	累计访问总计
A	2015-01	33	33
A	2015-02	10	43
B	2015-01	30	30
B	2015-02	15	45

准备数据：

A,2015-01,5

A,2015-01,15

B,2015-01,5

A,2015-01,8

B,2015-01,25

A,2015-01,5

A,2015-02,4

A,2015-02,6

B,2015-02,10

B,2015-02,5

create table t_access_time(username string,month string,salary int)
row format delimited fields terminated by ',';

load data local inpath '/home/hadoop/t_access_times.dat' into table t_access_time;

1、第一步，先求每个用户的月总金额

select username,month,sum(salary) from t_access_time group by username,month;

+-----------+----------+---------+--+

| username | month | salary |

+-----------+----------+---------+--+

| A | 2015-01 | 33 |

| A | 2015-02 | 10 |

| B | 2015-01 | 30 |

| B | 2015-02 | 15 |

+-----------+----------+---------+--+

2、第二步，将月总金额表 自己连接(自连接)

select * from
(select username,month,sum(salary) as salary from t_access_time group by username,month) TabA
inner join
(select username,month,sum(salary) as salary from t_access_time group by username,month) TabB
on TabA.username = TabB.username;

+-------------+----------+-----------+-------------+----------+-----------+--+

| a.username | a.month | a.salary | b.username | b.month | b.salary |

+-------------+----------+-----------+-------------+----------+-----------+--+

| A | 2015-01 | 33 | A | 2015-01 | 33 |

| A | 2015-01 | 33 | A | 2015-02 | 10 |

| A | 2015-02 | 10 | A | 2015-01 | 33 |

| A | 2015-02 | 10 | A | 2015-02 | 10 |

| B | 2015-01 | 30 | B | 2015-01 | 30 |

| B | 2015-01 | 30 | B | 2015-02 | 15 |

| B | 2015-02 | 15 | B | 2015-01 | 30 |

| B | 2015-02 | 15 | B | 2015-02 | 15 |

+-------------+----------+-----------+-------------+----------+-----------+--+

3、第三步，从上一步的结果中

进行分组查询，分组的字段是a.username a.month

求月累计值： 将b.month <= a.month的所有b.salary求和即可

select TabA.username,TabA.month,max(TabA.salary) as month_salary,sum(TabB.salary) as sum_salary
from
(select username,month,sum(salary) as salary from t_access_time group by username,month) TabA
inner join
(select username,month,sum(salary) as salary from t_access_time group by username,month) TabB
on TabA.username = TabB.username
where TabB.month<= TabA.month
group by TabA.username,TabA.month;

max(TabA.salary)不能直接写成TabA.salary，因为这个字段没有出现在group by中，也没有聚合函数，所以使用max表示。

结果：

A 2015-01 33 33

A 2015-02 10 43

B 2015-01 30 30

B 2015-02 15 45

参考http://www.w2b-c.com/article/150326（去掉-）

初接触，记下学习笔记，还有很多问题，望指导，谢谢。