认证来自sha1盐腌密码的用户登录

Question

我有一个非常简单的登录/用户注册脚本，它使用sha1和salt存储密码。我有密码和用户创建工作正常，并存储在数据库中的一切都很好，但是当我尝试使用凭据登录时，它不起作用。在搜索这个主题时，我似乎找不到任何东西。

这里是我的外接用户表单：

session_start(); 
include("includes/resume.config.php"); 

// make sure form fields have a value and strip them 
function check_input($data, $problem='') 
{ 
$data = trim($data); 
$data = stripslashes($data); 
$data = htmlspecialchars($data); 
if ($problem && strlen($data) == 0) 
{ 
    die($problem); 
} 
    return $data; 
} 

// get form values, escape them and apply the check_input function 
$name = $link->real_escape_string(check_input($_POST['name'], "Please enter a name!")); 
$email = $link->real_escape_string(check_input($_POST['email'], "Please enter an email!")); 
$password = $link->real_escape_string(check_input($_POST['password'], "Please enter a password!")); 

// generate a random salt for converting passwords into MD5 
$salt = bin2hex(mcrypt_create_iv(32, MCRYPT_DEV_URANDOM)); 
$saltedPW = $password . $salt; 
$hashedPW = sha1($saltedPW); 

mysqli_connect($db_host, $db_user, $db_pass) OR DIE (mysqli_error()); 
// select the db 
mysqli_select_db ($link, $db_name) OR DIE ("Unable to select db".mysqli_error($db_name)); 

// our sql query 
$sql = "INSERT INTO admins (name, email, password, salt) VALUES ('$name', '$email', '$hashedPW', '$salt');"; 

//save the updated information to the database   
mysqli_query($link, $sql) or die("Error in Query: " . mysqli_error($link)); 

if (!mysqli_error($link)) 
{ 
    header("Location: file_insert.php"); 
} 

这里是我的登录脚本：这就是不工作

function check_input($data, $problem='') 
{ 
$data = trim($data); 
$data = stripslashes($data); 
$data = htmlspecialchars($data); 
if ($problem && strlen($data) == 0) 
{ 
    die($problem); 
} 
    return $data; 
} 

if(isset($_POST['submitLogin'])) { //form submitted? 

// get form values, escape them and apply the check_input function 
$name = $link->real_escape_string(check_input($_POST['name'], "Please enter a name!")); 
$password = $link->real_escape_string(check_input($_POST['password'], "Please enter a password!")); 

$saltQuery = $link->query('SELECT salt FROM admins WHERE name = "'.$name.'"'); 

$salt = mysqli_fetch_assoc($saltQuery); 
$saltedPW = $password . $salt; 
$hashedPW = sha1($saltedPW); 

mysqli_connect($db_host, $db_user, $db_pass) OR DIE (mysqli_error()); 
// select the db 
mysqli_select_db ($link, $db_name) OR DIE ("Unable to select db".mysqli_error($db_name)); 

$validate_user = $link->query('SELECT id, name, password FROM admins WHERE name = "'.$name.'" AND password = "'.$hashedPW.'"'); 

if ($validate_user->num_rows == 1) { 
    $row = $validate_user->fetch_assoc(); 
    $_SESSION['id'] = $row['id']; 
    $_SESSION['loggedin'] = TRUE; 
    Header('Location: file_insert.php'); 
} else { 
    print "<center><p style='margin-top: 200px; font-weight: bold;'>Invalid Login Information</p>"; 
    print "<a href='admin-login.php'>Click here</a> to return to the login page.</center>"; 
} 
} 

Answer 1

可能有更多的事情，但肯定一个它不工作的原因是因为mysqli_fetch_assoc返回一个数组，并且您像字符串一样使用它。

由于此时$salt是一个数组，因此在调用$password . $salt时，PHP会抱怨数组要转换为字符串。结果是，您将字尾Array附加到导致不正确散列的密码。如果您有display_errors关闭和/或error_reporting设置为隐藏php.ini中的通知，那么您不会看到此消息。

如果更改：

$saltedPW = $password . $salt; 

到：

$saltedPW = $password . $salt['salt']; 

那么它应该工作。

此外，在将其插入到数据库之前，您应该转义$salt，因为它可能包含null，unprintable或single/double引号，因为它是随机生成的。

Answer 2

，如果你收到任何错误消息，但您发布的内容，我认为它是一个连接问题你没有说：

if(isset($_POST['submitLogin'])) { //form submitted? 

// Here, you didn't connect to database, but you are expecting to fetch salt! 
$saltQuery = $link->query('SELECT salt FROM admins WHERE name = "'.$name.'"'); 

$salt = mysqli_fetch_assoc($saltQuery); 

所以，你可能需要先连接到数据库：

if(isset($_POST['submitLogin'])) { //form submitted? 

mysqli_connect($db_host, $db_user, $db_pass) OR DIE (mysqli_error()); 
// select the db 
mysqli_select_db ($link, $db_name) OR DIE ("Unable to select db".mysqli_error($db_name)); 

// Here, you didn't connect to database, but you are expecting to fetch salt! 
$saltQuery = $link->query('SELECT salt FROM admins WHERE name = "'.$name.'"'); 

其次，mysqli_fetch_assoc与返回键名称为您在SELECT查询嵌入哪一个领域的一个数组，最终代码应该是这样的：

if(isset($_POST['submitLogin'])) { //form submitted? 

mysqli_connect($db_host, $db_user, $db_pass) OR DIE (mysqli_error()); 
// select the db 
mysqli_select_db ($link, $db_name) OR DIE ("Unable to select db".mysqli_error($db_name)); 

// Here, you didn't connect to database, but you are expecting to fetch salt! 
$saltQuery = $link->query('SELECT salt FROM admins WHERE name = "'.$name.'"'); 
$salt = mysqli_fetch_assoc($saltQuery); 
$saltedPW = $password . $salt["salt"]; 
$hashedPW = sha1($saltedPW);