Question

我有一个查询显示付款次数，按SQL Server 2012中的周数分组。

SELECT 
    DATEPART(wk, FPP.UPLOAD_DATE) AS "Date by week", 
    COUNT(DISTINCT FPP.ID) AS "Number of Bills")
FROM 
    FACT_PAY_PAYMENT FPP 
GROUP BY 
    DATEPART(wk, FPP.UPLOAD_DATE);

小问题，当我执行此代码时，我会收到周数，但我想要一周的开始日期。

例如，现在我收到以下输出：

 Date by week     Number of Bills 
 ---------------------------------
      40                 7

我想收到的是：

 Date by week     Number of Bills 
 ---------------------------------
 2015-09-28            7

Answer 1

首先，你必须确定什么是周SELECT @@DATEFIRST;的Datefirst，在我的计算机中Sunday是First day of week

DECLARE @dt DATE = '2015-09-08';
SELECT DATEADD(wk,DATEDIFF(wk,7,@dt),6) AS Datebyweek

结果

Datebyweek
2015-09-06 00:00:00.000

Answer 2

使用DATEDIFF()从开头获取周数，然后DATEADD()获取第一个日期。例如;

DECLARE @date DATETIME = GETDATE()

SELECT DATEADD(WEEK, DATEDIFF(WEEK, 0, @date) , 0) FirstDayOfWeek

您可以将其应用于您的查询，将@date替换为您的列名。就像;

SELECT 
    DATEADD(WEEK, DATEDIFF(WEEK, 0, FPP.UPLOAD_DATE) , 0) AS "Date by week", 
    COUNT(DISTINCT FPP.ID) AS "Number of Bills"
FROM 
    FACT_PAY_PAYMENT FPP 
GROUP BY 
    DATEADD(WEEK, DATEDIFF(WEEK, 0, FPP.UPLOAD_DATE) , 0);

Answer 3

请尝试

DECLARE @d AS DATETIME =GETDATE();

SELECT DATEPART(ISO_WEEK,@d) AS WeekIndex
      ,DATEADD(DAY,DATEPART(WEEKDAY,@d) * (-1) /* The following "+1" is depending on system's culture, look at DATEFIRST !!! */ + 1 ,@d) AS MondayForThisWeek

这将返回给定日期的本周指数和本周的周一。

了解系统文化的一些问题（特别是DATEFIRST !!!），年度结束时的周指数和一般日期（包括时间成分）。

Answer 4

以下功能对我有帮助：

CREATE FUNCTION [dbo].[fn_GetFirstDateOfWeek] (@YearNum int, @WeekNum int)
RETURNS datetime
AS
BEGIN 

DECLARE @FirstWeekDay As Date;

SET @FirstWeekDay = dateadd (week, @WeekNum, dateadd(year, @YearNum-1900, 0)) - 4 - datepart(dw, dateadd(week, @WeekNum, dateadd (year, @YearNum-1900, 0)) - 4) + 1

RETURN @FirstWeekDay
END

SQL Server，获取一周的日期编号

4 个答案: