Question

我正在处理的作业（电子商务课程）要求我使用php为用户生成新密码，并向用户发送包含新密码的电子邮件。我成功生成了密码，使用php mail（）从我学校的服务器发送电子邮件给自己（gmail帐户），但是代表密码的php变量始终为空。我一直在这里和其他网站寻找答案，但找不到我做错了什么。我希望解决这个特殊问题，我不打算使用PHPMailer或其他替代方案。此外，我不打算讨论更安全的方式来发送电子邮件，或讨论加密，只是想讨论这个特定问题以及它为什么或不起作用。提前感谢您的任何建议。

if ($mysqli->conect_errno) {
        die("Error: Could not connect to database." . $mysqli->connect_error);
    } else {
        echo "<p>Connected<br></p>";
    }

    $email = $_POST['email_input'];
    try {
        $password = reset_password($email, $mysqli);
        notify_password($email, $password, $mysqli);
        echo 'Your password has changed and has been emailed to you.<br>';
    }
    catch(Exception $e) {
        echo 'Your password could not be reset';
    }

    function reset_password($email, $mysqli){
        $new_password = randomString(8, 12);
        if ($new_password == false) {
            throw new Exception('could not generate new password');
        }
        $rand_number = rand(0, 999);
        $new_password .= $rand_number;
        echo "NEW PASSWORD: " .$new_password."\r\n";

        $query = "UPDATE registration 
        SET password = sha1('".$new_password."')
        WHERE email = '".$email."'";

        $result = $mysqli->query($query);
        if($result) {
            echo "<br>Password Reset<br>";
        }else {
            echo "An error has occured";
        }

    }
    function randomString($min_length, $max_length){
        return substr(str_shuffle("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"), $min_length, $max_length);       
    }

    function notify_password($email, $password, $mysqli){
        $query = "SELECT email FROM registration WHERE email='".$email."'";
        $result = $mysqli->query($query);
        if(!$result){
            throw new Exception('could not find email address');
        }else if ($result->num_rows == 0) {
            throw new Exception('Could not find email address:user not in database');    
        }else {
            $row = $result->fetch_object();
            $email = $row->email;
            $from = "From support@HelpFinder \r\n";
            $mesg = "Your password has been changed to ".$password."\r\n"."Please change it the next time you log in.\r\n";

            if(mail($email, 'HelpFinder Login Information', $mesg, $from)) {
                return true;
            }else {
                throw new Exception('Could not send email.');
            }
        }

    }

the email message that arrives

example from text book I'm learning from

Answer 1

首先，请在if声明之前检查您是否有双$结果。

$result = $result = $mysqli->query($query);

之后，您可以尝试将var_dump设置为$ password变量，以检查它是否已正确传递给notify_password函数。您还可以发布$ password变量定义，以便我们更深入地检查。

Answer 2

检查您是否正确发送$ password变量作为参数，可能是空的

php变量不会显示在电子邮件中

2 个答案: