При выводе в цикле, выводится по три раза вместо одного

Не могу понять где косяк.

PHP:

1. $order_list = mysql_query("SELECT tp.id, tp.model_name, tp.parent_brand, tp.image_main, tb.brand_name, tb.bid, tc.synonym, ts.product, ts.qty, ts.price, ts.type, ts.cart, ts.order_id FROM product_info tp, brands tb, categories tc, users tu, shopping_cart ts WHERE tp.parent_brand = tb.bid AND tp.parent_categorie = tc.id AND tp.id = ts.product AND ts.order_id = '$oid'") or die("Error2! ".mysql_error());
2.  
3. echo "<table>";
4.  
5. while ($order = mysql_fetch_array ($order_list))
6.     {
7.     $summ = $order['price'] * $order['qty'];
8.     $sum_sum += $summ;
9.     echo "<tr>
10.     <td><img src='../img_mid/".$order['image_main']."' width='100' height='150'></td>
11.     <td>".$order['synonym']." ".$order['brand_name']." ".$order['model_name']."</td>
12.     <td>".$order['type']."</td>
13.     <td>".$order['qty']."</td>
14.     <td>".$order['price']."</td>
15.     <td>".$summ."</td>
16.     </tr>";
17.  
18.     }
19.  
20. echo "</table>";
21. echo $sum_sum;

Выводит три раза одно и тоже

HTML:

1. <table><tr>
2.     <td><img src='../img_mid/pc_1024_1022.jpg' width='100' height='150'></td>
3.     <td>Комплект Pierre&nbsp;Cardin 1024/1022</td>
4.     <td></td>
5.     <td>1</td>
6.     <td>1860</td>
7.     <td>1860</td>
8.  
9.     </tr><tr>
10.     <td><img src='../img_mid/pc_1024_1022.jpg' width='100' height='150'></td>
11.     <td>Комплект Pierre&nbsp;Cardin 1024/1022</td>
12.     <td></td>
13.     <td>1</td>
14.     <td>1860</td>
15.     <td>1860</td>
16.  
17.     </tr><tr>
18.     <td><img src='../img_mid/pc_1024_1022.jpg' width='100' height='150'></td>
19.     <td>Комплект Pierre&nbsp;Cardin 1024/1022</td>
20.     <td></td>
21.     <td>1</td>
22.     <td>1860</td>
23.     <td>1860</td>
24.  
25.     </tr></table>5580
26.  
27. </td></tr></table>

 

Mr. T

Все верно!

но теперь пишет ошибку синтаксиса (((

PHP:

1.  
2. $order_list = mysql_query("SELECT tp.id, tp.model_name, tp.parent_brand, tp.image_main, tb.brand_name, tb.bid, tc.synonym, ts.product, ts.qty, ts.price, ts.type, ts.cart, ts.order_id FROM product_info tp left join brands tb left join categories tc left join users tu left join shopping_cart ts on tp.parent_brand = tb.bid AND tp.parent_categorie = tc.id AND tp.id = ts.product AND ts.order_id = '$oid'") or die("Error2! ".mysql_error());

 

Вроде разобрался с синтаксисом, но все-равно пишет ошибку

PHP:

1. $order_list = mysql_query("SELECT tp.id, tp.model_name, tp.parent_brand, tp.image_main, tb.brand_name, tb.bid, tc.synonym, ts.product, ts.qty, ts.price, ts.type, ts.cart, ts.order_id FROM product_info AS tp LEFT JOIN brands AS tb ON (tp.parent_brand = tb.bid) LEFT JOIN categories AS tc ON (tp.parent_categorie = tc.id) LEFT JOIN shopping_cart AS ts (tp.id = ts.product) WHERE ts.order_id = '$oid'") or die("Error2! ".mysql_error());

 

Denis.Y

Огромное спасибо за твою внимательность!!!