php class return array

Question

I tried to get followers from MySQL usingy this class

class get_followers {
    public $followers_arr = array();
    public function __construct($user_id) {
        $query = "select * from followsystem where following ='$user_id'";

        $q = mysql_query($query) or die(mysql_error());

        $count = mysql_num_rows($q);

        if ($count > 0) {
            while ($row = mysql_fetch_assoc($q)) {
                array_push($this->followers_arr, $row['userid']);
           }
        }

        return $this->followers_arr;
    }
}

Then I initialize this class

$fol = new get_followers($userid);
$fol_arr = json_encode($fol);
echo $fol_arr;

Then I get

{"followers_arr":["1234","456"]}

but what i want want just to get this

["1234","456"]

How is that works?

Try: print_r($fol->followers_arr); and remove the $fol_arr = json_encode($fol); just to: $fol_arr = $fol; — Rizier123
– Rizier123, Commented Jan 23, 2015 at 11:14
First of all it's not OOP even though you are creating object. Read some more about that. — Elon Than
– Elon Than, Commented Jan 23, 2015 at 11:14
$fol = new get_followers($userid); $fol_arr = json_encode($fol->followers_arr); echo $fol_arr; it works when i did this thank you all — Waleed Kasem
– Waleed Kasem, Commented Jan 23, 2015 at 11:20

Vlad Cazacu · Accepted Answer · 2015-01-23 11:22:39Z

I don't think you understand how constructors work. You can't return a value from a constructor because it's just used to instantiate the object. When you're doing $fol_arr = json_encode($fol); you're actually encoding the entire object, not it's return value.

If you really want to use a class to do this, you should add a method to the class and use that, like this:

class Followers {
    public $followers_arr = array();
    public $user_id = null;

    public function __construct($user_id) {
        $this->user_id = $user_id;            
    }

    public function get()
    {
        $query = "select * from followsystem where following ='{$this->user_id}'";

        $q = mysql_query($query) or die(mysql_error());

        $count = mysql_num_rows($q);

        if ($count > 0) {
            while ($row = mysql_fetch_assoc($q)) {
                array_push($this->followers_arr, $row['userid']);
           }
        }

        return $this->followers_arr;
    }
}

And use it like this:

$fol = new Followers($userid);
$fol_arr = json_encode($fol->get());
echo $fol_arr;

ziGi · Accepted Answer · 2015-01-23 11:22:27Z

The solution to your problem is to do $fol_arr = json_encode($fol->followers_arr);

Nonetheless, making a class in this case is completely obsolete, since you only make it as a wrapper for a single function you want to execute (called get_followers) Instead of making a class, you could simply make the following:

function get_followers($user_id) {
        $followers_arr = [];
        $query = "select * from followsystem where following ='$user_id'";

        $q = mysql_query($query) or die(mysql_error());

        $count = mysql_num_rows($q);

        if ($count > 0) {
            while ($row = mysql_fetch_assoc($q)) {
                array_push($followers_arr, $row['userid']);
           }
        }

        return $followers_arr;

}

$fol = get_followers($userid);
$fol_arr = json_encode($fol);
echo $fol_arr;

There is no need to put it in a class unless the class serves the purpose of combining a few functions and variables to create a behaviour.

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