2015-08-31 56 views
0

我正在使用本教程 http://www.wikihow.com/Create-a-Secure-Login-Script-in-PHP-and-MySQL 以保護對我的網頁的訪問,並且我想通過重置密碼功能對其進行修改。不幸的是我的重設密碼的插件商店的東西登錄表單/密碼重置

這使得帳戶不能登錄該代碼被執行用戶註冊和工程就像一個魅力

$random_salt = hash('sha512', uniqid(mt_rand(1, mt_getrandmax()), true)); 

    // Create salted password 
    $password = hash('sha512', $password . $random_salt); 

    // Insert the new user into the database 
    if ($insert_stmt = $mysqli->prepare("INSERT INTO members (username, email, password, salt) VALUES (?, ?, ?, ?)")) { 
     $insert_stmt->bind_param('ssss', $username, $email, $password, $random_salt); 
     // Execute the prepared query. 
     if (! $insert_stmt->execute()) { 
      header('Location: ../error.php?err=Registration failure: INSERT'); 
     } 
    } 
    header('Location: ./register_success.php'); 

登錄處理:

// hash the password with the unique salt. 
    $password = hash('sha512', $password . $salt); 

    if ($stmt->num_rows == 1) { 
     // If the user exists we check if the account is locked 
     // from too many login attempts 

     if (checkbrute($user_id, $mysqli) == true) { 
      // Account is locked 
      // Send an email to user saying their account is locked 
      return false; 
     } else { 
      // Check if the password in the database matches 
      // the password the user submitted. 
      if ($db_password == $password) { 

也適用精細。 (不是我的工作:)) ,現在我的一部分 - 隨機字符串發生器隨後通過散列

$characters = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'; 
$charactersLength = strlen($characters); 
$randomString = ''; 
for ($i = 0; $i < 8; $i++) { 
    $randomString .= $characters[rand(0, $charactersLength - 1)]; 
} 

$random_salt = hash('sha512', uniqid(mt_rand(1, mt_getrandmax()), true)); 

    // Create salted password 
$password = hash('sha512', $randomString . $random_salt); 

      if ($insert_stmt = $mysqli->prepare("UPDATE members SET password = ?, salt=? WHERE username= ?")){ 
     $insert_stmt->bind_param('sss', $password, $random_salt, $username); 



$insert_stmt->execute(); 
             $insert_stmt->close(); 
    } 

復位腳本成功修改數據庫用戶,存儲正確的鹽和正確的哈希密碼。 測試通過在屏幕上顯示隨機文本和鹽,比這裏 http://www.convertstring.com/cs/Hash/SHA512

任何意見如何跟蹤哈希處理或提示如何解決它,將不勝感激結合。

我有一個懷疑,可能有一些麻煩JS這在客戶端哈希密碼

function formhash(form, password) { 
    // Create a new element input, this will be our hashed password field. 
    var p = document.createElement("input"); 

    // Add the new element to our form. 
    form.appendChild(p); 
    p.name = "p"; 
    p.type = "hidden"; 
    p.value = hex_sha512(password.value); 

    // Make sure the plaintext password doesn't get sent. 
    password.value = ""; 

    // Finally submit the form. 
    form.submit(); 
} 

非常感謝您

+1

你真的應該使用PHP的[內置函數(http://jayblanchard.net/proper_password_hashing_with_PHP .html)來處理密碼安全性。如果您使用的PHP版本低於5.5,則可以使用'password_hash()'[兼容包](https://github.com/ircmaxell/password_compat)。 –

+3

修復什麼?你沒有告訴我們問題是什麼。 –

+0

...不幸的是我的重置密碼插件存儲的東西,使帳戶無法登錄.... – FrantisekNebojsa

回答

0

逆向工程從formhash()登錄處理會建議你應該不是散列$randomString,而是散列它的sha512版本。 所以在您的密碼重置插件,而不是這一行:

$password = hash('sha512', $randomString . $random_salt); 

使用這兩個:

$password = hash('sha512', $randomString); 
$password = hash('sha512', $password . $random_salt); 
+0

非常感謝。魔法! :) – FrantisekNebojsa