hive的典型应用场景

案例一：

需求：现有这么一批数据，现要求出：每个用户截止到每月为止的最大单月访问次数和累计到该月的总访问次数。
数据：

用户名，月份，访问次数A,2015-01,5A,2015-01,15B,2015-01,5A,2015-01,8B,2015-01,25A,2015-01,5A,2015-02,4A,2015-02,6B,2015-02,10B,2015-02,5A,2015-03,16A,2015-03,22B,2015-03,23B,2015-03,10B,2015-03,11

最终结果：

用户  月份      最大访问次数  总访问次数       当月访问次数A     2015-01          33              33               33A     2015-02          33              43               10A     2015-03          38              81               38B     2015-01          30              30               30B     2015-02          30              45               15B     2015-03          44              89                44

解决：

#step01 统计每个用户每月的总访问次数create view view_step01 as select name,month,sum(visitCount) total from t_user  group by name,month;#step02 （自连接，连接条件为name）create view view_step02 as    select t1.name aname,t1.month amonth,t1.total atotal,t2.name bname,t2.month bmonth,t2.total btotal    from view_step01 t1 join view_step01  t2 on t1.name =t2.name #step03 去除无用数据，每组找到小于等于自己月份的数据select bname,bmonth,max(btotal),sum(btotal),btotalfrom view_step02where unix_timestamp(amonth,'yyyy-MM')>=unix_timestamp(bmoth,'yyyy-MM')group by aname,amonth,atotal;

案例二：

#建表语句：

CREATE TABLE `course` (  `id` int(11) NOT NULL AUTO_INCREMENT PRIMARY KEY,  `sid` int(11) DEFAULT NULL,  `course` varchar(255) DEFAULT NULL,  `score` int(11) DEFAULT NULL ) ENGINE=InnoDB DEFAULT CHARSET=utf8;#插入数据INSERT INTO `course` VALUES (1, 1, 'yuwen', 43);INSERT INTO `course` VALUES (2, 1, 'shuxue', 55);INSERT INTO `course` VALUES (3, 2, 'yuwen', 77);INSERT INTO `course` VALUES (4, 2, 'shuxue', 88);INSERT INTO `course` VALUES (5, 3, 'yuwen', 98);INSERT INTO `course` VALUES (6, 3, 'shuxue', 65);

需求：所有数学课程成绩 大于 语文课程成绩的学生的学号
解决：（行列转换）

SELECT    t1.sid FROM    (SELECT    sid,    max( CASE `course` WHEN "yuwen" THEN score ELSE 0 END ) AS "yuwen",    max( CASE `course` WHEN "shuxue" THEN score ELSE 0 END ) AS "shuxue" FROM    `course` GROUP BY    sid     ) t1 WHERE    t1.yuwen < t1.shuxue;

案例三：

需求：比如：2010012325表示在2010年01月23日的气温为25度。现在要求使用hive，计算每一年出现过的最大气温的日期+温度。
数据：
年 温度
20140101 14
20140102 16
20140103 17
20140104 10
20140105 06
20120106 09
20120107 32
20120108 12
20120109 19
20120110 23
20010101 16
20010102 12
20010103 10
20010104 11
20010105 29
20130106 19
20130107 22
20130108 12
20130109 29
20130110 23
20080101 05

现在需要根据年月进行group by 但是最终的结果需要是20080101 05，也就是说，分组字段和最后保留的字段不相同，这时怎么办？

解决：

#Step1:CREATE VIEW view_step1 AS SELECTsubstr( tmp, 1, 4 ) AS YEAR,max( substr( tmp, 9, 2 ) ) AS tmp FROM    tmp GROUP BY    substr( tmp, 1, 4 );#Step2:SELECT    b.tmp,    a.tmp FROM    view_step1 a    JOIN tmp b ON a.YEAR = substr( b.tmp, 1, 4 )     AND a.tmp = substr( b.tmp, 9, 2 );

案例四：

数据

#表示有id为1,2,3的学生选修了课程a,b,c,d,e,f中其中几门：id course 1,a 1,b 1,c 1,e 2,a 2,c 2,d 2,f 3,a 3,b 3,c 3,e

需求：编写Hive的HQL语句来实现以下结果：表中的1表示选修，表中的0表示未选修。

解决（方案1）：

#行列转换select id max(case when course='a' then 1 else 0 and ) as a ,max(case when course='b' then 1 else 0 and ) as b ,max(case when course='c' then 1 else 0 and ) as c ,max(case when course='d' then 1 else 0 and ) as d ,max(case when course='e' then 1 else 0 and ) as e ,max(case when course='f' then 1 else 0 and ) as ffrom course  group by id;

解决（方案2）：

#collect_set函数#step01create view id_courses as select a.course acourse,b.course bcourse,b.id id(select collect_set(course) as course from course) a     join (selecet id ,colect_set(course) as course from course group by id) b#step02select id,case when array_contains(bcourse,acourse[0]) then 1 else 0 end as a ,case when array_contains(bcourse,acourse[1]) then 1 else 0 end as b ,case when array_contains(bcourse,acourse[2]) then 1 else 0 end as c ,case when array_contains(bcourse,acourse[3]) then 1 else 0 end as d ,case when array_contains(bcourse,acourse[4]) then 1 else 0 end as e ,case when array_contains(bcourse,acourse[5]) then 1 else 0 end as ffrom id_courses;