麻烦与SQL连接语句和引用项目在我的SQL语句

第一件事是我认为有错误在我的SQL连接语句(语法或其他)。我在语句中使用了2个表，一个叫做artists，另一个叫做prints。接下来，在我的php脚本中，我使用MYSQLI_ASSOC获取数组引用sql语句中的项目。我认为这里面也有问题。

    $sql = "SELECT CONCAT(a.first_name, a.middle_name, a.last_name) AS name p.print_name, p.price, p.size, p.description, p.what_genre, p.quantity_on_hand FROM artists AS a LEFT JOIN prints AS p USING (artist_id) ORDER BY $order_by $limit";

下面是第二部分:

    while ($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){
            $bg = ($bg=='#eeeeee' ? '#ffffff' : '#eeeeee');
            $list .= '<tr bgcolor="' . $bg . '">';
            $list .= '<td>' . $row['name'] . '</td><td>' . $row['print_name'] . '</td><td>' . $row['price'] . '</td><td>' . $row['size'] . '</td><td>' . $row['description'] . '</td><td>' . $row['what_genre'] . '</td><td>' . $row['quantity_on_hand'] . '</td></tr>';
        } 

下面是我得到的一些错误信息:

注意:查询:选择CONCAT (a。@ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @查看与MySQL服务器版本对应的手册，以便在第1行'p.print_name, p.price, p.size, p.description, p.what_genre, p.quantity_on_hand F'附近使用正确的语法

in D:'xampp'htdocs'larry'chapter192'view_users2.php on line 68
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in D:'xampp'htdocs'larry'chapter192'view_users2.php on line 110