Question

我想从/向电子邮件first@test.com选择所有电子邮件（ domain_name.emails.id = 1 ）。（作为收件人，作为发件人）

id | from            | message
1  | first@test.com  | A
2  | second@test.com | B /* Because second sent email to first */

我有：

SELECT
    c.*
FROM
    domain_name.emails AS a
    INNER JOIN domain_names AS b ON b.ID = a.domain_name_id
    INNER JOIN email.messages AS c ON c.from = ( a.NAME || '@' || b.NAME || '.' || b.domain_id )
    INNER JOIN email.message_recipients AS d ON d.recipient = ( a.NAME || '@' || b.NAME || '.' || b.domain_id )
WHERE
    a.ID = 1;

注意：＆＃34; INNER JOIN＆＃34;对不对？
我明白了：

id | from           | message
1  | first@test.com | A

表格（仅2个FK）

电子邮件架构类似于LOG

domain_name.emails

id | domain_name_id (domain_names.id FK) | name
1  | 1                                   | first

domain_names

id | domain_id | name
1  | com       | test

email.messages

id | from            | message
1  | first@test.com  | A
2  | second@test.com | B

email.message_recipients

id | message_id (email.messages.id FK) | recipient
1  | 1                                 | second@test.com
2  | 2                                 | first@test.com

Answer 1

我认为你可以使用这样的查询 - 未经测试 - ：

-- In fromTo I merge `TO`s data after `FROM`s data
;with fromTo as (
  select * 
  from email.messages
  union all
  select message_id, recipient, c.message
  from email.message_recipients d
  join email.messages c on c.id = d.message_id
)
select
    c.*
from
    domain_name.emails a
    join domain_names b on b.ID = a.domain_name_id
    join fromTo c on c.from = ( a.NAME || '@' || b.NAME || '.' || b.domain_id )        
where
    a.ID = 1;

[ SQL Fiddle Demo ]

Answer 2

您需要合并这两个包含电子邮件地址的表：

SELECT
    c.*
FROM domain_name.emails AS a
INNER JOIN domain_names AS b ON b.ID = a.domain_name_id
INNER JOIN (SELECT id AS message_id, "from" AS address FROM email.messages
            UNION
            SELECT message_id, recipient FROM email.message_recipients
           ) AS m ON m.address = ( a.NAME || '@' || b.NAME || '.' || b.domain_id )
INNER JOIN email.messages AS c ON c.id = m.message_id
WHERE
    a.ID = 1;

SQL语句 - 问题

2 个答案: