【Hadoop离线基础总结】Hive级联求和

Hive级联求和

---

• 建表

  CREATE TABLE t_salary_detail( 
  username string,
  month string,
  salary INT ) 
  ROW format delimited FIELDS TERMINATED BY ',';
  

• 导入数据

  LOAD DATA LOCAL inpath '/export/servers/hivedatas/click-part-r-00000' INTO TABLE t_salary_detail;
  

  +---------------------------+------------------------+-------------------------+--+
  | t_salary_detail.username  | t_salary_detail.month  | t_salary_detail.salary  |
  +---------------------------+------------------------+-------------------------+--+
  | 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                       |
  | A                         | 2015-03                | 7                       |
  | A                         | 2015-03                | 9                       |
  | B                         | 2015-03                | 11                      |
  | B                         | 2015-03                | 6                       |
  +---------------------------+------------------------+-------------------------+--+
  

---

• 思路
  累积求和其实就是通过inner join表本身来实现。首先要先明白 统计每个用户每个月总共获得多少小费 的hql语句怎么写

  SELECT username,month,sum(salary)
  FROM t_salary_detail
  GROUP BY username,month;
  

  +-----------+----------+------+--+
  | username  |  month   | _c2  |
  +-----------+----------+------+--+
  | A         | 2015-01  | 33   |
  | A         | 2015-02  | 10   |
  | A         | 2015-03  | 16   |
  | B         | 2015-01  | 30   |
  | B         | 2015-02  | 15   |
  | B         | 2015-03  | 17   |
  +-----------+----------+------+--+
  

  现在通过inner join连接自己

  SELECT a.*,b.*
  FROM (
  SELECT username,month,sum(salary)
  FROM t_salary_detail
  GROUP BY username,month ) a
  INNER JOIN (
  SELECT username,month,sum(salary)
  FROM t_salary_detail
  GROUP BY username,month ) b
  ON a.username = b.username;
  

  +-------------+----------+--------+-------------+----------+--------+--+
  | a.username  | a.month  | a._c2  | b.username  | b.month  | b._c2  |
  +-------------+----------+--------+-------------+----------+--------+--+
  | A           | 2015-01  | 33     | A           | 2015-01  | 33     |
  | A           | 2015-01  | 33     | A           | 2015-02  | 10     |
  | A           | 2015-01  | 33     | A           | 2015-03  | 16     |
  | A           | 2015-02  | 10     | A           | 2015-01  | 33     |
  | A           | 2015-02  | 10     | A           | 2015-02  | 10     |
  | A           | 2015-02  | 10     | A           | 2015-03  | 16     |
  | A           | 2015-03  | 16     | A           | 2015-01  | 33     |
  | A           | 2015-03  | 16     | A           | 2015-02  | 10     |
  | A           | 2015-03  | 16     | A           | 2015-03  | 16     |
  | B           | 2015-01  | 30     | B           | 2015-01  | 30     |
  | B           | 2015-01  | 30     | B           | 2015-02  | 15     |
  | B           | 2015-01  | 30     | B           | 2015-03  | 17     |
  | B           | 2015-02  | 15     | B           | 2015-01  | 30     |
  | B           | 2015-02  | 15     | B           | 2015-02  | 15     |
  | B           | 2015-02  | 15     | B           | 2015-03  | 17     |
  | B           | 2015-03  | 17     | B           | 2015-01  | 30     |
  | B           | 2015-03  | 17     | B           | 2015-02  | 15     |
  | B           | 2015-03  | 17     | B           | 2015-03  | 17     |
  +-------------+----------+--------+-------------+----------+--------+--+
  

  得到以上结果可以看出，我们可以利用b表的salary列来进行累积求和，因为我们要求每个用户每个月总共获得小费，到2月份是要将2月和前一个月的小费累计，也就是表中33,33+10,33+10+16这样

  SELECT a.*,b.*
  FROM (
  SELECT username,month,sum(salary)
  FROM t_salary_detail
  GROUP BY username,month ) a
  INNER JOIN (
  SELECT username,month,sum(salary)
  FROM t_salary_detail
  GROUP BY username,month ) b
  ON a.username = b.username
  WHERE b.month <= a.month;
  

  +-------------+----------+--------+-------------+----------+--------+--+
  | a.username  | a.month  | a._c2  | b.username  | b.month  | b._c2  |
  +-------------+----------+--------+-------------+----------+--------+--+
  | A           | 2015-01  | 33     | A           | 2015-01  | 33     |
  | A           | 2015-02  | 10     | A           | 2015-01  | 33     |
  | A           | 2015-02  | 10     | A           | 2015-02  | 10     |
  | A           | 2015-03  | 16     | A           | 2015-01  | 33     |
  | A           | 2015-03  | 16     | A           | 2015-02  | 10     |
  | A           | 2015-03  | 16     | A           | 2015-03  | 16     |
  | B           | 2015-01  | 30     | B           | 2015-01  | 30     |
  | B           | 2015-02  | 15     | B           | 2015-01  | 30     |
  | B           | 2015-02  | 15     | B           | 2015-02  | 15     |
  | B           | 2015-03  | 17     | B           | 2015-01  | 30     |
  | B           | 2015-03  | 17     | B           | 2015-02  | 15     |
  | B           | 2015-03  | 17     | B           | 2015-03  | 17     |
  +-------------+----------+--------+-------------+----------+--------+--+
  

  得到以上结果后，只需要按照Month分组，对salary列用SUM函数即可

  SELECT SUM(bSalSum)
  FROM (
  SELECT a.month AS aMonth,a.username AS aUser,a.salSum AS aSalSum,
  b.month AS bMonth,b.username AS bUser,b.salSum AS bSalSum
  FROM (
  SELECT month,username,SUM(salary) AS salSum
  FROM t_salary_detail
  GROUP BY username,month ) a
  INNER JOIN (
  SELECT month,username,SUM(salary) AS salSum
  FROM t_salary_detail
  GROUP BY username,month ) b
  ON a.username = b.username
  WHERE b.month <= a.month ) t
  GROUP BY aUser,aMonth;
  

  +------+--+
  | _c0  |
  +------+--+
  | 33   |
  | 43   |
  | 59   |
  | 30   |
  | 45   |
  | 62   |
  +------+--+