1

私は次のクエリを持っています：

$cuttinglist_products_query = 
  tep_db_query("select op.orders_products_id, op.orders_id, op.products_id, ".
                      "op.products_model, op.products_name, op.products_quantity, ".
                      "p.products_id from " . TABLE_ORDERS_PRODUCTS . " op " .
                  " left join " . TABLE_PRODUCTS . " p " .
                  " on (op.products_id = p.products_id) where orders_id = '" . 
                  (int)$cuttinglist['orders_id'] . "'");

これにより、2つのテーブルが同じIDで結合されます。次の3番目のテーブルに参加するにはどうすればよいですか。

"TABLE_ORDERS_PRODUCTS_ATTRIBUTES"

同じIDで：

"TABLE_ORDERS_PRODUCTS"

使用されるIDは次のとおりです。

"orders_products_id"
4

2 に答える 2

5

別の結合句を追加するだけです。

SELECT ...
FROM TABLE_ORDERS_PRODUCTS op
LEFT JOIN TABLE_PRODUCTS p ON op.products_id = p.products_id
LEFT JOIN TABLE_ORDERS_PRODUCTS_ATTRIBUTES pa ON op.products_id = pa.orders_products_id
WHERE ...
于 2012-09-03T16:26:49.130 に答える
0 
$sql  = " select ";
$sql .= "   op.orders_products_id,  op.orders_id,          op.products_id,  op.products_model, ";
$sql .= "   op.products_name,       op.products_quantity,  p.products_id ";
$sql .= " from ". TABLE_ORDERS_PRODUCTS ." op ";   
$sql .= " left join ". TABLE_PRODUCTS ." p on (op.products_id = p.products_id) ";
$sql .= " left join ". TABLE_ORDERS_PRODUCTS_ATTRIBUTES ." pa on (pa.orders_products_id = op.orders_products_id) ";
$sql .= " where op.orders_id = ";
$sql .= "'". (int)$cuttinglist['orders_id'] ."'";

$cuttinglist_products_query = tep_db_query( $sql );
于 2012-09-03T16:44:41.143 に答える