Question

我现在有一个会员系统，当有人注册一个帐户时，我需要通过他的电子邮件激活它。他将在收件箱中收到有效链接，例如： activate.php?email=ipoon2@outlook.com&email_code=b5b90ae21e31229878d681680db16bdf此链接有效，因此当我转到此链接时，他会成功激活该帐户。

您看到?email= ipoon2@outlook.com 所以当我将其更改为 ipodn2@outlook.com 并且email_code仍然相同时，他无法激活他的帐户。他需要收到We cannot find that email之类的错误，当他更改email_code时，他会收到类似problem activate your account的错误

这就是我所得到的问题当我更改电子邮件时，我没有收到任何错误。对于email_code

都没有

我有一个名为activate.php的文件，该代码包括：

<?php
} else if (isset($_GET['email'], $_GET['email_code']) === true) {
    $email  = urldecode(trim($_GET['email']));
    $email_code = trim($_GET['email_code']);
    $user = new User();
    if(User::email_exists($email) === false) {
        echo 'We cannot find that email'; // return error doesn't show up
    } else if (User::activate($email, $email_code) === false) {
        echo 'problem activate your account'; // return error doesn't show up
    }

}

?>

此外，我还有2个函数，类文件User.php

public function email_exists($email) {  
    require './config.php';     
    $email  = urldecode(trim($_GET['email']));  
    $sql_30 =   $db->query("SELECT COUNT(id) FROM users WHERE email = '$email'");   
    if ($sql_30->fetch_object() === true) {         
        return true;    
    } else if ($sql_30->fetch_object() === false) {         
        return false;   
    }   
}



public function activate($email, $email_code) {
    require './config.php';
    $email  = urldecode($email);
    $email_code = $db->real_escape_string($email_code);
    $sql_33 = $db->query("SELECT COUNT(`id`) FROM `users` WHERE `email` = '$email' AND `email_code` = '$email_code' AND `group` = 0");
    if ($sql_33->fetch_object()) {
        $db->query("UPDATE `users` SET `group` = 1 WHERE `email` = '$email' AND `email_code` = '$email_code'");
        return true;
    } else {
        return false;
    }
}

Answer 1

对我来说，你的email_exists（）和activate（）是错误的。

if ($sql_30->fetch_object() === true) {         
        return true;    
} else if ($sql_30->fetch_object() === false) {         
        return false;   
}

来自mysqli_result::fetch_object的php文档： Returns an object with string properties that corresponds to the fetched row or NULL if there are no more rows in resultset.所以你的测试必须是：

if ($sql_30->fetch_object() !== NULL) {         
        return true;    
} 
return false;

我想它应该可以解决你的问题。

没有收到错误

1 个答案: