我有一个网站,用户可以加入群组并发布与该群组相关的主题,我遇到的问题是,无论用户结果如何,它只显示"成员",即使在数据库中没有记录的测试帐户上,也可以有人请解释我做错了什么,谢谢.
<?php
$id = $_GET['gid'];
$user = $_SESSION['user_id'];
$iropen = "SELECT * FROM `group_users` WHERE user_id='$user' AND group_id='$id'";
$resultg = mysql_query($iropen);
$rows = mysql_fetch_array($resultg);
if ($rows['accepted'] = 1) {
echo 'member';
} else {
echo 'pending';
}
if ($resultg < 1) {
echo 'join';
}
?>
Run Code Online (Sandbox Code Playgroud) 我目前正在通过教程学习PHP,我正在尝试运行以下代码但是得到了一个无法解释的语法错误,其中没有其他人似乎与它们一起出现.
错误消息 解析错误:语法错误,意外';' 在第12行的..../users.php中
第12行=
return (mysql_result(mysql_query("SELECT `user_id` FROM `users` WHERE `username` = '$username'"), 0, 'user_id');
Run Code Online (Sandbox Code Playgroud)
整页代码
<?php
function user_exists($username) {
$username = sanitize($username);
return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username'"), 0) == 1) ? true : false;
}
function user_active($username) {
$username = sanitize($username);
return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username' AND `active` =1"), 0) ==1) ? true : false;
}
function user_id_from_username ($username) {
$username = sanitize($username);
return (mysql_result(mysql_query("SELECT `user_id` FROM `users` WHERE `username` …Run Code Online (Sandbox Code Playgroud)