面试用神sql--套路--累计报表

create table t_access_times(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_times;

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

1、第一步，先求个用户的月总金额
select username,month,sum(salary) as salary from t_access_times 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 自己连接
+-------------+----------+-----------+-------------+----------+-----------+--+
| 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 A.username,A.month,max(A.salary) as salary,sum(B.salary) as accumulate
from
(select username,month,sum(salary) as salary from t_access_times group by username,month) A
inner join
(select username,month,sum(salary) as salary from t_access_times group by username,month) B
on
A.username=B.username
where B.month <= A.month
group by A.username,A.month
order by A.username,A.month;

 

练习-练习题答案

查询全体学生的学号与姓名
　　hive> select Sno,Sname from student;

查询选修了课程的学生姓名
　　hive> select distinct Sname from student inner join sc on student.Sno=Sc.Sno;

----hive的group by 和集合函数

查询学生的总人数
　　hive> select count(distinct Sno)count from student;

计算1号课程的学生平均成绩
　　hive> select avg(distinct Grade) from sc where Cno=1;
查询各科成绩平均分
hive> select Cno,avg(Grade) from sc group by Cno;
查询选修1号课程的学生最高分数
　　select Grade from sc where Cno=1 sort by Grade desc limit 1;
(注意比较:select * from sc where Cno=1 sort by Grade
select Grade from sc where Cno=1 order by Grade)
　　
　　
求各个课程号及相应的选课人数
　　hive> select Cno,count(1) from sc group by Cno;

查询选修了3门以上的课程的学生学号
　　hive> select Sno from (select Sno,count(Cno) CountCno from sc group by Sno)a where a.CountCno>3;
或　hive> select Sno from sc group by Sno having count(Cno)>3;

----hive的Order By/Sort By/Distribute By
　　Order By ，在strict 模式下（hive.mapred.mode=strict),order by 语句必须跟着limit语句，但是在nonstrict下就不是必须的，这样做的理由是必须有一个reduce对最终的结果进行排序，如果最后输出的行数过多，一个reduce需要花费很长的时间。

查询学生信息，结果按学号全局有序
　　hive> set hive.mapred.mode=strict; <默认nonstrict>
hive> select Sno from student order by Sno;
FAILED: Error in semantic analysis: 1:33 In strict mode, if ORDER BY is specified, LIMIT must also be specified. Error encountered near token 'Sno'
　　Sort By，它通常发生在每一个redcue里，“order by” 和“sort by"的区别在于，前者能给保证输出都是有顺序的，而后者如果有多个reduce的时候只是保证了输出的部分有序。set mapred.reduce.tasks=<number>在sort by可以指定，在用sort by的时候，如果没有指定列，它会随机的分配到不同的reduce里去。distribute by 按照指定的字段对数据进行划分到不同的输出reduce中
　　此方法会根据性别划分到不同的reduce中 ，然后按年龄排序并输出到不同的文件中。

查询学生信息，按性别分区，在分区内按年龄有序
　　hive> set mapred.reduce.tasks=2;
　　hive> insert overwrite local directory '/home/hadoop/out'
select * from student distribute by Sex sort by Sage;

----Join查询,join只支持等值连接
查询每个学生及其选修课程的情况
　　hive> select student.*,sc.* from student join sc on (student.Sno =sc.Sno);
查询学生的得分情况。
　　hive>select student.Sname,course.Cname,sc.Grade from student join sc on student.Sno=sc.Sno join course on sc.cno=course.cno;

查询选修2号课程且成绩在90分以上的所有学生。
　　hive> select student.Sname,sc.Grade from student join sc on student.Sno=sc.Sno
where sc.Cno=2 and sc.Grade>90;
　　
----LEFT，RIGHT 和 FULL OUTER JOIN ,inner join, left semi join
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
　　hive> select student.Sname,sc.Cno from student left outer join sc on student.Sno=sc.Sno;
　　如果student的sno值对应的sc在中没有值，则会输出student.Sname null.如果用right out join会保留右边的值，左边的为null。
　　Join 发生在WHERE 子句之前。如果你想限制 join 的输出，应该在 WHERE 子句中写过滤条件——或是在join 子句中写。
　　
----LEFT SEMI JOIN Hive 当前没有实现 IN/EXISTS 子查询，可以用 LEFT SEMI JOIN 重写子查询语句

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);
可以被重写为：
SELECT a.key, a.val
FROM a LEFT SEMI JOIN b on (a.key = b.key)

查询与“刘晨”在同一个系学习的学生
　　hive> select s1.Sname from student s1 left semi join student s2 on s1.Sdept=s2.Sdept and s2.Sname='刘晨';

注意比较：
select * from student s1 left join student s2 on s1.Sdept=s2.Sdept and s2.Sname='刘晨';
select * from student s1 right join student s2 on s1.Sdept=s2.Sdept and s2.Sname='刘晨';
select * from student s1 inner join student s2 on s1.Sdept=s2.Sdept and s2.Sname='刘晨';
select * from student s1 left semi join student s2 on s1.Sdept=s2.Sdept and s2.Sname='刘晨';
select s1.Sname from student s1 right semi join student s2 on s1.Sdept=s2.Sdept and s2.Sname='刘晨';