转：SQL Server：获取当前日期是本月的第几周

来源：https://blog.csdn.net/wjiaoling136/article/details/84911215

 

 

第一种：

 

--获取当前日期是本月的第几周
--测试：select dbo.getMonthWeek(getdate()) 结果:10月的第2周
CREATE function [dbo].[getMonthWeek](@d datetime)
returns varchar(20)
as
begin
declare @returns varchar(20),
@monthfirstDay datetime,
@firstMondy datetime

select @monthfirstDay=left(convert(varchar,@d,23),7)+'-01'

if not exists(select 1 
from master.dbo.spt_values
where type=N'P' and number between 0 and datediff(d,@monthfirstDay,@d)
and datepart(dw,dateadd(d,number,@monthfirstDay))=2)
begin
select @monthfirstDay=dateadd(mm,-1,@monthfirstDay)
end

;with t as(select 'days'=dateadd(d,number,@monthfirstDay)
from master.dbo.spt_values
where type=N'P' and number<=7)
select @firstMondy=min([days]) 
from t
where datepart(dw,[days])=2

select @returns=rtrim(datepart(mm,@monthfirstDay))+'月的第'+rtrim(datediff(d,@firstMondy,@d)/7+1)+'周'

return @returns
end

 

第二种：

--获取当前日期是本月的第几周
--测试：select dbo.WeekOfMonth(getDate());结果:2
CREATE FUNCTION [dbo].[WeekOfMonth](@day datetime) 
RETURNS int 
AS 
begin 

----declare @day datetime 
declare @num int 
declare @Start datetime 
declare @dd int 
declare @dayofweek char(8) 
declare @dayofweek_num char(8) 
declare @startWeekDays int 
---set @day='2009-07-05' 
if datepart(dd,@day)=1 
return 1 
else 
set @Start= (SELECT DATEADD(mm, DATEDIFF(mm,0,@day), 0)) --一个月第一天的 
set @dayofweek= (datename(weekday,@Start)) ---得到本月第一天是周几 
set @dayofweek_num=(select (case @dayofweek when '星期一' then 2 
when '星期二' then 3 
when '星期三' then 4 
when '星期四' then 5 
when '星期五' then 6 
when '星期六' then 7 
when '星期日' then 1 
end)) 
set @dayofweek_num= 7-@dayofweek_num+1 ---得到本月的第一周一共有几天 
---print @dayofweek_num 
set @dd=datepart(dd,@day) ----得到今天是这个月的第几天 
--print @dd 
if @dd<=@dayofweek_num --小于前一周的天数 
return 1 
else 
set @dd=@dd-@dayofweek_num 
if @dd % 7=0 
begin 
set @num=@dd / 7 
return @num+1 

end 
else --if @dd % 7<>0 

set @num=@dd / 7 
set @num=@num+1+1 
return @num 
end 

 

 
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原文链接：https://blog.csdn.net/wjiaoling136/article/details/84911215