1、确定某年某月有多少天
实现原理:先利用DATEDIFF取得当前月的第一天,再将月份加一取得下月第一天,然后减去1分钟,再取日期的
天数部分,即为当月最大日期,也即当月天数
CREATE FUNCTION DaysInMonth ( @date datetime ) Returns int
AS
BEGIN
RETURN Day(dateadd(mi,-3,DATEADD(m, DATEDIFF(m,0,@date)+1,0)))
END
调用示例:
select dbo.DaysInMonth ('2006-02-03')
(2)计算哪一天是本周的星期一
SELECT DATEADD(week, DATEDIFF(week,'1900-01-01',getdate()), '1900-01-01') --返回2006-11-06 00:00:00.000
或
SELECT DATEADD(week, DATEDIFF(week,0,getdate()),0)
(3)当前季度的第一天
SELECT DATEADD(quarter, DATEDIFF(quarter,0,getdate()), 0)—返回2006-10-01 00:00:00.000
(4)一个季度多少天
declare @m tinyint,@time smalldatetime
select @m=month(getdate())
select @m=case when @m between 1 and 3 then 1
when @m between 4 and 6 then 4
when @m between 7 and 9 then 7
else 10 end
select @time=datename(year,getdate())+'-'+convert(varchar(10),@m)+'-01'
select datediff(day,@time,dateadd(mm,3,@time)) —返回92
实现原理:先利用DATEDIFF取得当前月的第一天,再将月份加一取得下月第一天,然后减去1分钟,再取日期的
天数部分,即为当月最大日期,也即当月天数
CREATE FUNCTION DaysInMonth ( @date datetime ) Returns int
AS
BEGIN
RETURN Day(dateadd(mi,-3,DATEADD(m, DATEDIFF(m,0,@date)+1,0)))
END
调用示例:
select dbo.DaysInMonth ('2006-02-03')
(2)计算哪一天是本周的星期一
SELECT DATEADD(week, DATEDIFF(week,'1900-01-01',getdate()), '1900-01-01') --返回2006-11-06 00:00:00.000
或
SELECT DATEADD(week, DATEDIFF(week,0,getdate()),0)
(3)当前季度的第一天
SELECT DATEADD(quarter, DATEDIFF(quarter,0,getdate()), 0)—返回2006-10-01 00:00:00.000
(4)一个季度多少天
declare @m tinyint,@time smalldatetime
select @m=month(getdate())
select @m=case when @m between 1 and 3 then 1
when @m between 4 and 6 then 4
when @m between 7 and 9 then 7
else 10 end
select @time=datename(year,getdate())+'-'+convert(varchar(10),@m)+'-01'
select datediff(day,@time,dateadd(mm,3,@time)) —返回92