Вопрос по if

Parse error: syntax error, unexpected $end in C:\wamp\www\login.php on line 32

и http://yandex[dot]ru - классный сайт. Заходил туда?

а Чемпион чето последнее время злой стал

<?php
include ('core/mysql.php');

class news {
var $num;
var $topic;
var $title;
var $text;
var $date;
var $img = array();
function __construct($num, $topic, $date, $img, $title, $text) {
$this->topic = $topic;
$this->title = $title;
$this->text = $text;
$this->date = $date;

list($this->img[0], $this->img[1], $this->img[2], $this->img[3], $this->img[4], $this->img[5])=explode("|", $img);

$this->num = $num;
}
}


function check_param($var)
    {
        $var = preg_replace("/[^a-z?-О0-9\.,_\n]/i", "", $var);
        return $var;
    }

function Load_News($sql, $t=null){
$arr_news = array();
if(isset($t))
$sql->sql_query('SELECT * FROM mscr_news WHERE id = '.$t.'ORDER BY mscr_news.date ASC');
else
$sql->sql_query('SELECT * FROM mscr_news ORDER BY mscr_news.date ASC');
$n = $sql->sql_numrows();
for($i=0; $i<$n; $i++)
{
$num = $sql->sql_fetchfield('id', $i);
$topic = $sql->sql_fetchfield('id', $i);
$text = $sql->sql_fetchfield('text', $i);
$date = $sql->sql_fetchfield('date', $i);
$title = $sql->sql_fetchfield('title', $i);
$img = $sql->sql_fetchfield('img', $i);

    $arr_news[$i] = new news($num, $topic, $date, $img, $title, $text);
    
}
return $arr_news;

$sql = new sql_db('localhost', 'root@localhost', '', 'mscr');
Load_News($sql,28);

for ($i = 0; $i < count($arr_news); $i++)
{
echo $arr_news[$i]." ";
}
?>