从周号获取周开始日期和周结束日期

98

我有一个查询，该查询计算数据库中成员的结婚日期。

SELECT 
  SUM(NumberOfBrides) AS [Wedding Count]
  , DATEPART( wk, WeddingDate) AS [Week Number]
  , DATEPART( year, WeddingDate) AS [Year]
FROM  MemberWeddingDates
GROUP BY DATEPART(year, WeddingDate), DATEPART(wk, WeddingDate)
ORDER BY SUM(NumberOfBrides) DESC

在结果集中表示每周的开始和结束时如何计算？

SELECT
  SUM(NumberOfBrides) AS [Wedding Count]
  , DATEPART(wk, WeddingDate) AS [Week Number]
  , DATEPART(year, WeddingDate) AS [Year]
  , ??? AS WeekStart
  , ??? AS WeekEnd
FROM  MemberWeddingDates
GROUP BY DATEPART(year, WeddingDate), DATEPART(wk, WeddingDate)
ORDER BY SUM(NumberOfBrides) DESC
sql  sql-server  date  sql-server-2005  week-number 
— 迪古鲁
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Answers:

161

您可以查找星期几，并在日期上加上日期以获取开始日期和结束日期。

DATEADD(dd, -(DATEPART(dw, WeddingDate)-1), WeddingDate) [WeekStart]

DATEADD(dd, 7-(DATEPART(dw, WeddingDate)), WeddingDate) [WeekEnd]

您可能还希望同时考虑从日期中剥离时间。

— 罗宾·戴
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3
请记住，设置DATEFIRST为7以外的任何值都会破坏此设置。
— Tomalak
5
它不会“破坏”它，它将使用datefirst将WeekStart =设置为DateFirst表示一周的第一天。您的版本始终将星期一和星期日作为一周的开始和结束，而不是将服务器设置为一周的开始和结束
— Robin Day
1
嗯...这是正确的点，+ 1。:)我要删除我的，然后（尽管是在脚射门，这是非常好的目标。g ^）。
— Tomalak
1
好吧，然后我再次删除它。我不确定在他们的正确思维中谁会承担除周一以外其他任何事情，无论如何还是本周的开始。要启动一周周日是没有意义的任何。:-)
— Tomalak
3
星期一（msdn.microsoft.com/zh-cn/library/ms181598.aspx）的“设置datefirst 1”
— Selrac 2014年
41

这是一个DATEFIRST不可知的解决方案：

SET DATEFIRST 4     /* or use any other weird value to test it */
DECLARE @d DATETIME

SET @d = GETDATE()

SELECT
  @d ThatDate,
  DATEADD(dd, 0 - (@@DATEFIRST + 5 + DATEPART(dw, @d)) % 7, @d) Monday,
  DATEADD(dd, 6 - (@@DATEFIRST + 5 + DATEPART(dw, @d)) % 7, @d) Sunday
— 托玛拉克
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11
很好，但是星期一对我不工作。我必须添加“ 0-”才能获得星期一。我的星期一代码现在是：DATEADD（dd，0-（@@ DATEFIRST + 5 + DATEPART（dw，@d））％7，@d）
— Warren
支持答案和沃伦斯评论。在不更改Warrens的情况下，SQL Server版本11.0.5058.0至少给出了错误的星期一日期。而是给我星期五。
— Morvael '16
18

您还可以使用以下命令：

  SELECT DATEADD(day, DATEDIFF(day, 0, WeddingDate) /7*7, 0) AS weekstart,
         DATEADD(day, DATEDIFF(day, 6, WeddingDate-1) /7*7 + 7, 6) AS WeekEnd
— 赫克拉维茨
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2

扩展@Tomalak的答案。该公式适用于除星期日和星期一以外的其他日子，但您需要对5的位置使用不同的值。实现所需价值的一种方法是

Value Needed = 7 - (Value From Date First Documentation for Desired Day Of Week) - 1

这是文档的链接：https : //msdn.microsoft.com/zh-cn/library/ms181598.aspx

这是一张为您布置的桌子。

          | DATEFIRST VALUE |   Formula Value   |   7 - DATEFIRSTVALUE - 1
Monday    | 1               |          5        |   7 - 1- 1 = 5
Tuesday   | 2               |          4        |   7 - 2 - 1 = 4
Wednesday | 3               |          3        |   7 - 3 - 1 = 3
Thursday  | 4               |          2        |   7 - 4 - 1 = 2
Friday    | 5               |          1        |   7 - 5 - 1 = 1
Saturday  | 6               |          0        |   7 - 6 - 1 = 0
Sunday    | 7               |         -1        |   7 - 7 - 1 = -1

但是您不必记住该表和公式，实际上您也可以使用稍有不同的表，主要的需求是使用一个值，该值将使剩余的天数正确。

这是一个工作示例：

DECLARE @MondayDateFirstValue INT = 1
DECLARE @FridayDateFirstValue INT = 5
DECLARE @TestDate DATE = GETDATE()

SET @MondayDateFirstValue = 7 - @MondayDateFirstValue - 1
SET @FridayDateFirstValue = 7 - @FridayDateFirstValue - 1

SET DATEFIRST 6 -- notice this is saturday

SELECT 
    DATEADD(DAY, 0 - (@@DATEFIRST + @MondayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate)  as MondayStartOfWeek
    ,DATEADD(DAY, 6 - (@@DATEFIRST + @MondayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate) as MondayEndOfWeek
   ,DATEADD(DAY, 0 - (@@DATEFIRST + @FridayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate)  as FridayStartOfWeek
    ,DATEADD(DAY, 6 - (@@DATEFIRST + @FridayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate) as FridayEndOfWeek


SET DATEFIRST 2 --notice this is tuesday

SELECT 
    DATEADD(DAY, 0 - (@@DATEFIRST + @MondayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate)  as MondayStartOfWeek
    ,DATEADD(DAY, 6 - (@@DATEFIRST + @MondayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate) as MondayEndOfWeek
   ,DATEADD(DAY, 0 - (@@DATEFIRST + @FridayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate)  as FridayStartOfWeek
    ,DATEADD(DAY, 6 - (@@DATEFIRST + @FridayDateFirstValue + DATEPART(dw,@TestDate)) % 7, @TestDate) as FridayEndOfWeek

此方法与DATEFIRST设置无关，因为我正在构建包含多个星期方法的日期维度，因此我需要此设置。

— 马特
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2

这是另一个版本。如果您的方案要求星期六是一周的第一天，星期五是星期的最后一天，则下面的代码将处理该问题：

  DECLARE @myDate DATE = GETDATE()
  SELECT    @myDate,
    DATENAME(WEEKDAY,@myDate),
    DATEADD(DD,-(CHOOSE(DATEPART(dw, @myDate), 1,2,3,4,5,6,0)),@myDate) AS WeekStartDate,
    DATEADD(DD,7-CHOOSE(DATEPART(dw, @myDate), 2,3,4,5,6,7,1),@myDate) AS WeekEndDate

查询的屏幕截图

— 阿杰·迪维迪（Ajay Dwivedi）
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1

下面的查询将提供从星期日到星期六的当前星期的开始和结束之间的数据

SELECT DOB FROM PROFILE_INFO WHERE DAY(DOB) BETWEEN
DAY( CURRENT_DATE() - (SELECT DAYOFWEEK(CURRENT_DATE())-1))
AND
DAY((CURRENT_DATE()+(7 - (SELECT DAYOFWEEK(CURRENT_DATE())) ) ))
AND
MONTH(DOB)=MONTH(CURRENT_DATE())
— 里雅
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1

让我们将问题分解为两部分：

1）确定星期几

该DATEPART(dw, ...)返回的数，1 ... 7，相对于DATEFIRST设定（文档）。下表总结了可能的值：

                                                   @@DATEFIRST
+------------------------------------+-----+-----+-----+-----+-----+-----+-----+-----+
|                                    |  1  |  2  |  3  |  4  |  5  |  6  |  7  | DOW |
+------------------------------------+-----+-----+-----+-----+-----+-----+-----+-----+
|  DATEPART(dw, /*Mon*/ '20010101')  |  1  |  7  |  6  |  5  |  4  |  3  |  2  |  1  |
|  DATEPART(dw, /*Tue*/ '20010102')  |  2  |  1  |  7  |  6  |  5  |  4  |  3  |  2  |
|  DATEPART(dw, /*Wed*/ '20010103')  |  3  |  2  |  1  |  7  |  6  |  5  |  4  |  3  |
|  DATEPART(dw, /*Thu*/ '20010104')  |  4  |  3  |  2  |  1  |  7  |  6  |  5  |  4  |
|  DATEPART(dw, /*Fri*/ '20010105')  |  5  |  4  |  3  |  2  |  1  |  7  |  6  |  5  |
|  DATEPART(dw, /*Sat*/ '20010106')  |  6  |  5  |  4  |  3  |  2  |  1  |  7  |  6  |
|  DATEPART(dw, /*Sun*/ '20010107')  |  7  |  6  |  5  |  4  |  3  |  2  |  1  |  7  |
+------------------------------------+-----+-----+-----+-----+-----+-----+-----+-----+

最后一列包含星期一至星期日星期几的理想星期几值*。通过仅查看图表，我们得出以下等式：

(@@DATEFIRST + DATEPART(dw, SomeDate) - 1 - 1) % 7 + 1

2）计算给定日期的星期一和星期日

这要归功于周几的价值。这是一个例子：

WITH TestData(SomeDate) AS (
    SELECT CAST('20001225' AS DATETIME) UNION ALL
    SELECT CAST('20001226' AS DATETIME) UNION ALL
    SELECT CAST('20001227' AS DATETIME) UNION ALL
    SELECT CAST('20001228' AS DATETIME) UNION ALL
    SELECT CAST('20001229' AS DATETIME) UNION ALL
    SELECT CAST('20001230' AS DATETIME) UNION ALL
    SELECT CAST('20001231' AS DATETIME) UNION ALL
    SELECT CAST('20010101' AS DATETIME) UNION ALL
    SELECT CAST('20010102' AS DATETIME) UNION ALL
    SELECT CAST('20010103' AS DATETIME) UNION ALL
    SELECT CAST('20010104' AS DATETIME) UNION ALL
    SELECT CAST('20010105' AS DATETIME) UNION ALL
    SELECT CAST('20010106' AS DATETIME) UNION ALL
    SELECT CAST('20010107' AS DATETIME) UNION ALL
    SELECT CAST('20010108' AS DATETIME) UNION ALL
    SELECT CAST('20010109' AS DATETIME) UNION ALL
    SELECT CAST('20010110' AS DATETIME) UNION ALL
    SELECT CAST('20010111' AS DATETIME) UNION ALL
    SELECT CAST('20010112' AS DATETIME) UNION ALL
    SELECT CAST('20010113' AS DATETIME) UNION ALL
    SELECT CAST('20010114' AS DATETIME)
), TestDataPlusDOW AS (
    SELECT SomeDate, (@@DATEFIRST + DATEPART(dw, SomeDate) - 1 - 1) % 7 + 1 AS DOW
    FROM TestData
)
SELECT
    FORMAT(SomeDate,                            'ddd yyyy-MM-dd') AS SomeDate,
    FORMAT(DATEADD(dd, -DOW + 1, SomeDate),     'ddd yyyy-MM-dd') AS [Monday],
    FORMAT(DATEADD(dd, -DOW + 1 + 6, SomeDate), 'ddd yyyy-MM-dd') AS [Sunday]
FROM TestDataPlusDOW

输出：

+------------------+------------------+------------------+
|  SomeDate        |  Monday          |    Sunday        |
+------------------+------------------+------------------+
|  Mon 2000-12-25  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Tue 2000-12-26  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Wed 2000-12-27  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Thu 2000-12-28  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Fri 2000-12-29  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Sat 2000-12-30  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Sun 2000-12-31  |  Mon 2000-12-25  |  Sun 2000-12-31  |
|  Mon 2001-01-01  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Tue 2001-01-02  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Wed 2001-01-03  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Thu 2001-01-04  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Fri 2001-01-05  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Sat 2001-01-06  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Sun 2001-01-07  |  Mon 2001-01-01  |  Sun 2001-01-07  |
|  Mon 2001-01-08  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Tue 2001-01-09  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Wed 2001-01-10  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Thu 2001-01-11  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Fri 2001-01-12  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Sat 2001-01-13  |  Mon 2001-01-08  |  Sun 2001-01-14  |
|  Sun 2001-01-14  |  Mon 2001-01-08  |  Sun 2001-01-14  |
+------------------+------------------+------------------+

*对于星期日至星期六，您只需稍微调整一下方程式，例如在某处加1。

— 萨尔曼A
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1

如果将星期日视为周开始日，则代码如下

Declare @currentdate date = '18 Jun 2020'

select DATEADD(D, -(DATEPART(WEEKDAY, @currentdate) - 1), @currentdate)

select DATEADD(D, (7 - DATEPART(WEEKDAY, @currentdate)), @currentdate)
— shyambabu
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0

我只是遇到了与此类似的情况，但是这里的解决方案似乎无济于事。所以我尝试自己弄清楚。我只计算周开始日期，周结束日期应具有相似的逻辑。

Select 
      Sum(NumberOfBrides) As [Wedding Count], 
      DATEPART( wk, WeddingDate) as [Week Number],
      DATEPART( year, WeddingDate) as [Year],
      DATEADD(DAY, 1 - DATEPART(WEEKDAY, dateadd(wk, DATEPART( wk, WeddingDate)-1,  DATEADD(yy,DATEPART( year, WeddingDate)-1900,0))), dateadd(wk, DATEPART( wk, WeddingDate)-1, DATEADD(yy,DATEPART( year, WeddingDate)-1900,0))) as [Week Start]

FROM  MemberWeddingDates
Group By DATEPART( year, WeddingDate), DATEPART( wk, WeddingDate)
Order By Sum(NumberOfBrides) Desc
— 用户名
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0

投票最多的答案可以正常工作，除了一年的第一周和最后一周。例如，如果WeddingDate的值为'2016-01-01'，则结果将为2015-12-27和2016-01-02，但正确的答案是2016-01-01和2016-01-02。

试试这个：

Select 
  Sum(NumberOfBrides) As [Wedding Count], 
  DATEPART( wk, WeddingDate) as [Week Number],
  DATEPART( year, WeddingDate) as [Year],
  MAX(CASE WHEN DATEPART(WEEK, WeddingDate) = 1 THEN CAST(DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate), 0) AS date) ELSE DATEADD(DAY, 7 * DATEPART(WEEK, WeddingDate), DATEADD(DAY, -(DATEPART(WEEKDAY, DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate), 0)) + 6), DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate), 0))) END) as WeekStart,
  MAX(CASE WHEN DATEPART(WEEK, WeddingDate) = DATEPART(WEEK, DATEADD(DAY, -1, DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate) + 1, 0))) THEN DATEADD(DAY, -1, DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate) + 1, 0)) ELSE DATEADD(DAY, 7 * DATEPART(WEEK, WeddingDate) + 6, DATEADD(DAY, -(DATEPART(WEEKDAY, DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate), 0)) + 6), DATEADD(YEAR, DATEDIFF(YEAR, 0, WeddingDate), 0))) END) as WeekEnd
FROM  MemberWeddingDates
Group By DATEPART( year, WeddingDate), DATEPART( wk, WeddingDate)
Order By Sum(NumberOfBrides) Desc;

结果如下： 在此处输入图片说明

它适用于所有第一周或其他一周。

— 狮子座
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0

Power BI Dax公式的星期开始日期和结束日期

WeekStartDate = [DateColumn] - (WEEKDAY([DateColumn])-1)
WeekEndDate = [DateColumn] + (7-WEEKDAY([DateColumn]))
— 维克拉姆·贾恩
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0

这是我的解决方案

    SET DATEFIRST 1; / *更改为先使用其他日期* /
    宣告@date DATETIME
    SET @date = CAST（“ 2/6/2019”作为日期）

    SELECT DATEADD（dd，0-（DATEPART（dw，@date）-1），@ date）[dateFrom]， 
            DATEADD（dd，6-（DATEPART（dw，@date）-1），@ date）[dateTo]
— 景达拉达
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0

通过自定义日期获取开始日期和结束日期 


   DECLARE @Date NVARCHAR(50)='05/19/2019' 
   SELECT
      DATEADD(DAY,CASE WHEN DATEPART(WEEKDAY, @Date)=1 THEN -6 ELSE 2 - DATEPART(WEEKDAY, @Date) END, CAST(@Date AS DATE)) [Week_Start_Date]
     ,DATEADD(DAY,CASE WHEN DATEPART(WEEKDAY, @Date)=1 THEN 0 ELSE  8 - DATEPART(WEEKDAY, @Date) END, CAST(@Date AS DATE)) [Week_End_Date]

— 拉姆·毛里亚
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尽管此代码可以解决问题，但提供说明确实有助于提高您的帖子质量。
— Shree
0

我还有其他方法，它是选择星期几和星期几当前时间：

DATEADD（d，-（DATEPART（dw，GETDATE（）-2），GETDATE（））是日期时间开始

和

DATEADD（day，7-（DATEPART（dw，GETDATE（）-1）），GETDATE（））是日期时间End

— NguyĐứnĐứcDuy
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0

另一种方法是：

declare @week_number int = 6280 -- 2020-05-07
declare @start_weekday int = 0 -- Monday
declare @end_weekday int = 6 -- next Sunday

select 
    dateadd(week, @week_number, @start_weekday), 
    dateadd(week, @week_number, @end_weekday)

说明：

  • @week_number是自初始日历日期“ 1900-01-01以来的星期数 ”。可以这样计算：select datediff(week, 0, @wedding_date) as week_number
  • 第一周的第一天@start_weekday：星期一为0，星期日为-1
  • 最后一天的@end_weekday：下周日为6，周六为5
  • dateadd(week, @week_number, @end_weekday)：将给定的星期数和天数添加到初始日历日期“ 1900-01-01 ”中
— hd84335
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0

这并非来自我，但无论如何，它都能完成工作：

SELECT DATEADD(wk, -1, DATEADD(DAY, 1-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --first day previous week
SELECT DATEADD(wk, 0, DATEADD(DAY, 1-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --first day current week
SELECT DATEADD(wk, 1, DATEADD(DAY, 1-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --first day next week

SELECT DATEADD(wk, 0, DATEADD(DAY, 0-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --last day previous week
SELECT DATEADD(wk, 1, DATEADD(DAY, 0-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --last day current week
SELECT DATEADD(wk, 2, DATEADD(DAY, 0-DATEPART(WEEKDAY, GETDATE()), DATEDIFF(dd, 0, GETDATE()))) --last day next week

我在这里找到的。

— 弗朗切斯科·曼托瓦尼（Francesco Mantovani）
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-3

不确定这有什么用，但是我最终在Netezza SQL上寻找一种解决方案，但在堆栈溢出时找不到解决方案。

对于IBM netezza，您可以使用以下命令（对于Week start mon，week end sun）：

选择next_day（WeddingDate，'SUN'）-6作为WeekStart，

next_day（WeddingDate，'SUN'）as WeekEnd

— 德库兰
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-4

对于访问查询，您可以使用以下格式作为字段

"FirstDayofWeek:IIf(IsDate([ForwardedForActionDate]),CDate(Format([ForwardedForActionDate],"dd/mm/yyyy"))-(Weekday([ForwardedForActionDate])-1))"

允许直接计算

— 拉比卜
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