Question

我想要做的是当在concessions_sold表中放置订单（使用插入）时，使用触发器更新库存表中销售的数量。我承认我吮吸PLSQL所以我不确定我是否正确地采用了这种方式。我得到的错误是：


SQL> insert into concessions_sold
  2  values(33, 104, '26-Apr-09', 50);
insert into concessions_sold
            *
ERROR at line 1:
ORA-01779: cannot modify a column which maps to a non key-preserved table

我的代码：


create or replace trigger LabEx5_1 after insert on Concessions_sold
for each row
begin
if inserting then
update
( 
select i.quantity 
from inventory i, concessions_sold cs, concession c
where i.inventory_id = c.inventory_id and c.concession_id = cs.concession_id
)
set quantity = :new.quantity;
end if;
end LabEx5_1;
/

Answer 1

您正在尝试更新触发器中的连接视图，这对于何时可以执行此操作有几个限制;有关详细信息，请参阅Oracle documentation。

这个更新应该做你想要实现的目标：

UPDATE inventory i 
 SET i.quantity = :new.quantity 
 WHERE i.inventory_id = 
  (SELECT c.inventory_id 
    FROM concessions c 
     WHERE c.concession_id = :new.concession_id)

Answer 2

首先，如果您使用“For Each Row”触发器，那么您不能在整个表上操作，只需要一行，所以

select i.quantity 
from inventory i, concession c
where i.inventory_id = c.inventory_id and c.concession_id = cs.concession_id

应改为

select i.quantity 
from inventory i, concessions_sold cs, concession c
where i.inventory_id = c.inventory_id and c.concession_id = :new.concession_id

其次，更新应该是这样的：

update inventory
set quantity = :new.quantity
where inventory_id = ( 
  select inventory_id from concession c where concession_id = :new.concession_id 
) ;

因此触发器应如下所示：

create or replace 
trigger LabEx5_1 after insert on Concessions_sold
for each row
begin
  if inserting then
    update inventory
      set quantity = :new.quantity
    where inventory_id = ( 
      select inventory_id from concession c 
      where concession_id = :new.concession_id 
    ) ;
  end if;
end LabEx5_1;

Answer 3

一些事情：

“插入后”和“如果（插入）”是重复的。删除“If（插入）”是不必要的，因为您的触发器仅限于AFTER INSERT。只需添加更多代码即可阅读。
看来你在卖东西时试图降低你的库存。我没有看到你真的这样做。
这是embeded查询没有与之关联的密钥。（这是你的错误信息来自）。

select i.quantity 
from inventory i, concessions_sold cs, concession c
where i.inventory_id = c.inventory_id and c.concession_id = cs.concession_id

您没有where子句，并且您的内联选择不会限制从库存表中返回的行。如果您确实这样做了，那么您将更新清单表中的每一行。

使您的实际更新条款有效。

  UPDATE (
    SELECT **i.inventory_id**, i.quantity   
    FROM Inventory i
      , Concessions_sold cs
      , Concessions c   
    WHERE i.inventory_id = c.inventory_id   
    AND c.concession_id = **:NEW.concession_id** 
  )
  SET quantity = :new.quantity

现在它的工作存在一些问题： 1.不需要链接表。 2.内联SQL使得这个UPDATE语句更难以阅读，因此将来更难修改。

我会更明确：


UPDATE Inventory
SET quatity = quanaity - :new.quanity
WHERE inventory_id IN (
  SELECT inventory_id
  FROM Conessions c
  JOIN c.concession_id = :NEW.concession_id
  )

显然，没有检查库存数量是否确实存在，您可能需要考虑其他因素。

所以你的新触发器看起来像这样。


CREATE or REPLACE TRIGGER LabEx5_1 AFTER INSERT ON Concessions_sold
FOR EACH ROW
BEGIN
  UPDATE Inventory
  SET quatity = quanaity - :NEW.quanity
  WHERE inventory_id IN (
    SELECT inventory_id
    FROM Conessions c
    JOIN c.concession_id = :NEW.concession_id
    );
END LabEx5_1;

PLSQL - 触发器：无法修改映射到非密钥保留表的列

3 个答案: