Oracle

Question

我有一个场景，我有两个 table

订单

ORDER_NO
LOCATION
ITEM
QTY_RECEIVED

发货

ORDER_NO
LOCATION
ITEM
QTY_RECEIVED

有些情况下 ORDER table 有记录但 SHIPMENT table 没有

我想要 ORDER table 中的所有行，其中数量不等于 SHIPMENT table 中的数量，并且将包括 ORDER 中但不在装运中的行。

我试过这样做：

SELECT 
    order_no, item, location, SUM(NVL(QTY_RECEIVED, 0)) 
FROM
    ORDERS ol                     
GROUP BY 
    ORDER_NO, ITEM, LOCATION
HAVING 
    SUM (NVL(ol.QTY_RECEIVED,0)) <>      

    (SELECT SUM(NVL(sk.QTY_RECEIVED, 0))
     FROM shipment s
     WHERE s.order_no = ol.order_no
       AND s.item (+)= ol.item
       AND s.location (+) = ol.location
     GROUP BY s.order_no, s.item, s.location);

但它没有给出正确的结果。

我应该怎么做？

Answer 1

您需要 LEFT JOIN 才能 return 结果，即使是 SHIPMENT table 的不存在值也是如此：

SELECT ol.order_no, ol.item, ol.location, 
       SUM(NVL(ol.QTY_RECEIVED, 0)) AS "Total Quantity Of Orders",
       SUM (NVL(s.QTY_RECEIVED,0))  AS "Total Quantity Of Shpm." 
  FROM orders ol 
  LEFT JOIN shipment s  
    ON s.order_no = ol.order_no
   AND s.item = ol.item
   AND s.location = ol.location
 GROUP BY ol.order_no, ol.item, ol.location
HAVING SUM (NVL(ol.QTY_RECEIVED,0)) <> SUM (NVL(s.QTY_RECEIVED,0))

Answer 2

在我看来，同一个商品可能有多个订单。如果是这种情况，您需要一种不同的方法：

select order_no, location, item, sum(o_qty), sum(s_qty)
from ((select order_no, location, item, qty_received as o_qty, 0 as s_qty
       from orders
      ) union all
      (select order_no, location, item, 0 as o_qty, qty_received as s_qty
       from shipment
      )
     ) os
group by order_no, location, item
having sum(o_qty) <> sum(s_qty);

Oracle - 具有连接条件的外部连接

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