Question

我有一个Postgres 9.3数据库，其中包含用户和附属机构表。

用户表格列

+----+-------+
|    |       |
+----+-------+
| id | email |
+----+-------+

关联列

+----+------------------+------------------+--------+
| id | referred_user_id | referrer_user_id | amount |
+----+------------------+------------------+--------+

我尝试了以下查询：

select
  users.email as referred_email,
  affiliates.amount
from affiliates
  JOIN users ON affiliates.referred_user_id = users.id

UNION ALL

select
  users.email as referrer_email,
  users.id
from affiliates
  JOIN users ON affiliates.referrer_user_id = users.id

它生成包含列的表：

+----------------+--------+
| referred_email | amount |
+----------------+--------+

但我想要一对一的表，例如：

+----------------+----------------+--------+
| referrer_email | referred_email | amount |
+----------------+----------------+--------+

我基本上用*_user_id和referrer_email替换每个referred_email，然后添加相应的amount。我能做些什么不同的事情？我以为UNION ALL将所有列加在一起。

Answer 1

我以为UNION ALL将所有专栏连在一起。

不，UNION ALL连接行。如果您想加入列，则需要使用JOIN。

这可能是这样的：

select
  rer.email as referrer_email,
  red.email as referred_email,
  affiliates.amount
from affiliates
  JOIN users rer ON affiliates.referrer_user_id = rer.id
  JOIN users red ON affiliates.referred_user_id = red.id

Answer 2

我猜你正在寻找这样的东西：

select u1.referred_email as email1, u2.referrer_email as email2, a.amount
from affiliates as a, users as u1, users as u2
where a.referred_user_id = u1.id
and a.referrer_user_id = u2.id

UNION ALL是获取此数据的正确方法吗？

2 个答案: