Question

我这里有一个用户定义函数，它返回一个日期时间。我根据原始日期设置了下一个日期的月，日和年。我想知道如何让函数返回日期时间格式，格式如下：mm / dd / yyyy。

ALTER FUNCTION [dbo].[udf_get_payout_date]
(
   -- Add the parameters for the function here
   @m_id as int
)
RETURNS Datetime
AS
BEGIN
-- Declare the return variable here
--the original date
DECLARE @m_createddate as Datetime 
--the date to be returned
DECLARE @p_date as Datetime
--the day part of the original date
DECLARE @dpart as int
--the month part of the original date
DECLARE @mdate as int
--this determines the day part of the given day ranges
DECLARE @pday as int
--the year part of the original date
DECLARE @ydate as int


--get the original date
SET @m_createddate = (SELECT [m_createddate] FROM rms_month_email where m_id = m_id)
SET @dpart = DATEPART(dd,@m_createddate)
SET @mdate = DATEPART(mm,@m_createddate)
SET @ydate = DATEPART(yyyy,@m_createddate)

--if the dpart ranges between 1 -> 15, the day of pay out will be on the 15th
if @dpart >= 1 and @dpart <= 15 
        set @pday = 15

 else 
-- if the dpart ranges between 15-> 16, the day of pay out will be on the 30th
        set @pday = 30
--add 1 to month part of the original date to signify that it will be on the next month
SET @mdate = @mdate + 1


if @mdate > 12 
    SET @mdate = 1 
    SET @ydate = @ydate + 1

    SET @p_date = @mdate+'/'+@pday + '/'+@ydate
-- Return the result of the function
RETURN @p_date

END

但我在值字段中得到null。你能告诉我的代码有什么问题吗？非常感谢。

Answer 1

通过单独打印变量来检查变量是否为空，因为当变量连接变量时，如果任何变量为null，则完整字符串将变为空。

Answer 2

select convert(varchar(10), cast('2015-10-12' as date), 101) select convert(varchar(10), cast(<column> as date), 101) from <table>

如何在用户定义函数中返回datetime

2 个答案: