如果在Oracle SQL中触发器的Else Condition

Question

我有两张桌子.他们是

CARD(cardid, credit, usertype,charge)

和

PAYMENTDEVICE(paydevid, paydevip,paydevdate, paydevtime, chargedcardid, mealtype). 

膳食类型可以是"客人"或"标准".我想在支付设备中插入新行时更新卡表中的信用.费用取决于使用类型.但如果用餐类型是客人,每个人都需要支付5美元.我尝试使用以下代码

CREATE OR REPLACE TRIGGER  "TRG_PAYMONEY" 
AFTER INSERT
ON PAYMENTDEVICE FOR EACH ROW

BEGIN
UPDATE CARD
WHERE CARDID = :NEW.CHARGEDCARDID
SET CREDIT = 
(CASE MEALTYPE

WHEN "STANDARD" THEN CREDIT - CHARGE
WHEN "GUEST" THEN CREDIT - 5
END);
END;

但是我得到了这个错误: PL/SQL: ORA-00971: missing SET keyword, PL/SQL: SQL Statement ignored.请问你能帮帮我吗？

Answer 1

请按照此语法进行更新,

update table_name set field1='value' where field2='value'

(即)

UPDATE CARD
SET CREDIT = 
(CASE MEALTYPE
WHEN "STANDARD" THEN CREDIT - CHARGE
WHEN "GUEST" THEN CREDIT - 5
END)
 WHERE CARDID = :NEW.CHARGEDCARDID;

有关更多信息,'update'声明如何工作参考此文档,

http://docs.oracle.com/cd/B19306_01/appdev.102/b14261/update_statement.htm

对于你的第二个错误试试这个,

UPDATE CARD
SET CREDIT = 
(CASE 
WHEN MEALTYPE="STANDARD" THEN CREDIT - CHARGE
WHEN MEALTYPE="GUEST" THEN CREDIT - 5
END MEALTYPE) 
 WHERE CARDID = :NEW.CHARGEDCARDID;

像这样的东西,

update card c
set c.credit=(select case when p.mealtype='STANDARD' then c.credit-c.charge
                 when p.mealtype='GUEST' then c.credit-5
            end credit from PAYMENTDEVICE p
           where c.cardid=p.chargedcardid)

fiddle_demo