php - Displaying array content from function? -

can tell me how display value of next code below?

in user.php file have next content:

class user {     protected $userid;     protected $useremail;     protected $userpassword;     public function __construct() {         $this->userid = preg_replace('#[^0-9]#i', '',              $_session['user_id']);         $this->useremail = preg_replace('#[^a-za-z0-9@_.-]#i', '',              $_session['user']);         $this->userpassword = preg_replace('#[^a-za-z0-9]#i', '',              $_session['user_password']);     }  public function userinfoquery() {     $sql = "select * users id =         '$this->userid' , email = '$this->useremail' ,          password = '$this->userpassword' limit 1";     $res = mysql_query($sql) or die(mysql_error());      $usermatch = mysql_numrows($res);      if ($usermatch == 1) {         while($row = mysql_fetch_array($res)) {             $userdata = array(                 $userfirstname = $row['firstname'],                 $userlastname = $row['lastname'],                 $userbirthdate = $row['birthdate'],                 $usersex = $row['sex'],                 $useremail = $row['email'],                 $usercountry = $row['country'],                 $userregion = $row['region']);             }         }          return $userdata;     } } 

in index php file when try:

$user = new user(); print_r($user->userinfoquery()); 

i have next results:

array ( [0] => firstname [1] =>      lastname [2] =>      1990-11-23 [3] =>      male [4] =>      mail [5] =>      srbija [6] => town )  

how can echo first , last names?

this:

array($userfirstname = $row['firstname']) 

assigns value of $row['firstname'] variable $userfirstname, puts result of assignment (the value of $row['firstname']) array. it's same writing:

$userfirstname = $row['firstname']; array($row['firstname']); 

to declare array key userfirstname, need write:

array('userfirstname' => $row['firstname']) 

from here, have normal array can access:

$userinfo = $user->userinfoquery(); echo $userinfo['userfirstname']; 

this seem clunky though, , honestly, you're not using objects here. should save data queried database properties of object, use getters access properties 1 one or together. how design proper object little beyond scope/point of answer though.