Question

我正在尝试让用户朋友列表显示在他们的个人资料页面上，但我可以显示的是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。

我的表格如此

USERS
user_id
username
profile

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

以下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

以上查询将返回好友ID

Answer 3

尝试使用

编辑命令

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

如果您收到任何错误，请注意？

如果您在语法中遇到任何错误，请更正。

Answer 4

假设fiends表中的user_id和friend_id是users表中user_id的外键，那么关系将是“用户可以拥有一个或多个朋友”。以下查询将为您提供特定用户的朋友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

尝试以下查询

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die（mysql_error（））;

试一下，然后给我们输出。

此致

无法获取好友列表显示

6 个答案: