我正在尝试确定ISO 8601格式中给定的一周日期(即YYYY-W ##)的日期。我的最终目标是将ISO 8601周日期转换为ISO 8601日历日期。我需要在TSQL中执行此操作(我正在使用SQL Server 2005),我不确定SQL Server 05中是否有任何内置允许这样做,但它有助于查看另一种语言的示例或在通用伪代码中
更新:
很抱歉,如果我的问题结构令人困惑。基本上,我有一个ISO8601周日期,我正在尝试转换为ISO8601日历日期。
示例(来自Wikipedia)
ISO 8601周日期:2012-W02(YYYY-W ##)
转换为...
ISO 8601日历日期:2012-01-09(YYYY-MM-DD)
由于我的周日期示例中未给出日期组件,因此可以假设一周的第一天。
答案 0 :(得分:2)
尝试以下解决方案:
declare
@date varchar(10)
,@convertedDate datetime
,@wk int
,@yr int
set @date = '2012-W02';
set @yr = parsename(replace(@date, '-W', '.'), 2)
set @wk = parsename(replace(@date, '-W', '.'), 1)
set @convertedDate = convert(varchar(10), dateadd(week, @wk, dateadd (year, @yr-1900, 0)) - 5 - datepart(dw, dateadd (week, @wk, dateadd (year, @yr-1900, 0))), 121)
select
'Year' = @yr
,'Week' = @wk
,'Date' = @convertedDate
<强>输出:强>
-----------------------------------------
| Year | Week | Date |
-----------------------------------------
| 2012 | 2 | 2012-01-09 00:00:00.000 |
-----------------------------------------
答案 1 :(得分:0)
从CREATE FUNCTION
documentation给出适当的ISO周功能(对所有不良习惯进行一些调整,恕我直言):
CREATE FUNCTION dbo.ISOWeek
(
@dt SMALLDATETIME
)
RETURNS TINYINT
AS
BEGIN
DECLARE @ISOweek TINYINT;
SET @ISOweek = DATEPART(WEEK, @dt) + 1
-DATEPART(WEEK, RTRIM(YEAR(@dt)) + '0104');
IF @ISOweek = 0
BEGIN
SET @ISOweek = dbo.ISOweek
(
RTRIM(YEAR(@dt)-1)+'12'+RTRIM(24 + DAY(@dt))
) + 1;
END
IF MONTH(@dt) = 12 AND DAY(@dt) - DATEPART(DAYOFWEEK, @dt) >= 28
BEGIN
SET @ISOweek = 1;
END
RETURN(@ISOweek);
END
GO
我们可以创建一个这样的表:
CREATE TABLE dbo.ISOWeekCalendar
(
[Date] SMALLDATETIME PRIMARY KEY,
ISOWeekNumber TINYINT,
[Year] INT,
ISOWeek CHAR(8)
);
CREATE UNIQUE INDEX iw ON dbo.ISOWeekCalendar(ISOWeek);
我们可以使用任何年份的数据来填充它,这使用ISO周1-52代表2000 - 2029:
DECLARE @StartDate SMALLDATETIME,
@EndDate SMALLDATETIME;
SELECT @StartDate = '20000102',
@EndDate = '20291229';
INSERT dbo.ISOWeekCalendar([Date])
SELECT TOP (DATEDIFF(DAY, @StartDate, @EndDate)+1) n
= DATEADD(DAY, ROW_NUMBER() OVER (ORDER BY s1.[object_id])-1, @StartDate)
FROM sys.all_objects AS s1
CROSS JOIN sys.all_objects AS s2
ORDER BY s1.[object_id];
现在我们可以更新数据。
-- delete all non-Mondays:
SET DATEFIRST 1;
DELETE dbo.ISOWeekCalendar WHERE DATEPART(WEEKDAY, [Date]) <> 1;
-- put the proper ISO week number:
UPDATE dbo.ISOWeekCalendar SET ISOWeekNumber = dbo.ISOWeek([Date]);
-- put the year:
UPDATE dbo.ISOWeekCalendar SET [Year] = DATEPART(YEAR, [Date]);
-- update to the correct year for fringe days:
UPDATE dbo.ISOWeekCalendar SET [Year] = [Year] + 1
WHERE ISOWeekNumber = 1 AND MONTH([Date]) = 12;
-- finally, build the calculated value for YYYY-W##:
UPDATE dbo.ISOWeekCalendar
SET ISOWeek = RTRIM([Year]) + '-W' + RIGHT('0' + RTRIM(ISOWeekNumber), 2);
请注意,上述内容只需进行一次。现在我们可以根据输入运行一个非常简单的查询:
SELECT [Date] FROM dbo.ISOWeekCalendar WHERE ISOWeek = '2012-W02';
结果:
Date
-------------------
2012-01-09 00:00:00
我们甚至可以创建一个执行此操作的函数:
CREATE FUNCTION dbo.ISOWeekDate(@ISOWeek CHAR(8))
RETURNS SMALLDATETIME
WITH SCHEMABINDING
AS
BEGIN
RETURN (SELECT [Date] FROM dbo.ISOWeekCalendar
WHERE ISOWeek = @ISOWeek);
END
GO
另一种功能:
CREATE FUNCTION dbo.ISOWeekFromDate(@Date SMALLDATETIME)
RETURNS CHAR(8)
WITH SCHEMABINDING
AS
BEGIN
RETURN (SELECT TOP (1) ISOWeek FROM dbo.ISOWeekCalendar
WHERE [Date] <= @Date
ORDER BY [Date] DESC);
END
GO
查询:
SELECT dbo.ISOWeekDate('2012-W02'), dbo.ISOWeekFromDate('20120110');
结果:
------------------- --------
2012-01-09 00:00:00 2012-W02
是的,它比复杂的查询更加前期工作,但我更喜欢易用性和更清晰的查询语义。