Question

我不确定我是否会采用正确的方式，所以我很欣赏现实检查。两张桌子;用户和Feed。

+--------+-----------+----------+----------+
| userid | firstname | lastname | username |
+--------+-----------+----------+----------+
|     63 | Chris     | Smith    | csmith   |
|     65 | Roger     | Smith    | rsmith   |
|     66 | Diane     | Smith    | dsmith   |
+--------+-----------+----------+----------+

+-----------+--------+-----------+----------------+
| messageid | userid | contactid | subject        |
+-----------+--------+-----------+----------------+
|         4 | 67     | 63        | Test message 1 |
|         5 | 67     | 63        | Test message 2 |
|         6 | 63     | 67        | Test message 3 |
|         7 | 63     | 67        | Test message 4 |
|         8 | 65     | 66        | Test message 5 |
|         9 | 65     | 66        | Test message 6 |
|        10 | 66     | 65        | Test message 7 |
|        11 | 66     | 65        | Test message 8 |
+-----------+--------+-----------+----------------+

当我在用户页面上生成消息时，我需要在用户ID（创建者）或contactid（接收者）中显示消息。将使用用户的用户ID。但是，我需要显示用户的用户名，而不是他们的用户ID。所以，这似乎是一个连接语句，我可以组合一个连接几乎所有这些信息。

select a.userid, b.username, a.contactid,
       a.subject, a.message, a.timestamp
from   feed as a,
       users as b
where a.userid=b.userid

结果：

+--------+----------+-----------+----------------+-------------------------------------+
| userid | username | contactid | subject        | message                             |
+--------+----------+-----------+----------------+-------------------------------------+
| 67     | Kimomaru | 63        | Test message 1 | This is a test, hello, hello, hello |
| 67     | Kimomaru | 63        | Test message 2 | This is a test, hello, hello, hello |
| 63     | csmith   | 67        | Test message 3 | This is a test, hello, hello, hello |
+--------+----------+-----------+----------------+-------------------------------------+

但是，我想在contactid之后添加一个列，显示收件人的用户名，该用户名将来自用户表中的用户名列。我怎么能这样做？

Answer 1

select a.userid, b.username, a.contactid, 
       c.username, a.subject, a.message, a.timestamp 
from feed as a, users as b , users as c
where a.userid=b.userid and a.contactid=c.userid

MySQL加入并创建一个新表（？）

1 个答案: