php calling variables from another script to assign for the session

Question

i have a login function in a seperate script(main.php) which is like the below

public function login($username,$password) {
$linkingcon = $this->getConnection();
$sqlquery = "SELECT ac.userID,ac.name,us.usertype FROM users us JOIN accounts ac ON us.userID = ac.userID WHERE us.username='$username' AND us.password='$password';";
$result = mysql_query($sqlquery , $linkingcon);
$this->throwMySQLExceptionOnError($linkingcon);
$row = mysql_fetch_array($result);
$survey = new stdClass();
if($row) {
    $res->userID = (int)$row['userID'];
    $res->name = $row['name'];
    $res->usertype = (int)$row['usertype'];
            $string = rand() . 'SurveyLand' . rand() . $username. $password;
        $_SESSION['SURVEYLAND_KEY'] = md5($string);
} else {
    $res = false;
}
return $res;

}

and im calling the above login function from another script but i am currently unable to call the "usertype" variable of the above function... below is the function that i have written to call the "usertype", can anybody check what is wrong with the below function

function login($parameters) {
$main = new Main();
$log = $main->login($parameters["username"], $parameters["password"]);
if($log != false){
    $_SESSION['usertype'] = $res->usertype;
    print_r($_SESSION);
    }
return $log;

}

what is the error you are getting? and what is "$pro" in your second code? — Shivek Parmar
– Shivek Parmar, Commented Aug 22, 2014 at 5:30
have you tried to print $res->usertype; over there? and is it print value? — Bhumi Shah
– Bhumi Shah, Commented Aug 22, 2014 at 5:38

Tyler Marien · Accepted Answer · 2014-08-22 05:38:44Z

1

Try changing $res to $log in the second method. $res is the variable name you use in the first method, but is out of scope in the second method. However you assign the response of the first method (ie. $res) to $log in the second method.

function login($parameters) {
  $main = new Main();
  $log = $main->login($parameters["username"], $parameters["password"]);
  if($log != false){
    $_SESSION['usertype'] = $log->usertype;
    print_r($_SESSION);
  }
  return $log;
}

answered Aug 22, 2014 at 5:38

Tyler Marien

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