2016-11-25 42 views
-3

我試圖讓一個登錄名和PHP中註冊系統和我不斷收到一個錯誤代碼時,我使用header("Location: index.php"),這是錯誤我得到:標題位置PHP不工作?

警告:不能更改頭信息 - 頭已經發出(輸出開始於C:\ Users \ Omg \ Desktop \ XAMPP \ htdocs \ Websites \ Social-Network \ register.php:43)在C:\ Users \ Omg \ Desktop \ XAMPP \ htdocs \ Websites \ Social-Network \ register .PHP上線141 的index.php

這裏是我的PHP代碼:

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"); 

} 

我正在使用一個網站來幫助我,我也有一個數據庫連接,但它沒有提到。任何幫助,將不勝感激。

+0

好 - 看看錯誤信息,以及它試圖告訴你什麼:在發送內容後,**不能**發送標題**! –

+0

我甚至在所有代碼之前發送它,但仍然出現錯誤 –

+0

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

回答

2

其實這個錯誤是相當明確的:頭部必須是你發送給客戶端的第一件東西,在主體之前。

所有回聲都是身體的一部分。

此外,它看起來像ajax調用和標準調用之間的混合。你到底想要達到什麼目標?