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PHP OOP MySQL注册系统MySQL\u NUM\u行错误_Php_Mysql_Oop - Fatal编程技术网

PHP OOP MySQL注册系统MySQL\u NUM\u行错误

PHP OOP MySQL注册系统MySQL\u NUM\u行错误,php,mysql,oop,Php,Mysql,Oop,好吧,我的注册是有效的,除了检测用户是否已经创建的代码不起作用之外 public function register($username, $password) { global $core, $db; $username = $core->encrypt($username); $password = $core->encrypt($password); $qry = $db->query("SELECT *

好吧,我的注册是有效的,除了检测用户是否已经创建的代码不起作用之外

public function register($username, $password) { 
  global $core, $db; 

  $username = $core->encrypt($username); 
  $password = $core->encrypt($password); 

  $qry = $db->query("SELECT * 
                       FROM `users` 
                      WHERE `username` = '$username' 
                        AND `password` = '$password'"); 
  $num = $db->num($qry); 

  if($num >= 1) { 
    echo 'This username has already been registered'; 
  } else { 
    $db->query("INSERT INTO `users` 
                  (username, password) 
                VALUES 
                  ('{$username}', '{$password}');"); 
  } 
}
这是我使用的代码,我认为这是查询中的某种形式的错误,但我不确定是什么-可能是一个小错误

MySQL错误:

>Warning: mysql_num_rows() expects parameter 1 to be resource, null given in C:\xampp\htdocs\housekeeping\_inc\db.inc.php on line 14

>$db->num code
        public function num($query) {
            mysql_num_rows($query); << line 14
        } 
>警告:mysql_num_rows()希望参数1是resource,在第14行的C:\xampp\htdocs\housing\\u inc\db.inc.php中给出空值
>$db->num代码
公共函数num($query){

mysql_num_rows($query);您忘记了num函数和qry函数中的return语句。它应该是:

public function num($query) {
    return mysql_num_rows($query);
}
您没有提供qry函数的代码,但应该是这样的:

public function qry($sql) {
    return mysql_query($sql);
}

您确定$db->query会返回结果(而不是true/false)。在函数内部,在全局变量之后,尝试“print_r($db);”并查看结果