Question

无论我输入什么（即使是正确的验证码），我总是会得到不匹配的输出。我已经尝试回显这些值（如果输入正确的代码，它们应该匹配）。我总是得到这样的东西：

6952304285049
-1247767175

我正在使用jquery-1.10.2.min.js（并在我的标题中链接了realperson.js文件） http://gfishdesigns.com/COMP2920/_COMPLETED/Assignment%202/SignUp.php

这是我的代码（我正在做其他验证）：

<?php
include 'Header.php';
include 'Database.php';
?>
<script type="text/javascript">
    $(function() {
        $('#defaultReal').realperson();
    });
</script>

<h1>Sign Up</h1>

<?php

if ($_POST){

    $username = $_POST['username'];
    $password = $_POST['password'];
    $check = '';

    //validate CAPTCHA
    function rpHash($value) { 
        $hash = 5381; 
        $value = strtoupper($value); 
        for($i = 0; $i < strlen($value); $i++) { 
        $hash = (($hash << 5) + $hash) + ord(substr($value, $i)); 
        } 
        return $hash; 
    } 
      if (rpHash($_POST['defaultReal']) == $_POST['defaultRealHash']) { ?>
        <p class="accepted">You have entered the "real person" value correctly and the form has been processed.</p>
<?php 

    //if username is not blank
    if($username != ''){

        //check if username exists already
        $query = "SELECT username FROM tbl_user;";
        $result = mysql_query($query) or die(mysql_error());
        while ($record = mysql_fetch_row($result))
        {
            foreach($record as $field)
            {
                if($field == $username){
                    //if user exists, dont let them add same user
                    $error_message_username = 'username already used; choose a unique name';
                }
                else{
                    $check = 'pass';
                }
            }
        }   
    }else{
        $error_message_username = 'username cannot be blank';
    }

    //if password is not blank
    if($password != ''){    
        $error_message_password = '';

        // encrypt password
        $encrypted_password = md5($password);   

        if($check == 'pass'){
            //set username and password into database
            $query = "INSERT INTO tbl_user VALUES('','".$username."','".$encrypted_password."');";
            $result = mysql_query($query) or die(mysql_error());
        }
    }else{
        $error_message_password = 'password cannot be blank';
    }



    } else { ?>
            <p class="rejected">You have NOT entered the CAPTCHA value correctly and the form has been rejected.</p>
    <?php 
echo rpHash($_POST['defaultReal']) . '<br/>';
echo $_POST['defaultRealHash'];

}

}    

?>

<form method="post" action="SignUp.php">
  <p>
    E-Mail:
    <input type="text" class="required email" id="username" name="username">
    <?php 
    if ( $error_message_username != '' ) {
        print "$error_message_username";
    }
    ?>
  </p>
  <p>
    Password:
    <input type="text" name="password">
            <?php 
    if ( $error_message_password != '' ) {
        print "$error_message_password";
    }
    ?>
  </p>
  <p>
    CAPTCHA:
    <input type="text" id="defaultReal" name="defaultReal">
  </p>
  <p>
    <button class="mybutton" type="submit" value="Sign Up">Sign Up</button>
  </p>
</form>

Answer 1

提供了两个版本的php rpHash函数，一个用于32位，一个用于64位PHP。运行phpinfo并确保使用此页http://keith-wood.name/realPerson.html中提供的正确版本的函数。这里使用的按位函数将在32位和64位机器上返回不同的值。查看此页面：http://www.php.net/manual/en/language.operators.bitwise.php

Answer 2

克雷格雅各布斯说，这是你指出的同样问题。我也遇到了同样的事情，并通过以下方式进行了更改来解决：

function rpHash($value) {
    $hash = 5381;
    $value = strtoupper($value);
    for($i = 0; $i < strlen($value); $i++) {
        $hash = (leftShift32($hash, 5) + $hash) + ord(substr($value, $i));
    }
    return $hash; }

function leftShift32($number, $steps) {
    $binary = decbin($number);
    $binary = str_pad($binary, 32, "0", STR_PAD_LEFT);
    $binary = $binary.str_repeat("0", $steps);
    $binary = substr($binary, strlen($binary) - 32);
    return ($binary{0} == "0" ? bindec($binary) :
        -(pow(2, 31) - bindec(substr($binary, 1)))); 
}

if (isset($_POST['submit'])) {

    .....
    if (rpHash($_POST['defaultReal']) != $_POST['defaultRealHash']) {
        echo "Invalid contact request, please try again with correct verification code...";
        exit;
    }
    .....
    .....
}

希望它也会帮助别人。

jQuery真人 - 总是失败

2 个答案: