SQL 判断两个时间段是否有交叉

费话不说，直接上代码

SQL 代码：

View Code

IF  EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'[dbo].[fun_GetTimeSlotDays]'))

DROP FUNCTION [dbo].fun_GetTimeSlotDays

GO



-- =============================================

-- Author:        <Mike.Jiang>

-- Create date: <2012-07-18>

-- Description:    <判断两个时间断是否有交叉,如果有则返回1,否则返回0>

-- =============================================

CREATE FUNCTION dbo.fun_GetTimeSlotDays(

@fromDate DATETIME,

@toDate DATETIME,

@startDate DATETIME,

@endDate DATETIME

)

RETURNS INT 

AS 

BEGIN

   DECLARE @ret INT;

   IF(DATEDIFF(DAY,@fromDate,@endDate)>=0 AND DATEDIFF(DAY,@endDate,@toDate)>=0 )

      SET @ret=1;

   IF(DATEDIFF(DAY,@startDate,@toDate)>=0 AND DATEDIFF(DAY,@toDate,@endDate)>=0 )

      SET @ret=1;

   IF (@ret is null)

      SET @ret=0;

   RETURN @ret;

END

GO

 

测试代码：

SELECT dbo.fun_GetTimeSlotDays('2012-03-01','2012-03-10','2012-02-10','2012-02-20');

SELECT dbo.fun_GetTimeSlotDays('2012-03-01','2012-03-10','2012-02-01','2012-03-01');

SELECT dbo.fun_GetTimeSlotDays('2012-03-01','2012-03-10','2012-03-01','2012-03-02');

SELECT dbo.fun_GetTimeSlotDays('2012-03-01','2012-03-10','2012-03-10','2012-03-11');

SELECT dbo.fun_GetTimeSlotDays('2012-03-01','2012-03-10','2012-03-11','2012-03-11');

测试结果：