PHP: Hashing code not working any longer? - php

Question

I used this hash function for a while (got it from the internet). The thing is it used to work immediately but now it is complaining about a paramter. The following is the code:
function generateHash($plainText, $salt = null)
{
if ($salt === null)
{
$salt = substr(md5(uniqid(rand(), true)), 0, SALT_LENGTH);
}
else
{
$salt = substr($salt, 0, SALT_LENGTH);
}
return $salt . sha1($salt . $plainText);
}
So I would use this code in the method call:
validateUserInput($userid, $pass);
and validateUserInput is:
function validateUserInput($username, $password)
{
//$username = mysql_real_escape_string($username);
//$password = mysql_real_escape_string($password);
if(!$username || !$password)
{
//$errors['credentials'] = 'Missing Credentials!';
//$_SESSION['errors_array'] = $errors;
//echo $errors['credentials'];
header("LOCATION:XXXXXXX.php");
}
$local_salt = generateHash($password);
//echo $local_salt;
$groupid;
if($username != null && $password !=null)
{
connectToServer();
$result = mysql_query("SELECT * FROM users WHERE hashkey = '{$local_salt}'");
while($row_access = mysql_fetch_array($result))
{
$groupid = $row_access['groupid'];
}
if(!isset($result))
{
$errors['not_found_user'] = 'No Users Found with Provided Credentials!';
//$_SESSION['errors_array'] = $errors;
$userfound = 0;
$_SESSION['user_available'] = $userfound;
}elseif(isset($result)){
$_SESSION['user_logged'] = array('username' => $username, 'password' => $password, 'salt' => $local_salt, 'groupid' => $groupid);
$userfound = 1;
//echo "stored";
$_SESSION['user_available'] = $userfound;
}
}
}
finally the error is:
Warning: substr() expects parameter 3 to be long, string given in /home/XXXX.php on line 64
This is pointing to the function generateHash()

Answer 1

The error itself tells you everything. The constant SALT_LENGTH is not a long. I suspect it's not defined at all, so PHP converts the bare string to a string ("SALT_LENGTH") and passes that to substr(), which complains.
That being said... This code is dangerously wrong:
if(!isset($result)): Really? This condition will always be false because $result will always be set (unless you run into a problem with mysql_query(), but that doesn't tell you anything about the valididty of the login). And since mysql_query() never returns null, no logins will ever be rejected.
This query:
SELECT * FROM users WHERE hashkey = '{$local_salt}'
Is invalid. $local_salt = generateHash($password);. From the generateHash function, if a salt is not given, one will be randomly created for you. Thus, every call to generateHash will generate a new hash, which means it can't be compared to anything in the database.
On the basis of the two (very) egregious mistakes above, I would throw away this piece of code for good.
The correct way to check for a valid hash when a salt is used is something like:
$_SESSION['user_logged'] = null;
// fetch hashed pw from db, where username is the submitted username
$result = mysqli_query("SELECT hash FROM users WHERE username = '{$username}'");
if ($result->num_rows != 0)
{
$row = $result->fetch_assoc();
$hash = $row['hash'];
$salt = substr($hash, 0, SALT_LENGTH); // extract salt
if (generateHash($password, $salt) == $hash)
{
// login successful.
$_SESSION['user_logged'] = $username; // don't store passwords here
}
}
// if $_SESSION['user_logged'] is not set, the login failed
if (!isset($_SESSION['user_logged']))
{
// you *don't* want to tell people which one (login or pw) is invalid
echo 'Invalid login or password';
}
Note: It's very important that the SALT_LENGTH is at most 32, or this won't work because of the way generateHash() is implemented.

Answer 2

Clearly SALT_LENGTH is not an integer. Find where it's defined and correct this.

Answer 3

Instead of making a new hashing function each time you write an application , you should utilize the one provided to you by php: crypt() ( i would recommend to go with any blowfish or sha256+ hash ).
And when selecting information from database, you should select it by username and then, with php, check if hash fits the password.