假设会计年度从 7 月/1 日到 6 月/30 日。
我必须计算周数。相应地在 SQL Server 2005 中。
请建议！
非常感谢！
问问题
7258 次
5 回答
1
试试这个并根据自己的需要使用变量@dt：
DECLARE @dt DATETIME = GETDATE()
SELECT WeekOfMonth = DATEPART(wk, @dt) - DATEPART(wk,DATEADD(m, DATEDIFF(M, 0, @dt), 0)) + 1
编辑：我的错，因为我错误地理解了这个问题，我的解决方案返回一个月的一周，而不是一年。
使用@Dems 答案的一部分并在此处更改我的答案是一个完整的工作测试，该测试基于临时表上可用的日期和财政年度的开始输出 3 列 WeekOfMonth、WeekOfYear 和 WeekOfFIscalYear。但我想对于一家特定的公司来说，一个财政年度的开始总是一样的。我只是添加了不同的日期和年份进行测试。
DECLARE @TT TABLE (auxVal INT,
auxdate DATETIME,
fiscal_year DATETIME
)
INSERT @TT
SELECT 100,'19120101 00:00:00','19120701' UNION ALL
SELECT 200,'18120615 00:00:00','18110701' UNION ALL
SELECT 100,'20121121 00:00:00','20120701' UNION ALL
SELECT 200,'20120101 00:00:00','20110701' UNION ALL
SELECT 100,'20150802 00:00:00','20140701' UNION ALL
SELECT 200,'20120330 00:00:00','20110701' UNION ALL
SELECT 322,'20110228 00:00:00','20100701'
SELECT DATEDIFF(week, DATEADD(MONTH, DATEDIFF(MONTH, 0, auxDate), 0), auxDate) + 1 WeekOfMonth,
DATEPART( wk, auxDate) WeekOfYear,
DATEDIFF(DAY, (DATEADD(YEAR, DATEDIFF(YEAR, fiscal_year, DATEADD(MONTH, -7, auxDate)), fiscal_year)), auxDate) / 7 + 1 WeekOfFiscalYear
FROM @TT
结果：
WeekOfMonth WeekOfYear WeekOfFiscalYear
-------------------------------------------
1 1 27
3 25 51
4 47 21
1 1 27
2 32 5
5 13 40
5 10 35
-------------------------------------------
于 2012-08-02T10:58:39.393 回答
1
如果您财政年度的第一周始终是 7 月 1 日至 7 月 7 日...
DECLARE
@inputDate DATETIME,
@fYearStart DATETIME,
@weekNumber INT
SET
@inputDate = getDate()
SET
@fYearStart = DATEADD(year, DATEDIFF(year, '20000101', DATEADD(month, -7, @inputDate)), '20000701')
SET
@weekNumber = DATEDIFF(day, @fYearStart, @inputDate) / 7 + 1
如果你的财务周总是周日到周六...
DECLARE
@inputDate DATETIME,
@fYearStart DATETIME,
@weekNumber INT
SET
@inputDate = getDate()
SET
@fYearStart = DATEADD(year, DATEDIFF(year, '20000101', DATEADD(month, -7, @inputDate)), '20000701')
SET
@weekNumber = DATEDIFF(WEEK, @fYearStart, @inputDate)
其中一项或多项应适用于您对week number.
于 2012-08-02T10:54:11.377 回答
0
无需创建表。只需将 @date 替换为您的列日期名称
declare @date datetime
set @date = '12/8/2016 00:00:00'
select case when datepart(month,@date) > 6 then
case when CEILING(datediff(day,'6/30/' + cast(datepart(year,@date) as nvarchar(4)),@date) / 7.0) > 52
then 52 else CEILING(datediff(day,'6/30/' + cast(datepart(year,@date) as nvarchar(4)),@date) / 7.0) end
else
case when CEILING(datediff(day,'6/30/' + cast(datepart(year,@date)-1 as nvarchar(4)),@date) / 7.0) > 52
then 52 else CEILING(datediff(day,'6/30/' + cast(datepart(year,@date)-1 as nvarchar(4)),@date) / 7.0) end
end WorkWeek
于 2016-08-02T04:16:02.537 回答
0
这个对我有用：
SELECT dt AS DayDate,
CONVERT(VARCHAR,DATENAME(weekday,dt)) AS [DayOfWk],
CONVERT(int,CONVERT(VARCHAR,DATEPART(yyyy,dt)) + RIGHT('00' + CONVERT(VARCHAR,DATEPART(wk,dt)),2)) AS [WeekID],
'Week ' + CONVERT(VARCHAR,DATEPART(wk,dt))AS [WeekDsc],
DATEPART(wk,dt) AS [WeekNbr],
DATEADD(dd, -(DATEPART(dw, dt)-1), dt) AS [WeekStartDay],
DATEADD(dd, 7-(DATEPART(dw, dt)), dt) AS [WeekEndDay],
CONVERT(VARCHAR,CASE WHEN DATEPART(mm,dt) > 6 THEN DATEPART(yyyy,dt) + 1 ELSE DATEPART(yyyy,dt) END) + RIGHT('00'+ CONVERT(VARCHAR,DATEDIFF(week,DATEADD(dd, 7-(DATEPART(dw, CASE WHEN DATEPART(mm,dt) > 6 THEN CONVERT(VARCHAR,DATEPART(yyyy,dt)) + '-07-01' ELSE CONVERT(VARCHAR,DATEPART(yyyy,dt) - 1) + '-07-01' END)), CASE WHEN DATEPART(mm,dt) > 6 THEN CONVERT(VARCHAR,DATEPART(yyyy,dt)) + '-07-01' ELSE CONVERT(VARCHAR,DATEPART(yyyy,dt) - 1) + '-07-01' END) + 1,DATEADD(dd, -(DATEPART(dw, dt)-1), dt)) + 2) ,2) AS [FinWkID]
FROM
(
SELECT DATEADD(day,increment,'2014-01-01') dt
FROM
(
SELECT top 5000 ROW_NUMBER() OVER (ORDER BY s1.[object_id]) - 1 AS increment
FROM sys.all_objects AS s1
CROSS JOIN sys.all_objects AS s2
ORDER BY s1.[object_id]
) SQ
) DATE_GENERATOR
于 2016-02-26T03:45:17.260 回答
-1
try this
select DATEPART(week,getdate())-DATEPART(week,'07-01-2012')
To get the week number starting from July 1st 2012
于 2012-08-02T10:53:47.213 回答