PHP return NULL from __construct - php

Question

I'm trying to write some classes which pull my data out of my database and create objects based on that data. I'm working with CodeIgniter.
The problem is that if an id is supplied to the __construct method but no row in the database table has that id, an object is returned but with all the properties set to NULL.
Is there a way that I can check this and return NULL instead of an object if there is no corresponding row?
class JS_Model extends CI_Model
{
protected $database_table_name;
protected $database_keys;
...
public function __construct($id = NULL){
if($id){
$this->getFromDatabase($id);
}
}
...
function getFromDatabase($id){
foreach($this->database_keys as $key){
$this->db->select($key);
}
$this->db->from($this->database_table_name);
$this->db->where('id', $id);
$this->db->limit(1);
$q = $this->db->get();
if ($q->num_rows() > 0){
foreach($q->result() as $property){
foreach($property as $key => $value){
$this->$key = $value;
}
}
} else {
// NEED TO SET THE OBJECT TO NULL FOR THIS CASE
}
}
...
}
Any help would be appreciated.

Answer 1

This is not possible, once __construct is called, you will receive an object instance.
The correct way to handle this is to throw an Exception from inside the constructor.
class JS_Model extends CI_Model
{
protected $database_table_name;
protected $database_keys;
public function __construct($id = NULL){
if($id){
$row = $this->getFromDatabase($id);
if (!$row) {
throw new Exception('requested row not found');
}
}
}
}
try {
$record = new JS_Model(1);
} catch (Exception $e) {
echo "record could not be found";
}
// from here on out, we can safely assume $record holds a valid record

Answer 2

why dont you make your constructor private and use a method such as GetInstance(); something like this. Syntax may be incorrect as i dont write much php anymore :-(.
class JS_Model extends CI_Model
{
protected $database_table_name;
protected $database_keys;
...
private function __construct($id = NULL){
}
public static function GetInstance($id)
{
$x = new JS_Model($id);
$x->getFromDatabase($id);
if(is_object($x))
{
return $x;
}
return null;
}
...
function getFromDatabase($id){
foreach($this->database_keys as $key){
$this->db->select($key);
}
$this->db->from($this->database_table_name);
$this->db->where('id', $id);
$this->db->limit(1);
$q = $this->db->get();
if ($q->num_rows() > 0){
foreach($q->result() as $property){
foreach($property as $key => $value){
$this->$key = $value;
}
}
} else {
return null;
}
}
...
}