Question

我在Hive中有两个表格，如下所示：

表1：

+----+----------+----------+-----------+
| id | subject  |   date   |     amount|
+----+----------+----------+-----------+
|  1 | Do this  | 10-10-13 |     20985 |
|  2 | Done this| 10-10-13 |     18657 |
|  3 | Dont do  | 12-12-13 |     22039 |
+----+----------+----------+-----------+

表2：

+----+----------+----------+-----------+
| id | subject  |   date   |     amount|
+----+----------+----------+-----------+
|  1 | Do this  | 10-10-13 |     10985 |
|  2 | Done this| 10-10-13 |     18657 |
|  3 | Dont do  | 12-12-13 |     22039 |
|  4 | Do this  | 10-10-13 |     10000 |
|  5 | Did this | 11-10-13 |     30000 |
+----+----------+----------+-----------+

当我在两个表上group by subject and date时，对于subject and date are Do this and 10-10-13的记录，我得到的总和（金额）相同。

现在，我想在table 2中找到要分组的记录，以匹配table 1中的分组量。

我想要的结果是

+----+----------+----------+-----------+
| id | subject  |   date   |     amount|
+----+----------+----------+-----------+
|  1 | Do this  | 10-10-13 |     10985 |
|  2 | Done this| 10-10-13 |     18657 |
|  3 | Dont do  | 12-12-13 |     22039 |
|  4 | Do this  | 10-10-13 |     10000 |
+----+----------+----------+-----------+

我如何实现自己想要的？

Answer 1

这是您想要的吗？

select t2.*
from (select t2.*,
             sum(amount) over (partition by date, subject) as total_amount
      from table2 t2
     ) t2 join
     table1 t1
     on t1.date = t2.date and t2.subject = t2.subject and t1.amount = t2.total_amount

在一个表中查找与另一表匹配的分组记录

1 个答案: