PHP Parse error: syntax error, unexpected ',' in
PHP Parse error: syntax error, unexpected ',' in
好吧,基本上我正在做一个关于 youtube 的注册和登录教程。它使用旧版本的 PHP,我尝试更新代码,但出现此错误:
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) ? $user_id : false;
}
?>
第 23
行是这条:return mysqli_affected_rows($con), 0, 'user_id';
必须是:return mysqli_affected_rows($con) ? 0 : 'user_id';
如果这是你的意思。
不能 return PHP 中的多个值。
好吧,基本上我正在做一个关于 youtube 的注册和登录教程。它使用旧版本的 PHP,我尝试更新代码,但出现此错误:
Parse error: syntax error, unexpected ',' in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\projects\Forum\forum\core\functions\users.php on line 23
users.php
<?php
function user_exists($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) ? $user_id : false;
}
?>
第 23
行是这条:return mysqli_affected_rows($con), 0, 'user_id';
必须是:return mysqli_affected_rows($con) ? 0 : 'user_id';
如果这是你的意思。
不能 return PHP 中的多个值。