Question

我在SQL中创建了一个函数，从ISO周中返回适当的日期，格式如＆＃34; 15W53＆＃34;。第一部分在＆＃34; W＆＃34;之前是年份数，下半年是周数。返回的日期应该返回该日期的一周开始。

因此，例如，15W53应返回12-28-2015，16W01应返回01-04-2016。但是，如果我按照以下示例运行此操作，我会从我读过的关于ISO周的内容中得到错误的结果。

我是否错误地创建了函数来解析日期？

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

更改回报中的-1以计算是否应将第一周视为第一周（如果超过3天）。像这样的东西

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

完整的代码，包括第一个星期天，可以改进，但有效...

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'

结果将是

----------------------- -----
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

从SQL Week中获取日期

1 个答案: