2013-02-09 78 views
0

我一直在开发近3周以后我自己的社交网络,我使用phpass使用下面的代码片段凑了密码到存储到我的数据库......Phpass散列密码检查

// END FORM DATA ERROR HANDLING 
// Begin Insertion of data into the database 

require("php_includes/PasswordHash.php"); 
$hasher = new PasswordHash(8,false); 
$p_hash = $hasher->HashPassword($p); 
if (strlen($p_hash)>=20){ 

    // Add user info into the database table for the main site table 
    $sql = "INSERT INTO users (username, email, password, gender, country, ip, signup, lastlogin, notescheck)  
      VALUES('$u','$e','$p_hash','$g','$c','$ip',now(),now(),now())"; 
    $query = mysqli_query($db_conx, $sql); 
    $uid = mysqli_insert_id($db_conx); 

小问题我已经是我的登录页面,在那里我有一个简单的表格,并在我的网页上这下面的代码块...

include_once("php_includes/check_login_status.php"); 
// If user is already logged in, header that weenis away 
if($user_ok == true){ 
    header("location: user.php?u=".$_SESSION["username"]); 
    exit(); 
} 
// AJAX CALLS THIS LOGIN CODE TO EXECUTE 
if(isset($_POST["e"])){ 
    // CONNECT TO THE DATABASE 
    include_once("php_includes/db_conx.php"); 
    // GATHER THE POSTED DATA INTO LOCAL VARIABLES AND SANITIZE 
    $e = mysqli_real_escape_string($db_conx, $_POST['e']); 
    require('php_includes/PasswordHash.php'); 
    $hasher = new PasswordHash(8, FALSE); 
    $hash = $hasher->HashPassword($p); 
    $checked = $hasher->CheckPassword($p, $hash); 
    $p = $_POST["p"]; 
    // GET USER IP ADDRESS 
    $ip = preg_replace('#[^0-9.]#', '', getenv('REMOTE_ADDR')); 
    // FORM DATA ERROR HANDLING 
    if($e == "" || $p == ""){ 
     echo "login_failed"; 
     exit(); 
    } else { 
    // END FORM DATA ERROR HANDLING 
     $sql = "SELECT id, username, password FROM users WHERE email='$e' AND activated='1' LIMIT 1"; 
     $query = mysqli_query($db_conx, $sql); 
     $row = mysqli_fetch_row($query); 
     $db_id = $row[0]; 
     $db_username = $row[1]; 
     $db_pass_str = $row[2]; 
     if($p != $db_pass_str){ 
      echo "login_failed"; 
      exit(); 
     } else { 
      // CREATE THEIR SESSIONS AND COOKIES 
      $_SESSION['userid'] = $db_id; 
      $_SESSION['username'] = $db_username; 
      $_SESSION['password'] = $db_pass_str; 
      setcookie("id", $db_id, strtotime('+30 days'), "/", "", "", TRUE); 
      setcookie("user", $db_username, strtotime('+30 days'), "/", "", "", TRUE); 
      setcookie("pass", $db_pass_str, strtotime('+30 days'), "/", "", "", TRUE); 
      // UPDATE THEIR "IP" AND "LASTLOGIN" FIELDS 
      $sql = "UPDATE users SET ip='$ip', lastlogin=now() WHERE username='$db_username' LIMIT 1"; 
      $query = mysqli_query($db_conx, $sql); 
      echo $db_username; 
      exit(); 
     } 
    } 
    exit(); 
} 

基本上什么发生的事情是努力当我收到一条错误信息登录!任何人有任何想法?

干杯的答案已经channged起来像这样...

?><?php 
// AJAX CALLS THIS LOGIN CODE TO EXECUTE 
if(isset($_POST["e"])){ 
    // CONNECT TO THE DATABASE 
    include_once("php_includes/db_conx.php"); 
    // GATHER THE POSTED DATA INTO LOCAL VARIABLES AND SANITIZE 
    $e = mysqli_real_escape_string($db_conx, $_POST['e']); 
    require('php_includes/PasswordHash.php'); 
    $hasher = new PasswordHash(8, FALSE); 
    $hash = $hasher->HashPassword($p); 
    $checked = $hasher->CheckPassword($p, $hash); 
    $hash = $_POST["p"]; 
    // GET USER IP ADDRESS 
    $ip = preg_replace('#[^0-9.]#', '', getenv('REMOTE_ADDR')); 
    // FORM DATA ERROR HANDLING 
    if($hash != $db_pass_str){//and not $p != $dp_pass_str 
     echo "login_failed"; 
     exit(); 
    } else { 
    // END FORM DATA ERROR HANDLING 
     $sql = "SELECT id, username, password FROM users WHERE email='$e' AND activated='1' LIMIT 1"; 
     $query = mysqli_query($db_conx, $sql); 
     $row = mysqli_fetch_row($query); 
     $db_id = $row[0]; 
     $db_username = $row[1]; 
     $db_pass_str = $row[2]; 
      // CREATE THEIR SESSIONS AND COOKIES 
      $_SESSION['userid'] = $db_id; 
      $_SESSION['username'] = $db_username; 
      $_SESSION['password'] = $db_pass_str; 
      setcookie("id", $db_id, strtotime('+30 days'), "/", "", "", TRUE); 
      setcookie("user", $db_username, strtotime('+30 days'), "/", "", "", TRUE); 
      setcookie("pass", $db_pass_str, strtotime('+30 days'), "/", "", "", TRUE); 
      // UPDATE THEIR "IP" AND "LASTLOGIN" FIELDS 
      $sql = "UPDATE users SET ip='$ip', lastlogin=now() WHERE username='$db_username' LIMIT 1"; 
      $query = mysqli_query($db_conx, $sql); 
      echo $db_username; 
      exit(); 
     } 
    exit(); 
} 
?> 

,但我仍然得到一个登录错误

+0

什么错误信息?数据库错误?或只是你的“Login_failed”? – 2013-02-09 11:41:54

+0

只是一个登录失败的错误消息! – 2013-02-09 11:42:55

+0

通过您的代码浏览,我只看到$ p设置为用户发布的密码,您将其与db中的假设为_hashed_的密码进行比较 – 2013-02-09 11:44:13

回答

1

你设置$ P作为POST密码(非散列)

$p = $_POST["p"]; 

看起来应该然后就:

if($hash != $db_pass_str){   //and not $p != $dp_pass_str 
    echo "login_failed"; 

即比较存储的密码与输入的密码和散列(这样比较两个散列)

+0

已将其更改,但仍然没有快乐!感谢您的帮助Damien! – 2013-02-09 12:01:58