从 SQL 中的 ISO 周获取日期

Question

我在 SQL 中创建了一个函数，以从“15W53”之类的格式返回 ISO 周的正确日期。其中“W”之前的第一部分是年份编号，后半部分是周编号。返回的日期应返回该日期的星期开始。

例如，15W53 应返回 12-28-2015，16W01 应返回 01-04-2016。但是，如果我为以下示例运行此程序，从我所读到的关于 ISO 周的内容中，我会得到不正确的结果。

我是否错误地创建了解析日期的函数？

SET DATEFIRST 1; 
SELECT dbo.[GetDateFromISOweek]('15W52') AS Correct, '15W52'  -- Returns: 2015-12-21
SELECT dbo.[GetDateFromISOweek]('15W53') AS Correct, '15W53'  -- Returns: 2015-12-28
SELECT dbo.[GetDateFromISOweek]('16W01') AS Incorrect, '16W01'-- Returns: 2015-12-28
SELECT dbo.[GetDateFromISOweek]('16W02') AS Incorrect, '16W02'-- Returns: 2016-01-04
SELECT dbo.[GetDateFromISOweek]('16W03') AS Incorrect, '16W03'-- Returns: 2016-01-11

功能：

CREATE FUNCTION [dbo].[GetDateFromISOweek] (@Input VARCHAR(10))  
RETURNS DATETIME  
WITH EXECUTE AS CALLER  
AS  
BEGIN  
    DECLARE @YearNum CHAR(4) 
    DECLARE @WeekNum VARCHAR(2)

    SET @YearNum = SUBSTRING(@Input,0,CHARINDEX('W',@Input,0))
    SET @WeekNum = SUBSTRING(@Input,CHARINDEX('W',@Input,0)+1,LEN(@Input))

    RETURN(DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + @YearNum) + (@WeekNum-1), 7));
END; 

Answer 1

更改您的回报中的 -1 以计算是否应考虑一年中的第一周（如果有超过 3 天）。 像这样的

case when DATEDIFF ( day ,  convert(datetime,'01/01/'+ @YearNum),@FirstDay )>=3 then 1 else 0 end

完整的代码，包括第一个星期日，可以改进，但工作......

CREATE FUNCTION [dbo].[GetDateFromISOweek] (@Input VARCHAR(10))  
RETURNS DATETIME  
WITH EXECUTE AS CALLER  
AS  
BEGIN  
    DECLARE @YearNum CHAR(4) 
    DECLARE @WeekNum VARCHAR(2)
    declare @FirstDay datetime

    SET @YearNum = cast(SUBSTRING(@Input,0,CHARINDEX('W',@Input,0)) as int)+2000
    SET @WeekNum = SUBSTRING(@Input,CHARINDEX('W',@Input,0)+1,LEN(@Input))
    set @FirstDay=DATEADD(DAY, (@@DATEFIRST - DATEPART(WEEKDAY, DATEADD(YEAR, @YearNum - 1900, 0)) +  (8 - @@DATEFIRST) * 2) % 7, DATEADD(YEAR, @YearNum - 1900, 0))-1

    RETURN(DATEADD(wk, DATEDIFF(wk, 6, '1/1/' + @YearNum) + (@WeekNum-case when DATEDIFF ( day ,  convert(datetime,'01/01/'+ @YearNum),@FirstDay )>=3 then 1 else 0 end), 7));
END; 
go
SET DATEFIRST 1; 
SELECT dbo.[GetDateFromISOweek]('15W52'),'15W52' union
SELECT dbo.[GetDateFromISOweek]('15W53'),'15W53' union
SELECT dbo.[GetDateFromISOweek]('16W01'), '16W01' union
SELECT dbo.[GetDateFromISOweek]('16W02'), '16W02' union
SELECT dbo.[GetDateFromISOweek]('16W03'), '16W03'

结果是

----------------------- -----
2015-12-21 00:00:00.000 15W52
2015-12-28 00:00:00.000 15W53
2016-01-04 00:00:00.000 16W01
2016-01-11 00:00:00.000 16W02
2016-01-18 00:00:00.000 16W03