0

以下代码用于注册表单，由 WYSIWYG Web builder 自动生成。它一直在说password并且confirmpassword不一样。我对PHP一无所知，所以我找不到错误。

我需要帮助：

<?php $mysql_server = 'db4free.net';
$mysql_username = '----';

$mysql_password = '----';
$mysql_database = '----';
$mysql_table = 'userdb';
$success_page = './userpage.php';
$error_message = "";
if ($_SERVER['REQUEST_METHOD'] == 'POST' && $_POST['form_name'] == 'signupform')
{
   $newusername = $_POST['username1'];
   $newemail = $_POST['email'];
   $newpassword = $_POST['password1'];
   $confirmpassword = $_POST['confirmpassword'];
   $newfullname = $_POST['fullname'];
   $code = 'NA';
   if ($newpassword!= $confirmpassword)
   {
      $error_message = 'Password and Confirm Password are not the same!';
   }
   else
   if (!preg_match("/^[A-Za-z0-9_!@$]{1,50}$/", $newusername))
   {
      $error_message = 'Username is not valid, please check and try again!';
   }
   else
   if (!preg_match("/^[A-Za-z0-9_!@$]{1,50}$/", $newpassword))
   {
      $error_message = 'Password is not valid, please check and try again!';
   }
   else
   if (!preg_match("/^[A-Za-z0-9_!@$.' &]{1,50}$/", $newfullname))
   {
      $error_message = 'Fullname is not valid, please check and try again!';
   }
   else
   if (!preg_match("/^.+@.+\..+$/", $newemail))
   {
      $error_message = 'Email is not a valid email address. Please check and try again.';
   }
   if (empty($error_message))
   {
      $db = mysql_connect($mysql_server, $mysql_username, $mysql_password);
      if (!$db)
      {
         die('Failed to connect to database server!<br>'.mysql_error());
      }
      mysql_select_db($mysql_database, $db) or die('Failed to select database<br>'.mysql_error());
      $sql = "SELECT username FROM ".$mysql_table." WHERE username = '".$newusername."'";
      $result = mysql_query($sql, $db);
      if ($data = mysql_fetch_array($result))
      {
         $error_message = 'Username already used. Please select another username.';
      }
   }
   if (empty($error_message))
   {
      $crypt_pass = md5($newpassword);
      $newusername = mysql_real_escape_string($newusername);
      $newemail = mysql_real_escape_string($newemail);
      $newfullname = mysql_real_escape_string($newfullname);
      $sql = "INSERT `".$mysql_table."` (`username`, `password`, `fullname`, `email`, `active`, `code`) VALUES ('$newusername', '$crypt_pass', '$newfullname', '$newemail', 1, '$code')";
      $result = mysql_query($sql, $db);
      mysql_close($db);
      $subject = 'Your new account';
      $message = 'A new account has been setup.';
      $message .= "\r\nUsername: ";
      $message .= $newusername;
      $message .= "\r\nPassword: ";
      $message .= $newpassword;
      $message .= "\r\n";
      $header  = "From: webmaster@yourwebsite.com"."\r\n";
      $header .= "Reply-To: webmaster@yourwebsite.com"."\r\n";
      $header .= "MIME-Version: 1.0"."\r\n";
      $header .= "Content-Type: text/plain; charset=utf-8"."\r\n";
      $header .= "Content-Transfer-Encoding: 8bit"."\r\n";
      $header .= "X-Mailer: PHP v".phpversion();
      mail($newemail, $subject, $message, $header);
      header('Location: '.$success_page);
      exit;
   }
}
?>
4

2 回答 2

0

密码字段的名称是“密码”。用户名字段相同。

您应该更改以下行：

$newusername = $_POST['username1'];
$newpassword = $_POST['password1'];

至：

$newusername = $_POST['username'];
$newpassword = $_POST['password'];
于 2013-06-20T20:06:36.920 回答
0

字段“password1”的名称实际上是“password”。只有 id 是“password1”，但在发布数据中没有使用。

于 2013-06-20T20:01:39.407 回答