好吧,基本上我正在YouTube上注册和登录教程。它正在使用旧版本的PHP,并且我尝试更新代码,但是出现此错误:
Parse error: syntax error, unexpected ',' in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\projects\Forum\forum\core\functions\users.php on line 23
users.php
<?php
function user_exists($username, $con) {
$data = $username;
$username = sanitize($data, $con);
$username = $data;
mysqli_query($con, "SELECT `user_id` FROM `users` WHERE `username` = '$username'");
return(mysqli_affected_rows($con) == 1) ? true : false;
}
function user_active($username, $con) {
$data = $username;
$username = sanitize($data, $con);
$username = $data;
mysqli_query($con, "SELECT `user_id` FROM `users` WHERE `username` = '$username' AND `active` = 1");
return(mysqli_affected_rows($con) == 1) ? true : false;
}
function user_id_from_username ($username, $con) {
$data = $username;
$username = sanitize($data, $con);
$username = $data;
mysqli_query($con, "SELECT `user_id` FROM `users` WHERE `username` = '$username'");
return mysqli_affected_rows($con), 0, 'user_id';
}
function login($username, $password, $con) {
$user_id = user_id_from_username($username, $con);
$data = $username;
$username = sanitize($data, $con);
$username = $data;
$password = md5($password);
return (mysqli_affected_rows(mysqli_query($con, "SELECT `user_id` FROM `users` WHERE `username` = '$username' AND `password` = '$password'"), 0) == 1) ? $user_id : false;
}
?>
最佳答案
行23
是这个:return mysqli_affected_rows($con), 0, 'user_id';
必须是:return mysqli_affected_rows($con) ? 0 : 'user_id';
(如果这是您的意思)。
无法在PHP中返回多个值。
关于php - PHP解析错误: syntax error, unexpected ',' in,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29241208/