2016-11-25 45 views
-3

我试图让一个登录名和PHP中注册系统和我不断收到一个错误代码时,我使用header("Location: index.php"),这是错误我得到:标题位置PHP不工作?

警告:不能更改头信息 - 头已经发出(输出开始于C:\ Users \ Omg \ Desktop \ XAMPP \ htdocs \ Websites \ Social-Network \ register.php:43)在C:\ Users \ Omg \ Desktop \ XAMPP \ htdocs \ Websites \ Social-Network \ register .PHP上线141 的index.php

这里是我的PHP代码:

if(!empty($_POST)) { 
    if(empty($_POST['username'])){ 
    echo("Please enter a Display Name<br />"); 
    } 
    if(empty($_POST['password'])){ 
    echo("Please enter a password<br />"); 
    } 

    if(!filter_var($_POST['email'], FILTER_VALIDATE_EMAIL)){ 
    echo("Please enter a valid email address<br />"); 
    } 

    $query = " 
     SELECT 
      1 
     FROM users 
     WHERE 
      username = :username 
    "; 

    $query_params = array(
    ':username' => $_POST['username'] 
); 
    try{ 
    $stmt = $db->prepare($query); 
    $result = $stmt->execute($query_params); 
    } 
    catch(PDOException $ex) { 
    die("Failed to run query: "); 
    } 
    $row = $stmt->fetch(); 

    if($row){ 
    echo("This name is taken by another person<br />"); 
    } 
    $query = " 
     SELECT 
      1 
     FROM users 
     WHERE 
      email = :email 
    "; 

    $query_params = array(
    ':email' => $_POST['email'] 
); 
    try{ 
    $stmt = $db->prepare($query); 
    $result = $stmt->execute($query_params); 
    } 
    catch(PDOException $ex) { 
    echo("Failed to run query: "); 
    } 

    $row = $stmt->fetch(); 

    if($row){ 
    echo("This email is already taken!<br />"); 
    } 

    $query = " 
     INSERT INTO users (
     username, 
     password, 
     salt, 
     email 
    ) VALUES (
     :username, 
     :password, 
     :salt, 
     :email 
    ) 
    "; 

    $salt = dechex(mt_rand(0, 2147483647)) . dechex(mt_rand(0, 2147483647)); 

    $password = hash('sha256', $_POST['password'] . $salt); 

    for($round = 0; $round < 65536; $round++) { 
    $password = hash('sha256', $password . $salt); 
    } 

    $query_params = array(
    ':username' => $_POST['username'], 
    ':password' => $password, 
    ':salt' => $salt, 
    ':email' => $_POST['email'] 
); 

    try{ 
    $stmt = $db->prepare($query); 
    $result = $stmt->execute($query_params); 
    }catch(PDOException $ex) { 
    echo("Failed to run query: "); 
    } 

    header("Location: index.php"); 



    die("index.php"); 

} 

我正在使用一个网站来帮助我,我也有一个数据库连接,但它没有提到。任何帮助,将不胜感激。

+0

好 - 看看错误信息,以及它试图告诉你什么:在发送内容后,**不能**发送标题**! –

+0

我甚至在所有代码之前发送它,但仍然出现错误 –

+0

以下是网站:[http://forums.devshed.com/php-faqs-stickies-167/program-basic-secure-login-system-使用-php-mysql-891201.html](链接) –

回答

2

其实这个错误是相当明确的:头部必须是你发送给客户端的第一件东西,在主体之前。

所有回声都是身体的一部分。

此外,它看起来像ajax调用和标准调用之间的混合。你到底想要达到什么目标?