根据第三列中的类型代码将列分为两列

Question

我的SQL很生锈。我正在尝试转换此表：

+----+-----+--------------+-------+
| ID | SIN |   CONTACT    | TYPE  |
+----+-----+--------------+-------+
|  1 | 737 | [e-mail]  | email |
|  2 | 760 | 250-555-0100 | phone |
|  3 | 737 | 250-555-0101 | phone |
|  4 | 800 | 250-555-0102 | phone |
|  5 | 850 | [e-mail]  | email |
+----+-----+--------------+-------+

进入此表：

+----+-----+--------------+-------------+
| ID | SIN |    PHONE     |    EMAIL    |
+----+-----+--------------+-------------+
|  1 | 737 | 250-555-0101 | [e-mail] |
|  2 | 760 | 250-555-0100 |             |
|  4 | 800 | 250-555-0102 |             |
|  5 | 850 |              | [e-mail] |
+----+-----+--------------+-------------+

我写了这个查询：

SELECT *
  FROM (SELECT *
          FROM people
         WHERE TYPE = 'phone') phoneNumbers
       FULL JOIN (SELECT *
                    FROM people
                   WHERE TYPE = 'email') emailAddresses
          ON phoneNumbers.SIN = emailAddresses.SIN;

产生：

+----+-----+--------------+-------+------+-------+-------------+--------+
| ID | SIN |   CONTACT    | TYPE  | ID_1 | SIN_1 |  CONTACT_1  | TYPE_1 |
+----+-----+--------------+-------+------+-------+-------------+--------+
|  2 | 760 | 250-555-0100 | phone |      |       |             |        |
|  3 | 737 | 250-555-0101 | phone |    1 |   737 | [e-mail] | email  |
|  4 | 800 | 250-555-0102 | phone |      |       |             |        |
|    |     |              |       |    5 |   850 | [e-mail] | email  |
+----+-----+--------------+-------+------+-------+-------------+--------+

我知道我可以选择所需的列，但SIN列不完整。我似乎记得我应该第三次加入表以获得完整的SIN列，但是我不记得怎么做。

如何生成目标表（ID，SIN，PHONE，EMAIL）？

编辑和澄清：到目前为止，我非常感谢我收到的答案，但是作为SQL新手，我不熟悉您所使用的技术（条件语句，条件聚合和数据透视）。使用JOIN和SELECT不能做到这一点吗？请原谅我对此事的无知。（并不是我对高级技术不感兴趣，但我不想太快地采取行动。）

Answer 1

解决此问题的一种方法是条件聚合：

select min(ID), SIN, 
       max(case when type = 'phone' then contact end) as phone,
       max(case when type = 'email' then contact end) as email
from people t
group by sin;