Question

我尝试做的就是声明一个名为@ReservationID的变量，将其初始化为Reservation表中存在某些条件的值，并使用该值作为参数插入新的Folio表条目。当我去打印我的Folio表时，新的表条目不存在。我已经检查并运行了SELECT语句，并根据条件提供了正确的值。我没有正确初始化或传递变量吗？我在这里错过了什么？任何帮助将不胜感激。

DECLARE @ReservationID smallint;
SET @ReservationID = (SELECT ReservationID FROM Reservation WHERE ReservationDate = CONVERT(varchar, GETDATE()) AND CreditCardID = 1)

SET IDENTITY_INSERT Folio OFF
INSERT INTO Folio (ReservationID, GuestID, RoomID, QuotedRate, CheckinDate, Nights, Status, Comments, DiscountID)
VALUES(@ReservationID, 500, 20, 35.50, '07/24/2016', 3, 'R', NULL, 1);

Answer 1

您可能无法正确初始化或传递变量。

请参阅MSDN Here以将变量分配给表达式的值。

您需要更正SET＆amp; IDENTITY_INSERT如下，

DECLARE @ReservationID smallint;
select @ReservationID = ReservationID FROM Reservation 
WHERE ReservationDate = CONVERT(varchar, GETDATE()) AND CreditCardID = 1

SET IDENTITY_INSERT Folio ON
INSERT INTO Folio (ReservationID, GuestID, RoomID, QuotedRate,  CheckinDate,Nights, Status, Comments, DiscountID)
VALUES(@ReservationID, 500, 20, 35.50, '07/24/2016', 3, 'R', NULL, 1);

以下是我的快速测试代码

create table Reservation(ReservationID smallint identity, CreditCardID int) 

insert into Reservation values(1),(2),(3),(4),(5)

SET IDENTITY_INSERT Reservation ON

DECLARE @ReservationID smallint;

select @ReservationID =  ReservationID FROM Reservation

insert into Reservation (ReservationID, CreditCardID)
values(@ReservationID,100)

SQL变量作为INSERT INTO值

1 个答案: