Parse error: syntax error, unexpected T_VARIABLE, expecting T_FUNCTION

2
votes

Could someone tell me what I'm doing wrong?

I want to display the users online on specific rooms only.

the code below is the function that calls my online.php this is under my chat.php when I load the page this function also loads.

function whos_online() {
  if ( window.XMLHttpRequest ) {
    xmlhttp = new XMLHttpRequest();
  } else { 
    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
  }
  xmlhttp.open("GET", "online.php?room=<?php $_SESSION['room']?>", false);
  xmlhttp.send();
  document.getElementById("whos_online").innerHTML = xmlhttp.responseText; 
}

ONLINE.PHP

this is the content of my online.php

<link rel="stylesheet" type="text/css" href="style.css" />
<?php

    session_start();
    include 'db.inc.php';

    class WhosOnline{
        $rn = $_GET['room'];
        protected $get_status_query = "SELECT * FROM `online_users` WHERE `room` = '{$rn}'";
        public function DisplayUsers(){
            $get_current_status = mysql_query( $this->get_status_query );
            if( mysql_num_rows( $get_current_status ) != 0 ) {
                while( $row_status = mysql_fetch_array( $get_current_status ) ) {
                    if( $_SESSION['username'] == true ) {
                        echo "<div class='online_margin'>
                                <b>".base64_decode($row_status['username'])."</b>
                              </div>
                              <hr style='border: 0; border-top: solid 1px #D8D8D8;margin: 5px 10px 5px 10px;' />";
                    }
                }
            }
        }
    }

    $Online = new WhosOnline;
    $Online->DisplayUsers();
?>

Any help?

3
phpajaxsyntax-error
You made a type somewhere, look in the neighborhood of the error and see if there is a syntax error. – Zombaya
what do you mean by type?. the error points out to $rn = $_GET['room']; and i dont know why. – user1232117
This is not how you should do it. Ever. – Ignacio Vazquez-Abrams
Should have been typo. And you can not have code inside a class, only within a method of a class. – Zombaya

3 Answers

2
votes
$rn = $_GET['room'];
protected $get_status_query = "SELECT * FROM `online_users` WHERE `room` =     '{$rn}'";

This is a bad habit that you need to break RIGHT NOW.

protected function get_status_query($rn) {
  return "SELECT * FROM `online_users` WHERE `room` =     '". sanitize($rn) . "'";
};

Implementation of sanitize() is left to the reader.

2
votes

you could not initialized any variable directly in class , try this

public $rn;
protected $get_status_query;

public __construct(){
      $this->rn = $_GET['room'];
      $this->get_status_query = "SELECT * FROM `online_users` WHERE `room` = '{$this->rn}'";
}
1
votes

Ok, even this gives an error:

class WhosOnline{
    public $rn = $_GET['room'];
}

This also gives an error:

$v = "Hi there";
class WhosOnline{
    public $rn = $v;
}

The error is because you're trying to set a variable based on another variable in the class definition. You could do this in the constructor. Or you can set class members based on CONSTANTS (as you were doing with the query string). But why not rewrite your WhosOnline method like this:

public function DisplayUsers(){
    $get_current_status = mysql_query(
        "SELECT * FROM `online_users` WHERE `room` = '" 
            . mysql_real_escape_string($_GET['room']) . "'");
    if(mysql_num_rows($get_current_status)!=0){
        while($row_status = mysql_fetch_array($get_current_status)){
            if($_SESSION['username']==true){
                echo "<div class='online_margin'>   <b>".base64_decode($row_status['username'])."</b></div><hr style='border: 0; border-top:  solid 1px #D8D8D8;margin: 5px 10px 5px 10px;' />";
            }
        }
    }
}

This will also remove any potential errors you might have with $this-> references missing.