Question

在这种情况下，我需要一种避免来自oracle join的重复值的方法。

第一个表格包含有关某个人的一般信息。

+-----------+-------+-------------+
| ID        | Name  | Birtday_date|
+-----------+-------+-------------+
| 1         | Byron | 12/10/1998  |
| 2         | Peter | 01/11/1973  |
| 4         | Jose  | 05/02/2008  |
+-----------+-------+-------------+

第二个表包含有关第一个表中人员电话的信息。

+-------+----------+----------+----------+
| ID    |ID_Person |CELL_TYPE | NUMBER   |
+-------+- --------+----------+----------+
| 1221  | 1        | 3        | 099141021|
| 2221  | 1        | 2        | 099091925|
| 3222  | 1        | 1        | 098041013|
| 4321  | 2        | 1        | 088043153|
| 4561  | 2        | 2        | 090044313|
| 5678  | 4        | 1        | 092049013|
| 8990  | 4        | 2        | 098090233|
+----- -+----------+----------+----------+

第三张表包含有关第一张表中人员电子邮件的信息。

+------+----------+----------+---------------+
| ID   |ID_Person |EMAIL_TYPE| Email         |
+------+- --------+----------+---------------+
| 221  | 1        | 1        |jdoe@aol.com   |
| 222  | 1        | 2        |jdoe1@aol.com  |
| 421  | 2        | 1        |xx12@yahoo.com |
| 451  | 2        | 2        |dsdsa@gmail.com|
| 578  | 4        | 1        |sasaw1@sdas.com|
| 899  | 4        | 2        |cvcvsd@wew.es  |
| 899  | 4        | 2        |cvsd@www.es    |
+------+----------+----------+---------------+

我能够产生这样的结果，您可以检查此链接http://sqlfiddle.com/#!4/8e326/1

+-----+-------+-------------+----------+----------+----------+----------------+
| ID  | Name  | Birtday_date| CELL_TYPE|  NUMBER  |EMAIL_TYPE|EMAIL|
+-----+-------+-------------+----------+----------+----------+----------------+
| 1   | Byron | 12/10/1998  | 3        | 099141021|1         |jdoe@aol.com    |   
| 1   | Byron | 12/10/1998  | 2        | 099091925|2         |jdoe1@aol.com   |  
| 1   | Byron | 12/10/1998  | 1        | 099091925|          |                |  
| 2   | Peter | 01/11/1973  | 1        | 088043153|1         |xx12@yahoo.com  |
| 2   | Peter | 01/11/1973  | 2        | 090044313|2         |dsdsa@gmail.com |
| 4   | Jose  | 05/02/2008  | 1        | 092049013|1         |sasaw1@sdas.com |
| 4   | Jose  | 05/02/2008  | 2        | 098090233|2         |cvcvsd@wew.es   |
+-----+-------+-------------+----------+----------+----------+----------------+

如果您检查表ID_Person = 4的用户的电子邮件中的数据仅显示三个电子邮件中的两个，则此情况的问题是此人拥有更多具有手机号码的电子邮件，并且只会显示相同的号码手机号码。

我期望的结果是这样的。

    +-----+-------+-------------+----------+----------+----------+----------------+
    | ID  | Name  | Birtday_date| CELL_TYPE|  NUMBER  |EMAIL_TYPE|EMAIL|
    +-----+-------+-------------+----------+----------+----------+----------------+
    | 1   | Byron | 12/10/1998  | 3        | 099141021|1         |jdoe@aol.com    |   
    | 1   | Byron | 12/10/1998  | 2        | 099091925|2         |jdoe1@aol.com   | 
    | 1   | Byron | 12/10/1998  | 1        | 099091925|          |                | 
    | 2   | Peter | 01/11/1973  | 1        | 088043153|1         |xx12@yahoo.com  |
    | 2   | Peter | 01/11/1973  | 2        | 090044313|2         |dsdsa@gmail.com |
    | 4   | Jose  | 05/02/2008  | 1        | 092049013|1         |sasaw1@sdas.com |
    | 4   | Jose  | 05/02/2008  | 2        | 098090233|2         |cvcvsd@wew.es   |
    | 4   | Jose  | 05/02/2008  |          |          |2         |cvsd@www.es     |
    +-----+-------+-------------+----------+----------+----------+----------------+

这是我需要呈现数据的方式。

Answer 1

我不明白为什么您的查询如此复杂，因此添加了简单的完全外部联接，它似乎可以正常工作：

select distinct p.id, p.name,
case when Lag(CELL) over(partition by p.id order by p.id,pe.id) = CELL then null else cell_type end as cell_type,
case when Lag(CELL) over(partition by p.id order by p.id,pe.id) = CELL then null else CELL end as CELL,
EMAIL_TYPE as EMAIL_TYPE, EMAIL as EMAIL
from person p full outer join phones pe on p.id = pe.id 
full outer join emails e
on p.id = e.id and pe.cell_type = e.email_type;

从oracle join确定唯一值？

1 个答案: