Hive Full Outer Join为相同的Join Key返回多行

我正在对同一列上的 4 个表进行完全外连接。 我想为连接列中的每个不同值仅生成 1 行。

输入是：

employee1
+---------------------+-----------------+--+
| employee1.personid  | employee1.name  |
+---------------------+-----------------+--+
| 111                 | aaa             |
| 222                 | bbb             |   
| 333                 | ccc             | 
+---------------------+-----------------+--+
employee2
+---------------------+----------------+--+
| employee2.personid  | employee2.sal  |
+---------------------+----------------+--+
| 111                 | 2              |
| 200                 | 3              |
+---------------------+----------------+--+
employee3
+---------------------+------------------+--+
| employee3.personid  | employee3.place  |
+---------------------+------------------+--+
| 111                 | bbsr             |
| 300                 | atl              |
| 200                 | ny               |
+---------------------+------------------+--+
employee4
+---------------------+---------------+--+
| employee4.personid  | employee4.dt  |
+---------------------+---------------+--+
| 111                 | 2019-02-21    |
| 300                 | 2019-03-18    |
| 400                 | 2019-03-18    |
+---------------------+---------------+--+

预期结果 每个 personid 一条记录，所以总共应该有 6 条记录（111,222,333,200,300,400） 喜欢：

+-----------+---------+--------+----------+-------------+--+
| personid  | f.name  | u.sal  | v.place  |   v_in.dt   |
+-----------+---------+--------+----------+-------------+--+
| 111       | aaa     | 2      | bbsr     | 2019-02-21  |
| 200       | NULL    | 3      | ny       | NULL        |
| 222       | bbb     | NULL   | NULL     | NULL        |
| 300       | NULL    | NULL   | atl      | 2019-03-18  |
| 333       | ccc     | NULL   | NULL     | NULL        |
| 400       | NULL    | NULL   | NULL     | 2019-03-18  |
+-----------+---------+--------+----------+-------------+--+

我得到的结果是：

+-----------+---------+--------+----------+-------------+--+
| personid  | f.name  | u.sal  | v.place  |   v_in.dt   |
+-----------+---------+--------+----------+-------------+--+
| 111       | aaa     | 2      | bbsr     | 2019-02-21  |
| 200       | NULL    | 3      | NULL     | NULL        |
| 200       | NULL    | NULL   | ny       | NULL        |
| 222       | bbb     | NULL   | NULL     | NULL        |
| 300       | NULL    | NULL   | atl      | NULL        |
| 300       | NULL    | NULL   | NULL     | 2019-03-18  |
| 333       | ccc     | NULL   | NULL     | NULL        |
| 400       | NULL    | NULL   | NULL     | 2019-03-18  |
+-----------+---------+--------+----------+-------------+--+

使用的查询：

select coalesce(f.personid, u.personid, v.personid, v_in.personid) as personid,f.name,u.sal,v.place,v_in.dt
from employee1 f FULL OUTER JOIN employee2 u on f.personid=u.personid
FULL OUTER JOIN employee3 v on f.personid=v.personid
FULL OUTER JOIN employee4 v_in on f.personid=v_in.personid;

请建议如何生成预期结果。

full outer join很棘手，因为你必须采取以前的NULL考虑在内。但你可以这样做：

select coalesce(f.personid, u.personid, v.personid, v_in.personid) as personid,f.name,u.sal,v.place,v_in.dt
from employee1 f FULL OUTER JOIN
     employee2 u
     on f.personid = u.personid FULL OUTER JOIN
     employee3 v
     on v.personid in (f.person_id, u.person_id) FULL OUTER JOIN
     employee4 v_in
     on v_in.personid in (f.person_id, u.person_id, v.person_id);

在支持的数据库中using for joins（而不是on）这更简单。我认为 Hive 不支持using， 尽管。