Php 如何连接三个表并添加一个；及；操作人员

作为一名菜鸟，我有点不知道如何做到这一点。我正试着加入三张桌子。这是表结构：

Accounts Table: 

1) id
2) User_id (Id given to user upon sign up)
3) Full_name 
4) Username
5) Email
6) Password

Contacts Table: 

1) My_id (This is the user who added friend_id)
2) Contact_id (this is the contact who was added by my_id)
3) Status (status of relationship)

Posts Table: 

1) Post_id
2) User_posting (this is the user who posted it)
3) Post_name
4) User_allowed (this is where its specified who can see it)

以下是我的代码结构：

<?php

$everybody = "everybody";
$sid = "user who is logged in.";

$sql = <<<SQL
SELECT DISTINCT contacts.contact_id, accounts.full_name, posts.post_id, posts.post_name
FROM contacts, accounts, posts
WHERE
    (contacts.my_id = '$sid'
      AND contacts.contact_id = accounts.user_id)
    OR (contacts.my_id = '$sid'
      AND contacts.contact_id = accounts.user_id)
    OR (posts.user_posting = '$sid'
      AND contacts.contact_id = accounts.user_id
      AND posts.user_allowed = '$everybody')
    OR (posts.user_id = '$sid'
    AND contacts.user_id = accounts.user_id
    AND posts.user_allowed = '$everybody')

LIMIT $startrow, 20;


$query = mysql_query($sql) or die ("Error: ".mysql_error());

$result = mysql_query($sql);

if ($result == "")
{
echo "";
}
echo "";


$rows = mysql_num_rows($result);

if($rows == 0)
{
print("");

}
elseif($rows > 0)
{
while($row = mysql_fetch_array($query))
{

$contactid = htmlspecialchars($row['contact_id']);
$name = htmlspecialchars($row['full_name']);
$postid = htmlspecialchars($row['post_id']);
$postname = htmlspecialchars($row['post_name']);

// If $dob is empty
if (empty($contactid)) {

$contactid = "You haven't added any friends yet.";
}


print("$contactid <br>$postname by $name <br />");
}

}

应该是这样的（未彻底检查）：

假设您只想显示朋友的帖子（而不是自己的）：

编辑：解释查询

查询
SELECT
s用于输出的特定字段
FROM
posts

表，因为结果应该包含每个post的一行

该基本表是

与
帐户
表联接
以获取过帐用户和用户的数据

另外，

JOIN

ed与

contacts

表一起使用，这样我们就可以筛选出只有发帖用户也是相关用户添加的朋友

允许的值上有一个附加筛选器，并且

出于分页目的，我们有一个由

$startrow

定义的偏移量

---

您应该使用正确的

JOIN

语法。您期望的输出是什么？您期望的输出是什么？您的输出有什么问题？你期望什么？为什么要检查相同的条件两次

（contacts.my_id='$sid'和contacts.contact_id=accounts.user_id）或（contacts.my_id='$sid'和contacts.contact_id=accounts.user_id）

由于注释被锁定，我建议您去阅读更多关于PHP和MySQL的内容。你们似乎对这两个方面都缺乏一些基本的了解。嘿，我试过你们的，结果显示错误：未知列“accounts.id”在“on子句”中“accounts表上难道并没有id字段吗？”？你能提供你的数据库结构吗？是的我能。。。但我无法理解为什么它没有发现它。数据库结构如我在上面发布的一样。请添加mysql表定义。好的，滚动调试会话现在停止。我如何查看前面的注释？我需要他们帮我查询。

SELECT DISTINCT contacts.contact_id, 
                accounts.full_name,
                posts.post_id, 
                posts.post_name 
FROM contacts
INNER JOIN accounts ON accounts.id = contacts.id
INNER JOIN posts ON posts.User_posting = accounts.id /* (where'd get posts.user_id btw ) */
WHERE contacts.my_id = $sid
      AND posts.user_allowed = $everybody
LIMIT $startrow, 20

$everybody = "everybody";
$sid = user who is logged in.  

$sql = "SELECT a.User_id, a.Full_name, p.Post_id, p.Post_name
FROM posts p
INNER JOIN accounts a ON a.User_id = p.User_posting
INNER JOIN contacts c ON c.My_id = '$sid' AND c.Contact_id = a.User_id
WHERE p.User_allowed = '$everybody'
LIMIT $startrow, 20";