Question

如果日期差异已经大于或等于30，我编写了一个返回1的函数。

这是我的代码：

DECLARE @flag int
DECLARE @isactive bit
DECLARE @date  Datetime

SET @flag = 0

SELECT @isactive = [m_isactive] 
FROM dbo.rms_month_email 
WHERE m_id = 3

SELECT @date = [m_createddate] 
FROM dbo.rms_month_email 
WHERE m_id = 3

IF (@isactive = 1 AND DATEDIFF(dd,@date,GETDATE()) = 30)
   SET  @flag = 1

RETURN @flag

但它产生了这个错误：

Msg 178，Level 15，State 1，Line 16
带有返回值的RETURN语句不能在此上下文中使用。

这是什么意思？

CREATE FUNCTION [dbo].[udf_isBonus]
(@m_id AS INT)
RETURNS INT
AS
BEGIN
    -- Declare the return variable here
    DECLARE @flag AS int
    DECLARE @isactive AS bit
    DECLARE @date AS Datetime

    SET @flag = 0

    SELECT @isactive = [m_isactive] FROM dbo.rms_month_email WHERE m_id = @m_id
    SELECT @date = [m_createddate] FROM dbo.rms_month_email WHERE m_id = @m_id

    IF (@isactive = 1 AND DATEDIFF(dd,@date,GETDATE()) = 30)
         SET @flag = 1

    RETURN @flag
END
GO

这是此功能的完整代码。

Answer 1

您需要了解Return仅适用于函数或存储过程。如果您没有在函数或存储过程中使用它，那么它只是一批select语句。因此，您需要使用select代替return。

<强> SQL DEMO

修改

将其修改为函数后的编辑工作完全正常。

<强> SQL DEMO

Answer 2

尝试使用CASE代替IF并删除SET @flag = 0：

USE [petc_rms]
GO

 /****** Object:  UserDefinedFunction [dbo].[udf_isBonus]    Script Date:     9/29/2015 3:35:49 PM ******/
SET ANSI_NULLS ON
GO

SET QUOTED_IDENTIFIER ON
GO


CREATE FUNCTION [dbo].[udf_isBonus]
(
 -- Add the parameters for the function here

    @m_id AS INT

)
RETURNS INT
AS
BEGIN
-- Declare the return variable here
DECLARE @flag AS int
DECLARE @isactive AS bit
DECLARE @date AS Datetime


    SELECT @isactive = [m_isactive] FROM dbo.rms_month_email WHERE m_id = @m_id
    SELECT @date = [m_createddate] FROM dbo.rms_month_email WHERE m_id = @m_id

    SET @flag = CASE WHEN @isactive = 1 AND DATEDIFF(dd,@date,GETDATE()) = 30 THEN 1 ELSE 0 END    

Return @flag

END

GO

查看如何CREATE FUNCTION

在此上下文中不能使用返回值

2 个答案: