Question

我已经尝试了一切来找出mysqli_query失败的原因。任何人都能明白我做错了什么。有可能我不再连接到dababase了?? !!先感谢您！

  function email_exists($email){
      $email = sanitize($email);
    $db = new mysqli('localhost','root','','secured_login');
    if($db->connect_errno){
        $connect_error = 'Sorry, we are experiencing connection problems.'; 
        die ($connect_error);
    }
    return (mysql_result(mysqli_query($db, "SELECT COUNT(`user_id`) FROM `users` WHERE `email` = '$email'"), 0) == 1) ? true : false;
}

错误

Warning: mysql_result() expects parameter 1 to be resource, object given in....

使用mysqli_fetch_row（）的替代解决方案; ＆lt; -----以下替代有效吗？

  function email_exists($email){
    $email = sanitize($email);
    $db = new mysqli('localhost','root','','secured_login');
        if($db->connect_errno){
            $connect_error = 'Sorry, we are experiencing connection problems.'; 
            die ($connect_error);
        }
    $query = "SELECT COUNT(`user_id`) FROM `users` WHERE `email` = '$email'";
    if ($result = mysqli_query($db, $query)){
        while ($result= mysqli_fetch_row($result)){
            return ($result);
        }           
    }
}

感谢任何反馈！

Answer 1

以最简单的形式，您应该看到类似于以下内容的内容，

function email_exists($email){
    $email = sanitize($email);
    $db = new mysqli('localhost','root','','secured_login');
    if($db->connect_errno){
        $connect_error = 'Sorry, we are experiencing connection problems.'; 
        die ($connect_error);
    }
    $query = $db->query("SELECT `user_id` FROM `users` WHERE `email` = '$email'");
    return ($query->num_rows > 1) ? true : false;
}

请记住清理您的输入，或者甚至更好地使用prepared statements。

Answer 2

您将mysql与mysqli混为一谈，为什么？

我认为这是一个错字 - 但将mysql_result更改为mysqli_result

为什么mysqli_query失败 - 当它返回资源时返回对象

2 个答案: