php - 验证链接未激活账户

Question

所以我在注册后发送了一个链接来验证帐户，该链接包含用户的电子邮件地址和 32 个字符的代码，例如:

                $to      = $email;
                $subject = 'Signup | Verification';
                $message = '

                Thanks for signing up!
                Your account has been created, you can login with the following credentials after you have activated your account by pressing the url below.

                ------------------------
                Username: '.$username.'
                Password: '.$password.'
                ------------------------

                Please click this link to activate your account:
                localhost:8888/website/verify.php?email='.$email.'&hash='.$hash.'
                '; 

                $headers = 'From:myemail@email.com' . "\r\n"; 
                mail($to, $subject, $message, $headers); 

一切似乎都很好，我收到了带有如下链接的电子邮件:

http://localhost:8888/website/verify.php?email=myemail@email.com&hash=fe646d38bc2145ca6c3cf77d52820cd0

当我点击链接并尝试激活帐户时，问题就来了。我可以正常访问 Verify.php，但我一直收到 Invalid Approach，我无法将 Validation 设置为 1。

    <?php include "includes/base.php"; ?>

    <?php

        if(isset($_GET['Email']) && !empty($_GET['Email']) AND isset($_GET['Hash']) && !empty($_GET['Hash'])){
            $email = mysql_escape_string($_GET['Email']); 
            $hash = mysql_escape_string($_GET['Hash']); 
            $search = mysql_query("SELECT Email, Hash, Validation FROM users WHERE Email = '".$email."' AND Hash = '".$hash."' AND Validation = 0") or die(mysql_error()); 
            $match  = mysql_num_rows($search);


            if($match > 0){
                mysql_query("UPDATE users SET Validation = 1 WHERE Email = '".$email."' AND Hash = '".$hash."' AND Validation = 0") or die(mysql_error());
                echo "Your account has been activated, you can now login";
            }else{
                echo "The url is either invalid or you already have activated your account.";
            }

        }else{
            echo "Invalid approach, please use the link that has been sent to your email.";
        }


    ?>

Answer 1

1) 此代码不安全，因为它存在 SQL 注入(inject)问题。使用 prepared statements请记住，不再支持 mysql_* 函数，它们是 depriated

2) 关于您的代码，我发现您的 GET 请求的“email”和“hash”都是小写的，但在 PHP 代码中您使用 $_GET['Email'] 和 $_GET['Hash']。你需要改变这个:

 if(isset($_GET['Email']) && !empty($_GET['Email']) AND isset($_GET['Hash']) && !empty($_GET['Hash'])){
            $email = mysql_escape_string($_GET['Email']); 
            $hash = mysql_escape_string($_GET['Hash']); 

为此

 if(isset($_GET['email']) && !empty($_GET['email']) AND isset($_GET['eash']) && !empty($_GET['eash'])){
            $email = mysql_escape_string($_GET['email']); 
            $hash = mysql_escape_string($_GET['eash']); 

或将您的 GET 请求更改为下一个:

http://localhost:8888/website/verify.php?Email=myemail@email.com&Hash=fe646d38bc2145ca6c3cf77d52820cd0