如何从星期一分开一周

Question

我对如何按月划分周有疑问。实际上我的一周从星期一开始，而在 sql server 中的一周从星期日开始，这就产生了问题。我必须编写代码，因此如果用户给出的开始日期和结束日期的日期差大于 30 且小于 90，则日期应该以天和周为单位进行拆分。这样做是这样的- 首先，我必须找到开始日期是什么，如果开始日期是星期一，那么除了首先我必须找到我发现的上一个星期一是什么时候-

declare @enddate date = '2015-08-01';               
select convert (date, DATEADD(wk, DATEDIFF(wk, 0,@enddate),0))

但是如果开始日期是星期天，这将不起作用为此我添加了一个变量来检查开始日是星期天，如果是星期天，然后我将其设置为 1，然后我将代码更改为 - （仅适用于星期天)

declare @enddate date = '2015-08-02';               
select convert (date, DATEADD(wk, DATEDIFF(wk, 6,@enddate),0))

对我来说主要问题是如何在几周内分手。我按给定的代码划分了几周。 这是示例日期

declare @enddate date = '2015-11-03';      
SELECT 'Week ' + CAST(DATEPART(WEEK,@ENDDATE) 
- DATEPART(WEEK,DATENAME(YEAR,@ENDDATE) + '/' 
+ DATENAME(MONTH,@ENDDATE) + '/' + '1') AS NVARCHAR)+ ' ' + DATENAME(MONTH,@ENDDATE)

输出是-

Week 0 November

11 月 1 日至 7 日之间的所有日期都显示第 0 周

谁能帮助我为什么显示第 0 周而不是第 2 周，因为我的一周从星期一开始，2015 年 11 月从星期日开始，所以只有一天应该是第 1 周，而从 11 月 2 日到 11 月 8 日应该是第 2 周 欢迎任何帮助，如果您不理解某些内容，请发表评论

已编辑 - 需要一个逻辑，以便我能够将一周从周一拆分到周日，但我无法做到 -

如果月份在几周之间发生变化，例如一个月的 30 号是星期二，那么本周应该只在此处结束，下个月 1 将在星期三现在该月的新一周从星期三开始到星期日，这将是第一周月。

在我的案例逻辑中，我得到了它，但特别是 2015 年 11 月，我无法得到它。

需要输出-

Start date -2015-09-07        End date-2015-11-01

Output 
Start date                                      End date
Week 2 September                            Week 1 November



Start date -2015-09-07              End date-2015-11-05

Output -

Start date                                      End date
Week 2 September                             Week 1 November
2015-11-02                                   2015-11-05

Start date -2015-09-07              End date-2015-11-08

Output -

Start date                                      End date
Week 2 September                             Week 2 November

Start date -2015-09-08              End date-2015-11-05

Start date                                      End date
2015-09-08                                   2015-09-13
Week 3 September                             Week 1 November
2015-11-02                                   2015-11-05

Answer 1

问题在于您计算月份星期的方式

这应该会给你想要的结果：

    SET DATEFIRST 1;
go
    alter function WeekOfMonth (@Day Date) returns int
    as 
    begin  
      return  DATEPART(wk, @Day) - DATEPART(wk, CAST(DATEADD(month, DATEDIFF(month, 0, @day), 0) AS DATE)) + 1
    end
    go
    declare @startdate date = '2015-09-07'; 
    declare @enddate date = '2015-11-03'; 

-使用 UDF

SELECT  @startdate as "Start date",  'Week ' + cast(dbo.WeekOfMonth(@startdate) as NVARCHAR) + ' ' + DATENAME(MONTH,@ENDDATE)

-不使用 UDF

SELECT  @enddate as "End date",  'Week ' + cast(DATEPART(wk, @enddate) - DATEPART(wk, CAST(DATEADD(month, DATEDIFF(month, 0, @enddate), 0) AS DATE)) + 1 as NVARCHAR) + ' ' + DATENAME(MONTH,@ENDDATE)

Answer 2

这给出了正确的输出：

declare @data table(id int identity(0, 1), startdate date, enddate date)
insert into @data(startdate, enddate) values
    ('20150906', '20151101')
    , ('20150907', '20151102')
    , ('20150907', '20151103')
    , ('20150907', '20151104')
    , ('20150907', '20151105')
    , ('20150907', '20151108')
    , ('20150908', '20151108')
    , ('20150908', '20151109')


SET DATEFIRST 1

-- days before weeks
Select id, type = 0, startdate, enddate
    , [Start Date] = Cast(startdate as varchar(10)), [End Date] = Cast(DATEADD(DAY, (7-DATEPART(WEEKDAY, startdate))%7, startdate) as varchar(10))
From @data
Where DATEPART(WEEKDAY, startdate) <> 1
Union All
-- Weeks
Select id, type = 1, startdate, enddate
    , 'Week ' + Cast((
        DATEPART(WEEK, startdate) - DATEPART(WEEK, DATEADD(MONTH, DATEDIFF(MONTH, 0, startdate), 0)) + 1 + Case When DATEPART(WEEKDAY, startdate) <> 1 then 1 else 0 end
    ) as varchar(10)) + ' ' + DATENAME(MONTH,startdate)
    , 'Week ' + Cast(DATEPART(WEEK, enddate) - DATEPART(WEEK, DATEADD(MONTH, DATEDIFF(MONTH, 0, enddate), 0)) + 1 as varchar(10)) + ' ' + DATENAME(MONTH,enddate)
From @data
Union All
-- Days after weeks
Select id, type = 2, startdate, enddate
    , cast(DATEADD(DAY, -(DATEPART(WEEKDAY, enddate)+6)%7, enddate) as varchar(10))
    , cast(enddate as varchar(10))
From @data
Where DATEPART(WEEKDAY, enddate) <> 7
Order By id, type

输出：

id  type    startdate   enddate     Start Date          End Date
0   0       2015-09-06  2015-11-01  2015-09-06          2015-09-06
0   1       2015-09-06  2015-11-01  Week 2 September    Week 1 November
1   1       2015-09-07  2015-11-02  Week 2 September    Week 2 November
1   2       2015-09-07  2015-11-02  2015-11-02          2015-11-02
2   1       2015-09-07  2015-11-03  Week 2 September    Week 2 November
2   2       2015-09-07  2015-11-03  2015-11-02          2015-11-03
3   1       2015-09-07  2015-11-04  Week 2 September    Week 2 November
3   2       2015-09-07  2015-11-04  2015-11-02          2015-11-04
4   1       2015-09-07  2015-11-05  Week 2 September    Week 2 November
4   2       2015-09-07  2015-11-05  2015-11-02          2015-11-05
5   1       2015-09-07  2015-11-08  Week 2 September    Week 2 November
6   0       2015-09-08  2015-11-08  2015-09-08          2015-09-13
6   1       2015-09-08  2015-11-08  Week 3 September    Week 2 November
7   0       2015-09-08  2015-11-09  2015-09-08          2015-09-13
7   1       2015-09-08  2015-11-09  Week 3 September    Week 3 November
7   2       2015-09-08  2015-11-09  2015-11-09          2015-11-09