我有以下查询:
$cuttinglist_products_query =
tep_db_query("select op.orders_products_id, op.orders_id, op.products_id, ".
"op.products_model, op.products_name, op.products_quantity, ".
"p.products_id from " . TABLE_ORDERS_PRODUCTS . " op " .
" left join " . TABLE_PRODUCTS . " p " .
" on (op.products_id = p.products_id) where orders_id = '" .
(int)$cuttinglist['orders_id'] . "'");
这将两个表连接在一起,具有相同的id。我如何加入名为的第3个表:
"TABLE_ORDERS_PRODUCTS_ATTRIBUTES"
具有相同的ID:
"TABLE_ORDERS_PRODUCTS"
使用的ID是:
"orders_products_id"
答案 0 :(得分:5)
只需添加另一个连接子句:
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 ...
答案 1 :(得分: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 );