php - 无法显示好友列表

Question

我试图让用户的朋友列表显示在他们的个人资料页面上，但我只能显示用户个人资料的名称，而不是他们的任何朋友或个人资料图片。

我的桌子也是这样

USERS
user_id
username
profile

FRIENDS
user_id
friend_id
accepted (ENUM 0 for no, 1 for accepted).

这是我正在使用的代码，我还没有开始将我的网站转换为 MYSQLI，所以是的，我知道代码目前很容易受到攻击。

<?php
$user_id = $_SESSION['user_id'];

$sql = "SELECT friends.id, friends.user_id, friends.friend_id, friends.accepted, users.user_id, users.username, users.profile
                FROM `friends`
                LEFT JOIN `users` on users.user_id = friends.user_id OR users.user_id = friends.friend_id
                WHERE friends.friend_id = $user_id OR friends.user_id = $user_id
                AND users.user_id != $user_id AND friends.accepted = '1' OR friends.friend_id = $user_id AND friends.accepted = '1'";

$result = mysql_query($sql);
?>
<img src="<? echo $profile; ?>"><? echo $username; ?>

感谢您的任何提示。

Answer 1

试试这个。

<?php
$friends_result = mysql_query("
    SELECT
        users.`username` AS username
    FROM
        friends
    LEFT JOIN
        users
    ON
        users.`user_id` = friends.`user_id`
    OR
        users.`user_id` = friends.`friend_id`
    AND
        users.`user_id` != '" . $user_id . "'
    WHERE
        friends.`user_id` = '" . $user_id . "'
    OR
        friends.`friend_id` = '" . $user_id . "'");


while($friends_array = mysql_fetch_array($friends_result))
{
    echo $friends_array['username'] . "<br />";
}
?>

Answer 2

尝试使用此查询

SELECT friend_id as id FROM `friends` where user_id=$user_id and accepted=1 union select user_id as id from friends where friend_id=$user_id and accepted=1

上面的查询将返回好友 id

Answer 3

尝试编辑命令

$result = mysql_query($sql) or die(mysql_error());

看看你是否有任何错误？

如果您在语法中遇到任何错误，请更正它。

Answer 4

假设fiends表中的user_id和friend_id是users表中user_id的外键，那么关系就是“一个用户可以有一个或多个朋友”。以下查询将为您提供特定用户的好友 ID，

SELECT friend_id from friends f
JOIN users u ON f.user_id = u.user_id
WHERE u.user_id = $user_id AND f.accepted = 1

Answer 5

试试下面的查询

SELECT users.id as userid, firstname, lastname, username,profile_id,friends.accepted FROM users RIGHT JOIN (SELECT friend_id as id,accepted FROM `friends` where user_id=$user_id and accepted=1  union select user_id as id,status from friends where friend_id=$user_id and accepted=1) friends  ON users.id = friends.id;

Answer 6

$result = mysql_query($sql) 或死(mysql_error());

试试看，然后给我们输出。

问候