我需要创建在特定日期开始和结束的年历。它不是从1月1日开始，也不是在12月31日结束，而是在每年不同的日期（我以编程方式声明/设置它们）。这是学校（学年）的日历。目前明年将于10月1日开始，9月30日结束。我需要的是连续的周数，周开始和结束日期以及所有日期。
因此，对于2011/2012学年的第一周（10月1日至9月30日），我会得到week number = 1，week start date = 2011-10-01和week end date = 2011-10-02（基本上每周有两行） ）等等。
我遇到的问题是，在使用native Microsoft Date and Time functions时，我只能获得“默认年份范围”中的周数：
DATEPART(ww, [Date])
如果我在10月1日通过，鞋面将返回40.但我的结果应为1。
有任何建议如何做到这一点？我目前拥有的表格包括多年的所有日期（默认年份从1990年到2100年），默认周数和月份。我想从中选择日期范围（这将是我的学年开始和结束日期），并以某种方式分配正确的工作日期和开始和结束日期。
这不是一些学校项目或家庭作业对我的工作而言：）
答案 0 :(得分：1)
在再次阅读你的问题后，我提出了这个问题
declare @start datetime
declare @end datetime
set @start = '2011-10-01'
set @end = '2012-9-30'
;with cte as
(
select @start firstday, @start + 6 - DATEDIFF(day, 0, @start) %7 lastday, 1 week
union all
select lastday + 1, case when @end < lastday + 7 then @end else lastday + 7 end, week + 1
from cte
where lastday < @end
)
select cast(firstday as date) firstday, cast(lastday as date) lastday, week from cte
option(MAXRECURSION 0)
结果：
firstday lastday week
---------- ---------- ----
2011-10-01 2011-10-02 1
2011-10-03 2011-10-09 2
2011-10-10 2011-10-16 3
2011-10-17 2011-10-23 4
....
2012-09-17 2012-09-23 52
2012-09-24 2012-09-30 53
旧的消化
declare @start datetime
declare @end datetime
set @start = '2011-10-01'
set @end = '2012-9-30'
;with cte as
(
select @start calendardate, 1 week
union all
select calendardate + 1, week + case when DATEDIFF(day, 0, calendardate) %7 = 6 then 1 else 0 end
from cte
where calendardate < @end
)
select cast(calendardate as date) calendardate, week from cte
option(MAXRECURSION 0)
结果：
calendardate week
------------ -----------
2011-10-01 1
2011-10-02 1
2011-10-03 2
2011-10-04 2
2011-10-05 2
2011-10-06 2
2011-10-07 2
2011-10-08 2
2011-10-09 2
2011-10-10 3
2011-10-11 3
2011-10-12 3
.....
2012-09-29 53
2012-09-30 53
答案 1 :(得分：1)
[t-clausen.dk]的答案已经足够了，但你需要EndOfWeek和BeginOf Week：
DECLARE @start DATETIME
DECLARE @end DATETIME
SET @start = '2011-10-01'
SET @end = '2012-9-30';
WITH cte(calendardate, week, beginofweek, endofweek)
AS (SELECT @start calendardate,
CAST(1 AS INT) week,
Dateadd(DAY, 0, Datediff(DAY, 0, @start) -
Datediff(DAY, 0, @start) %
7)
beginofweek,
Dateadd(DAY, 6, Datediff(DAY, 0, @start) -
Datediff(DAY, 0, @start) %
7)
UNION ALL
SELECT calendardate + 1,
week + CASE
WHEN Datediff(DAY, 0, calendardate) %7 = 6 THEN 1
ELSE 0
END,
Dateadd(DAY, 0, Datediff(DAY, 0, calendardate + 1) -
Datediff(DAY, 0, calendardate + 1) % 7)
beginofweek,
Dateadd(DAY, 6, Datediff(DAY, 0, calendardate) -
Datediff(DAY, 0, calendardate) % 7)
FROM cte
WHERE calendardate < @end)
SELECT CAST(calendardate AS DATE) calendardate,
week,
CAST(beginofweek AS DATE) beginofweek,
CAST(endofweek AS DATE) endofweek
FROM cte
OPTION( MAXRECURSION 0)
结果：
calendardate week beginofweek endofweek
------------ ----------- ----------- ----------
2011-10-01 1 2011-09-26 2011-10-02
2011-10-02 1 2011-09-26 2011-10-02
2011-10-03 2 2011-10-03 2011-10-02
2011-10-04 2 2011-10-03 2011-10-09
2011-10-05 2 2011-10-03 2011-10-09
2011-10-06 2 2011-10-03 2011-10-09
2011-10-07 2 2011-10-03 2011-10-09
2011-10-08 2 2011-10-03 2011-10-09
2011-10-09 2 2011-10-03 2011-10-09
2011-10-10 3 2011-10-10 2011-10-09
2011-10-11 3 2011-10-10 2011-10-16
2011-10-12 3 2011-10-10 2011-10-16
2011-10-13 3 2011-10-10 2011-10-16
2011-10-14 3 2011-10-10 2011-10-16
2011-10-15 3 2011-10-10 2011-10-16
2011-10-16 3 2011-10-10 2011-10-16
2011-10-17 4 2011-10-17 2011-10-16
2011-10-18 4 2011-10-17 2011-10-23
...
2012-09-21 52 2012-09-17 2012-09-23
2012-09-22 52 2012-09-17 2012-09-23
2012-09-23 52 2012-09-17 2012-09-23
2012-09-24 53 2012-09-24 2012-09-23
2012-09-25 53 2012-09-24 2012-09-30
2012-09-26 53 2012-09-24 2012-09-30
2012-09-27 53 2012-09-24 2012-09-30
2012-09-28 53 2012-09-24 2012-09-30
2012-09-29 53 2012-09-24 2012-09-30
2012-09-30 53 2012-09-24 2012-09-30