Question

我有两个存储电子邮件信息的表：

EMAIL
EMAIL_ADDRESS

EMAIL：

电子邮件ID
时间戳
以及我们不关心的其他信息

EMAIL_ADDRESS：

ID（外键引用EMAIL.ID）
EMAIL_ADDRESS
TYPE（to，from）

假设我在EMAIL中有6行 - 查询应该返回地址的ID，时间戳。

目前我有这个：

SELECT ea.EMAIL_ADDRESS, e.ID, e.sent_date
FROM EMAIL_ADDRESS ea, CHANN_EMAIL e
WHERE e.ID=ea.id
AND ea.TYPE in ('to','from')

这将返回12行，格式为： - ，ID，日期 -from，ID，日期

查询是什么，所以我将有6行： -to，from，ID，date

Answer 1

您必须将EMAIL_ADDRESS表与两个视图区分开来：

SELECT eat.EMAIL_ADDRESS as to ,ea.EMAIL_ADDRESS as from, e.ID, e.sent_date
FROM EMAIL_ADDRESS ea, CHANN_EMAIL e,EMAIL_ADDRESS eat
WHERE e.ID=ea.id and e.ID=eat.id
AND ea.TYPE in ('from') AND eat.TYPE in ('to')

Answer 2

尝试使用GROUP BY e.ID或GROUP BY ea.id

Answer 3

示例数据：电子邮件表：

| email_id |     timestamp          |
-------------------------------
| 1        | 2014-02-14 17:30:32.803|
| 2        | 2014-02-24 17:30:32.803|

email_address表：

| id | email_add | type |
-------------------------
| 1  |d@gmail.com| to   |
| 1  |c@gmail.com| from |
| 2  |i@gmail.com| to   |
| 2  |k@gmail.com| from |

查询：

SELECT tab.email_id, MAX([to]) AS [to], MAX([from]) AS [from], MAX(tab.timestamp) AS [time] FROM
(SELECT e.email_id, 
CASE WHEN type= 'to' THEN ea.email_add ELSE NULL END AS [to],
CASE WHEN type= 'from' THEN ea.email_add ELSE NULL END AS [from], e.timestamp
FROM email e
INNER JOIN email_address ea
ON e.email_id = ea.id) tab
GROUP BY tab.email_id

结果：

|email_id|    to     |  from     |   time                |
----------------------------------------------------------
|   1    |d@gmail.com|c@gmail.com|2014-02-14 17:30:32.803|
|   2    |i@gmail.com|k@gmail.com|2014-02-24 17:30:32.803|

在单个查询中选择多个列值 - oracle

3 个答案: