我正在使用一个 SQL Server 函数，它返回 abigInt并在触发器中使用它来为类型的列分配一个值bigint。但是，当我运行触发器时，会发生溢出异常 ( Arithmetic overflow error converting expression to data type int.)，即它会将其视为int，而不是bigint
功能是：
ALTER FUNCTION [dbo].[longIntDateTime] ()
RETURNS bigint
AS
BEGIN
-- Declare the return variable here
DECLARE @ResultVar bigint;
DECLARE @now Datetime;
set @now = getdate();
SET @ResultVar=DATEPART(YYYY,@now)*100000000 + DATEPART(MM,@now)*1000000 + DATEPART(DD,@now)*10000 + DATEPART(HH,@now)*100;
-- DATEPART(HH,@now)*100 + DATEPART(MI,@now);
-- Return the result of the function
RETURN (@ResultVar);
END
触发器是：
ALTER TRIGGER [dbo].[employeesInsert]
ON [dbo].[employees]
AFTER INSERT
AS
BEGIN
-- SET NOCOUNT ON added to prevent extra result sets from
-- interfering with SELECT statements.
SET NOCOUNT ON;
-- Insert statements for trigger here
UPDATE employees
SET changeTimeStamp = dbo.longIntDateTime()
FROM inserted INNER JOIN employees On inserted._id = employees._id
END
表定义是：
CREATE TABLE [dbo].[employees](
[_id] [int] IDENTITY(1,1) NOT NULL,
[employee_name] [varchar](50) NOT NULL,
[password] [varchar](50) NOT NULL,
[isActive] [int] NOT NULL,
[isDeleted] [int] NOT NULL,
[changeTimeStamp] [bigint] NOT NULL,
CONSTRAINT [PK_employees] PRIMARY KEY CLUSTERED ([_id] ASC)
)
ALTER TABLE [dbo].[employees]
ADD CONSTRAINT [DF_employees_isActive] DEFAULT ((0)) FOR [isActive]
ALTER TABLE [dbo].[employees]
ADD CONSTRAINT [DF_employees_isDeleted] DEFAULT ((0)) FOR [isDeleted]
GO
如果我从函数的第一个 yyyy 部分中取出两个“0”，则触发器会成功，但按原样，它会失败。
显然，产生的值小于一个大整数。
有任何想法吗？
安东