如何触发一个表改变另一个表列的值

Question

我已经创建了三个表。

CREATE TABLE Clients
(
    ClientID INT IDENTITY(1,1) PRIMARY KEY,
    First_Name VARCHAR(50) NOT NULL,
    Last_Name VARCHAR(50) NOT NULL,
)
CREATE TABLE Reservation
(
    ReservationID INT IDENTITY(1,1) PRIMARY KEY,
    ClientID INT FOREIGN KEY (ClientID) REFERENCES Clients(ClientID),
    Reservation_paid VARCHAR(3) DEFAULT 'NO',
)
CREATE TABLE Payment
(
    Payment_ID INT IDENTITY(1,1) PRIMARY KEY,
    ClientID INT FOREIGN KEY (ClientID) REFERENCES Clients(ClientID),
    ReservationID INT FOREIGN KEY (ReservationID) REFERENCES Reservation(ReservationID),
)

每当客户支付预订时，我想在预订表中将 Reservation_paid 列的值更改为 YES，并且我想使用触发器自动完成。

示例：如果 Reservation 表中的 ClientID 自动存在于 Payment 表中，则 Reservation_paid 的值将设置为 YES。 提前谢谢你。

Answer 1

CREATE TRIGGER trgAfterInsert ON [dbo].[Payment] 
FOR INSERT
AS
declare @ClientID int;
select @ClientID =i.ClientID from inserted i;   

if update(ClientID)
    UPDATE Reservation set Reservation_paid='Yes' WHERE  
    ClientID=@ClientID;

--PRINT 'AFTER INSERT trigger fired.'

Answer 2

插入后触发器应该做这样的事情

UPDATE R 
SET    Reservation_paid = 'Yes' 
FROM   reservation R 
WHERE  EXISTS (SELECT 1 
               FROM   INSERTED I 
               WHERE  I.clientid = R.clientid 
                      AND I.reservationid = R.reservationid) 

Answer 3

CREATE TRIGGER trgAfterInsert ON [dbo].[Payment] 
FOR INSERT
AS
declare @ClientID int;
select @ClientID =i.ClientID from inserted i;   
insert into Reservation(ClientID,Reservation_paid)    
values(@ClientID,'Yes');

--PRINT 'AFTER INSERT trigger fired.'

去

Answer 4

在表Payment 的ClientId 列上的任何insert 或update 之后，编写一个将作用于表Reservation 的触发器。然后将ClientID 与Reservation 表的ClientID 列匹配，并将对应的Reservation_paid 更新为YES。

编辑： 触发器会是这样的

CREATE TRIGGER `UpdateReservation_paid` AFTER INSERT OR UPDATE ON `Payment`
 FOR EACH ROW BEGIN
 AS 
begin
 update Reservation 
 SET Reservation_paid='YES'
 Where NEW.ClientID = Reservation.ClientID 
       and NEW.ReservationID = Reservation.ReservationID   
end