请问一段简单的脚本错误……

Question

CREATE PROCEDURE [PostVote]
@PID nvarchar(100),
@UserID nvarchar(100),
@IP nvarchar(15),
@return int =0 output
AS
BEGIN
declare @time datetime
select @time=[Time] from [Vote] Where [UserID]=@UserID
select @time=isnull(@time,dateadd(hh,-2,getdate())
if datediff(hh,getdate(),@time)>1 --这个地方提示语法错误，请问该怎么写？
begin
Update [Photos] Set [Vote]=[Vote]+1 Where ID=@PID
Insert into [Vote](PhotoID,UserID,IP) Values(@PID,@UserID,@IP)
select @return=1
end
END
GO

Answer 1

--确保@time是Datetime型,试试这个if datediff(hh,convert(datetime,getdate()),convert(datetime,@time))>1

Answer 2

CREATE PROCEDURE [PostVote]
    @PID nvarchar(100),
    @UserID nvarchar(100),
    @IP nvarchar(15),
    @return int =0 output
AS
BEGIN
    declare @time datetime
    select @time=[Time] from [Vote] Where [UserID]=@UserID
    select @time=isnull(@time,dateadd(hh,-2,getdate()))
    if datediff(hh,getdate(),@time)>1 --这个地方提示语法错误，请问该怎么写？
    begin
        Update [Photos] Set [Vote]=[Vote]+1 Where ID=@PID
        Insert into [Vote](PhotoID,UserID,IP) Values(@PID,@UserID,@IP)
        select @return=1
    end
END
GO

Answer 3

select @time=isnull(@time,dateadd(hh,-2,getdate()))
少了一边括号

Answer 4

CREATE PROCEDURE [PostVote]
    @PID nvarchar(100),
    @UserID nvarchar(100),
    @IP nvarchar(15),
    @return int =0 output
AS
BEGIN
    declare @time datetime
    select @time=[Time] from [Vote] Where [UserID]=@UserID
    set @time=isnull(@time,dateadd(hh,-2,getdate()))--
if (datediff(hh,getdate(),@time)>1) --
    begin
        Update [Photos] Set [Vote]=[Vote]+1 Where ID=@PID
        Insert into [Vote](PhotoID,UserID,IP) Values(@PID,@UserID,@IP)
        select @return=1
    end
END
GO

Answer 5

select @time=isnull(@time,dateadd(hh,-2,getdate())是这一句的问题要这样才行
select @time=isnull(@time,cast(dateadd(hh,-2,getdate()) as datetime))