Question

我在HIVE中有两个表格，如下所示。

Table A

+-----+---------+-----------+------------------------+------------------------+
| id  | event   | c_name    |              c_date    |          test_time     |
+-----+---------+-----------+------------------------+------------------------+
| 1   | click   | abc       | 2018-07-02 22:36:32.0  | 2018-06-22 22:36:32.0  |
| 2   | click   | abc 123   | 2018-07-01 22:36:32.0  | 2018-06-01 22:36:32.0  |
| 2   | click   | abc       | 2018-07-02 23:46:32.0  | 2018-07-02 23:46:32.0  |
| 3   | done    | abc 345   | 2018-07-22 23:56:32.0  | 2018-07-22 22:36:32.0  |
| 4   | done    | 123 abc   | 2018-08-22 22:36:32.0  | 2018-08-12 22:36:32.0  |
| 1   | click   | abc 123   | 2018-07-01 22:36:32.0  | 2018-07-01 22:36:32.0  |
+-----+---------+-----------+------------------------+------------------------+

Table B

+-----+---------+------------------------+
| id  | event   |          test_time     |
+-----+---------+------------------------+
| 1   | signup  | 2018-07-01 20:36:32.0  |
| 2   | signup  | 2018-07-02 23:36:32.0  |
| 3   | signup  | 2018-08-02 20:36:32.0  |
| 4   | signup  | 2018-09-02 20:36:32.0  |
+-----+---------+------------------------+

从table A中，我想从id中c_name之前的c_date，test_time，table B中查找每个id基于test_time。

Expected result

+-----+-----------+------------------------+
| id  | c_name    |            c_date      |
+-----+-----------+------------------------+
| 1   | abc       | 2018-07-02 22:36:32.0  |
| 2   | abc 123   | 2018-07-01 22:36:32.0  |
| 3   | abc 345   | 2018-07-22 23:56:32.0  |
| 4   | 123 abc   | 2018-08-22 22:36:32.0  |
+-----+-----------+------------------------+

我已经尝试过以下方法，但没有得到正确的结果。

select a.c_name, a.c_date, b.id from table A a left outer join table B b where a.id = b.id and a.test_time < b.test_time

如何获得预期的结果

Answer 1

使用LEFT JOIN，请尝试以下操作：

SELECT ta.id, 
       ta.c_name, 
       ta.c_date 
FROM Table_A AS ta 
LEFT JOIN Table_B AS tb 
  ON tb.id = ta.id 
WHERE tb.test_time > ta.test_time

根据Hive另一个表中的时间戳在表中查找记录

1 个答案: