相关疑难解决方法(0)

我有以下代码:

 $final = array();
    foreach ($words as $word) {
        $query = "SELECT Something";
        $result = $this->_db->fetchAll($query, "%".$word."%");
        foreach ($result as $row)
        {
            $id = $row['page_id'];
            if (!empty($final[$id][0]))
            {
                $final[$id][0] = $final[$id][0]+3;
            }
            else
            {
                $final[$id][0] = 3;
                $final[$id]['link'] = "/".$row['permalink'];
                $final[$id]['title'] = $row['title'];
            }
        } 
    }

代码SEEMS工作正常,但我收到此警告:

Warning: Cannot use a scalar value as an array in line X, Y, Z (the line with: $final[$id][0] = 3, and the next 2).

谁能告诉我如何解决这个问题？