Question

我创建了三个表。

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),
)

每当客户确实支付预订时，我想在Reservation表中将Reservation_paid列的值更改为YES，我想用触发器自动执行。

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。

Answer 1

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 2

在表Reservation的{{1}}列的insert或update之后，编写一个可在表ClientId上使用的触发器。然后将Payment与ClientID表的ClientID列进行匹配，并将相应的Reservation更新为YES。

编辑：触发器将是这样的

Reservation_paid

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.'

GO

如何触发表来更改另一个表列的值

4 个答案: