实际上我已经在sql server 2008中创建了一个用户定义的表类型。结构如下所示。
我将它作为参数传递给函数,该函数也返回该类型的表类型。我正面临这个问题,而我在函数中声明了该类型的变量,在其中插入了一些数据并返回该参数。
表类型结构是:
Create Type OfferWithSubscription as Table
(
OfferID int,
OfferUserID int,
OfferImage varchar(200),
OfferExactPrice Decimal(18,2),
OfferContent varchar(max),
OfferTitle varchar(100),
StartDate datetime,
EndDate datetime,
StartTime datetime,
StopTime datetime,
ShowToUser bit,
SubID int,
SubLevel varchar(100)
)
功能,我想创建的是:
CREATE FUNCTION FN_ShowOffer
(
@Gold int,
@Silver int,
@Bronze int,
@table dbo.OfferWithSubscription Readonly )
RETURNS dbo.OfferWithSubscription
AS
BEGIN
DECLARE @ReturnTable AS dbo.OfferWithSubscription;
Declare @Case as varchar(20)
if(@Gold=0 and @Silver=1 and @Bronze=0 )
begin
set @Case='1S'
end
if(@Case='1S')
Begin
insert into @ReturnTable
select OfferID, OfferUserID, OfferImage,
OfferExactPrice, OfferContent,
OfferTitle, StartDate, EndDate,
StartTime, StopTime, ShowToUser,
SubID, SubLevel
from @table
where SubID=4
End
RETURN (
@ReturnTable
)
END
答案 0 :(得分:4)
您只需要扩展下面的类型。
仅供参考 - Can T-SQL function return user-defined table type?
CREATE FUNCTION FN_ShowOffer
(
@Gold int,
@Silver int,
@Bronze int,
@table dbo.OfferWithSubscription Readonly )
RETURNS @ReturnTable Table
(
OfferID int,
OfferUserID int,
OfferImage varchar(200),
OfferExactPrice Decimal(18,2),
OfferContent varchar(max),
OfferTitle varchar(100),
StartDate datetime,
EndDate datetime,
StartTime datetime,
StopTime datetime,
ShowToUser bit,
SubID int,
SubLevel varchar(100)
)
AS
BEGIN
Declare @Case as varchar(20)
if(@Gold=0 and @Silver=1 and @Bronze=0 )
begin
set @Case='1S'
end
if(@Case='1S')
Begin
insert into @ReturnTable
select OfferID,OfferUserID,OfferImage,OfferExactPrice,OfferContent,OfferTitle,
StartDate,EndDate,StartTime,StopTime,ShowToUser,SubID,SubLevel from @table where SubID=4
End
RETURN
END
进一步澄清,这完全兼容并可分配给该表类型的变量,例如SQL Fiddle
declare @t OfferWithSubscription
insert @t
select * from fn_showoffer(1,2,3,@t)