我已经看到了关于如何解决此错误的多个答案,但我不知道如何将它应用到我的代码:警告:mysql_result()期望参数1是资源,给定的对象?
function user_exists($username) {
$username = sanitize($username);
global $con;
return (mysql_result(mysqli_query($con, "SELECT COUNT(`user_id`) FROM `users` WHERE `username` ='$username' "), 0) == 1) ? true : false;
}
我觉得它有什么根据这里的答案与mysql_result
做:mysql_result() expects parameter 1 to be resource, object given。我将如何让它在我的代码上工作?
编辑:
function login($username, $password){
$user_id = user_id_from_username($username);
global $con;
$username = sanitize($username);
$password = md5($password);
return (mysqli_num_rows(mysqli_query($con, "SELECT COUNT(`user_id`) FROM `users` WHERE `username`='$username' AND `password`='$password'"))==1) ? $user_id : false;
}
不嵌入这样的功能,它使无法正确调试 – rtfm
[我可以在PHP中混合MySQL API?](https://stackoverflow.com/questions/17498216/can-i-mix-mysql -apis-in-php)...你在混合'mysqli_'和'mysql_' ... – Scuzzy
强制性:停止使用mysql_ * – rtfm