获取每月每周的开始和结束日期
我需要在给定的月/年中获得每周的开始和结束日期。 （总是给出月份和年份 - 就像2017年3月）。
示例，2017年1月：
1 week : '2017-01-01' - '2017-01-01'
2 week: '2017-01-02' - '2017-01-08'
3 week: '2017-01-09' - '2017-01-15'
4 week: '2017-01-16' - '2017-01-22'
5 week: '2017-01-23' - '2017-01-29'
6 week: '2017-01-30' - '2017-01-31'
我已经know如何获得给定月/年的周数：
select *,
DATEDIFF(WEEK, DATEADD(day,-1,StartAt), DATEADD(day,-1,EndAt)) +1
as NumWeeks
但是如何获得给定月/年的每周开始/结束日期？
6 个答案:
答案 0 :(得分：1)
select
DateValue
, WeekStart =convert(date,(
case when datepart(week,DateValue) =1
then convert(date, (datename(year,DateValue) +'0101'))
else dateadd(day,@@datefirst-datepart(weekday,DateValue)-(@@datefirst-1),DateValue)
end) )
, WeekEnd =convert(date,(
case when datepart(week,DateValue) =53
then convert(date, (datename(year,DateValue) +'1231'))
else dateadd(day,(@@datefirst)-datepart(weekday,DateValue)+(7-@@datefirst),DateValue)
end) )
from dates
rextester演示：http://rextester.com/KYKS44588
测试设置：
set datefirst 1;
declare @fromdate date = '20161227', @thrudate date = '20201231';
;with n as (select n from (values(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) t(n))
, dates as (
select top (datediff(day, @fromdate, @thrudate)+1)
[DateValue]=convert(date,dateadd(day,row_number() over(order by (select 1))-1,@fromdate))
from n as deka cross join n as hecto cross join n as kilo cross join n as [tenK]
order by 1
)
, test as (
select
convert(varchar(10),DateValue,120) as Date
, WeekStart = convert(varchar(10),convert(date,(
case when datepart(week,DateValue) =1
then convert(date, (datename(year,DateValue) +'0101'))
else dateadd(day,@@datefirst-datepart(WeekDay,DateValue)-(@@datefirst-1),DateValue)
end) ),120)
, WeekEnd = convert(varchar(10),convert(date,(
case when datepart(week,DateValue) =53
then convert(date, (datename(year,DateValue) +'1231'))
else dateadd(day,(@@datefirst)-datepart(WeekDay,DateValue)+(7-@@datefirst),DateValue)
end) ),120)
, week= datepart(week,DateValue)
from dates
)
select *
, dayname= datename(weekday,date)
, weekstartdayname= datename(weekday,weekstart)
, weekenddayname= datename(weekday,weekend)
from test
where week > 51
or week < 3
order by 1
结果：
+------------+------------+------------+------+-----------+------------------+----------------+
| Date | WeekStart | WeekEnd | week | dayname | weekstartdayname | weekenddayname |
+------------+------------+------------+------+-----------+------------------+----------------+
| 2016-12-27 | 2016-12-26 | 2016-12-31 | 53 | Tuesday | Monday | Saturday |
| 2016-12-28 | 2016-12-26 | 2016-12-31 | 53 | Wednesday | Monday | Saturday |
| 2016-12-29 | 2016-12-26 | 2016-12-31 | 53 | Thursday | Monday | Saturday |
| 2016-12-30 | 2016-12-26 | 2016-12-31 | 53 | Friday | Monday | Saturday |
| 2016-12-31 | 2016-12-26 | 2016-12-31 | 53 | Saturday | Monday | Saturday |
| 2017-01-01 | 2017-01-01 | 2017-01-01 | 1 | Sunday | Sunday | Sunday |
| 2017-01-02 | 2017-01-02 | 2017-01-08 | 2 | Monday | Monday | Sunday |
| 2017-01-03 | 2017-01-02 | 2017-01-08 | 2 | Tuesday | Monday | Sunday |
| 2017-01-04 | 2017-01-02 | 2017-01-08 | 2 | Wednesday | Monday | Sunday |
| 2017-01-05 | 2017-01-02 | 2017-01-08 | 2 | Thursday | Monday | Sunday |
| 2017-01-06 | 2017-01-02 | 2017-01-08 | 2 | Friday | Monday | Sunday |
| 2017-01-07 | 2017-01-02 | 2017-01-08 | 2 | Saturday | Monday | Sunday |
| 2017-01-08 | 2017-01-02 | 2017-01-08 | 2 | Sunday | Monday | Sunday |
| 2017-12-18 | 2017-12-18 | 2017-12-24 | 52 | Monday | Monday | Sunday |
| 2017-12-19 | 2017-12-18 | 2017-12-24 | 52 | Tuesday | Monday | Sunday |
| 2017-12-20 | 2017-12-18 | 2017-12-24 | 52 | Wednesday | Monday | Sunday |
| 2017-12-21 | 2017-12-18 | 2017-12-24 | 52 | Thursday | Monday | Sunday |
| 2017-12-22 | 2017-12-18 | 2017-12-24 | 52 | Friday | Monday | Sunday |
| 2017-12-23 | 2017-12-18 | 2017-12-24 | 52 | Saturday | Monday | Sunday |
| 2017-12-24 | 2017-12-18 | 2017-12-24 | 52 | Sunday | Monday | Sunday |
| 2017-12-25 | 2017-12-25 | 2017-12-31 | 53 | Monday | Monday | Sunday |
| 2017-12-26 | 2017-12-25 | 2017-12-31 | 53 | Tuesday | Monday | Sunday |
| 2017-12-27 | 2017-12-25 | 2017-12-31 | 53 | Wednesday | Monday | Sunday |
| 2017-12-28 | 2017-12-25 | 2017-12-31 | 53 | Thursday | Monday | Sunday |
| 2017-12-29 | 2017-12-25 | 2017-12-31 | 53 | Friday | Monday | Sunday |
| 2017-12-30 | 2017-12-25 | 2017-12-31 | 53 | Saturday | Monday | Sunday |
| 2017-12-31 | 2017-12-25 | 2017-12-31 | 53 | Sunday | Monday | Sunday |
| 2018-01-01 | 2018-01-01 | 2018-01-07 | 1 | Monday | Monday | Sunday |
| 2018-01-02 | 2018-01-01 | 2018-01-07 | 1 | Tuesday | Monday | Sunday |
| 2018-01-03 | 2018-01-01 | 2018-01-07 | 1 | Wednesday | Monday | Sunday |
| 2018-01-04 | 2018-01-01 | 2018-01-07 | 1 | Thursday | Monday | Sunday |
| 2018-01-05 | 2018-01-01 | 2018-01-07 | 1 | Friday | Monday | Sunday |
| 2018-01-06 | 2018-01-01 | 2018-01-07 | 1 | Saturday | Monday | Sunday |
| 2018-01-07 | 2018-01-01 | 2018-01-07 | 1 | Sunday | Monday | Sunday |
| 2018-01-08 | 2018-01-08 | 2018-01-14 | 2 | Monday | Monday | Sunday |
| 2018-01-09 | 2018-01-08 | 2018-01-14 | 2 | Tuesday | Monday | Sunday |
| 2018-01-10 | 2018-01-08 | 2018-01-14 | 2 | Wednesday | Monday | Sunday |
| 2018-01-11 | 2018-01-08 | 2018-01-14 | 2 | Thursday | Monday | Sunday |
| 2018-01-12 | 2018-01-08 | 2018-01-14 | 2 | Friday | Monday | Sunday |
| 2018-01-13 | 2018-01-08 | 2018-01-14 | 2 | Saturday | Monday | Sunday |
| 2018-01-14 | 2018-01-08 | 2018-01-14 | 2 | Sunday | Monday | Sunday |
| 2018-12-24 | 2018-12-24 | 2018-12-30 | 52 | Monday | Monday | Sunday |
| 2018-12-25 | 2018-12-24 | 2018-12-30 | 52 | Tuesday | Monday | Sunday |
| 2018-12-26 | 2018-12-24 | 2018-12-30 | 52 | Wednesday | Monday | Sunday |
| 2018-12-27 | 2018-12-24 | 2018-12-30 | 52 | Thursday | Monday | Sunday |
| 2018-12-28 | 2018-12-24 | 2018-12-30 | 52 | Friday | Monday | Sunday |
| 2018-12-29 | 2018-12-24 | 2018-12-30 | 52 | Saturday | Monday | Sunday |
| 2018-12-30 | 2018-12-24 | 2018-12-30 | 52 | Sunday | Monday | Sunday |
| 2018-12-31 | 2018-12-31 | 2018-12-31 | 53 | Monday | Monday | Monday |
| 2019-01-01 | 2019-01-01 | 2019-01-06 | 1 | Tuesday | Tuesday | Sunday |
| 2019-01-02 | 2019-01-01 | 2019-01-06 | 1 | Wednesday | Tuesday | Sunday |
| 2019-01-03 | 2019-01-01 | 2019-01-06 | 1 | Thursday | Tuesday | Sunday |
| 2019-01-04 | 2019-01-01 | 2019-01-06 | 1 | Friday | Tuesday | Sunday |
| 2019-01-05 | 2019-01-01 | 2019-01-06 | 1 | Saturday | Tuesday | Sunday |
| 2019-01-06 | 2019-01-01 | 2019-01-06 | 1 | Sunday | Tuesday | Sunday |
| 2019-01-07 | 2019-01-07 | 2019-01-13 | 2 | Monday | Monday | Sunday |
| 2019-01-08 | 2019-01-07 | 2019-01-13 | 2 | Tuesday | Monday | Sunday |
| 2019-01-09 | 2019-01-07 | 2019-01-13 | 2 | Wednesday | Monday | Sunday |
| 2019-01-10 | 2019-01-07 | 2019-01-13 | 2 | Thursday | Monday | Sunday |
| 2019-01-11 | 2019-01-07 | 2019-01-13 | 2 | Friday | Monday | Sunday |
| 2019-01-12 | 2019-01-07 | 2019-01-13 | 2 | Saturday | Monday | Sunday |
| 2019-01-13 | 2019-01-07 | 2019-01-13 | 2 | Sunday | Monday | Sunday |
| 2019-12-23 | 2019-12-23 | 2019-12-29 | 52 | Monday | Monday | Sunday |
| 2019-12-24 | 2019-12-23 | 2019-12-29 | 52 | Tuesday | Monday | Sunday |
| 2019-12-25 | 2019-12-23 | 2019-12-29 | 52 | Wednesday | Monday | Sunday |
| 2019-12-26 | 2019-12-23 | 2019-12-29 | 52 | Thursday | Monday | Sunday |
| 2019-12-27 | 2019-12-23 | 2019-12-29 | 52 | Friday | Monday | Sunday |
| 2019-12-28 | 2019-12-23 | 2019-12-29 | 52 | Saturday | Monday | Sunday |
| 2019-12-29 | 2019-12-23 | 2019-12-29 | 52 | Sunday | Monday | Sunday |
| 2019-12-30 | 2019-12-30 | 2019-12-31 | 53 | Monday | Monday | Tuesday |
| 2019-12-31 | 2019-12-30 | 2019-12-31 | 53 | Tuesday | Monday | Tuesday |
| 2020-01-01 | 2020-01-01 | 2020-01-05 | 1 | Wednesday | Wednesday | Sunday |
| 2020-01-02 | 2020-01-01 | 2020-01-05 | 1 | Thursday | Wednesday | Sunday |
| 2020-01-03 | 2020-01-01 | 2020-01-05 | 1 | Friday | Wednesday | Sunday |
| 2020-01-04 | 2020-01-01 | 2020-01-05 | 1 | Saturday | Wednesday | Sunday |
| 2020-01-05 | 2020-01-01 | 2020-01-05 | 1 | Sunday | Wednesday | Sunday |
| 2020-01-06 | 2020-01-06 | 2020-01-12 | 2 | Monday | Monday | Sunday |
| 2020-01-07 | 2020-01-06 | 2020-01-12 | 2 | Tuesday | Monday | Sunday |
| 2020-01-08 | 2020-01-06 | 2020-01-12 | 2 | Wednesday | Monday | Sunday |
| 2020-01-09 | 2020-01-06 | 2020-01-12 | 2 | Thursday | Monday | Sunday |
| 2020-01-10 | 2020-01-06 | 2020-01-12 | 2 | Friday | Monday | Sunday |
| 2020-01-11 | 2020-01-06 | 2020-01-12 | 2 | Saturday | Monday | Sunday |
| 2020-01-12 | 2020-01-06 | 2020-01-12 | 2 | Sunday | Monday | Sunday |
| 2020-12-21 | 2020-12-21 | 2020-12-27 | 52 | Monday | Monday | Sunday |
| 2020-12-22 | 2020-12-21 | 2020-12-27 | 52 | Tuesday | Monday | Sunday |
| 2020-12-23 | 2020-12-21 | 2020-12-27 | 52 | Wednesday | Monday | Sunday |
| 2020-12-24 | 2020-12-21 | 2020-12-27 | 52 | Thursday | Monday | Sunday |
| 2020-12-25 | 2020-12-21 | 2020-12-27 | 52 | Friday | Monday | Sunday |
| 2020-12-26 | 2020-12-21 | 2020-12-27 | 52 | Saturday | Monday | Sunday |
| 2020-12-27 | 2020-12-21 | 2020-12-27 | 52 | Sunday | Monday | Sunday |
| 2020-12-28 | 2020-12-28 | 2020-12-31 | 53 | Monday | Monday | Thursday |
| 2020-12-29 | 2020-12-28 | 2020-12-31 | 53 | Tuesday | Monday | Thursday |
| 2020-12-30 | 2020-12-28 | 2020-12-31 | 53 | Wednesday | Monday | Thursday |
| 2020-12-31 | 2020-12-28 | 2020-12-31 | 53 | Thursday | Monday | Thursday |
+------------+------------+------------+------+-----------+------------------+----------------+
日历和数字表格引用：
- Generate a set or sequence without loops - 1 - Aaron Bertrand
- Generate a set or sequence without loops - 3 - Aaron Bertrand
- The "Numbers" or "Tally" Table: What it is and how it replaces a loop - Jeff Moden
- Creating a Date Table/Dimension in SQL Server 2008 - David Stein
- Calendar Tables - Why You Need One - David Stein
- Creating a date dimension or calendar table in SQL Server - Aaron Bertrand
- TSQL Function to Determine Holidays in SQL Server - Tim Cullen
- F_TABLE_DATE - Michael Valentine Jones
答案 1 :(得分：1)
这样的东西？
DECLARE @Dates TABLE (DateId INT IDENTITY, Dt Date);
DECLARE @STart Date = DATEADD(Year, DATEDIFF(Year, 0, GETDATE()),0)
SET NOCOUNT ON;
WHILE @STart <= DATEADD(DAY, DATEDIFF(DAY, 0, GETDATE()), 0)
BEGIN
INSERT INTO @Dates (Dt) VALUES (@STart)
SELECT @Start = DATEADD(DAY, 1, @STart)
END
<强>更新
SELECT
DateId
, Dt
, DATEADD(WEEK, DATEDIFF(DAY, 0, Dt)/7, 0) AS WeekBeginningMondayOf
, DATEADD(DAY, 6, DATEADD(WEEK, DATEDIFF(DAY, 0, Dt)/7, 0)) AS WeekEndingSundayOf
, DENSE_RANK() OVER(PARTITION BY MONTH(Dt) ORDER BY DATEADD(WEEK, DATEDIFF(DAY, 0, Dt)/7, 0)) AS WeekInMonth
FROM @Dates
答案 2 :(得分：1)
我认为你在寻找的是 - ＆gt; someone asked on stackoverflow与你正在寻找的相似。
这是最终版本，希望按照您的期望 我建立在上面的链接上希望这是你正在寻找的（apologies new to stackoverflow，unformatted t-sql下面）
DECLARE @sDate DATETIME,
@eDate DATETIME
SET @sDate = '2017-03-01'
SET @eDate = DATEADD(DAY,-1, CAST(Cast(DatePart(YEAR,@sdate) AS varchar(4))
+'-' + Cast((DatePart(MONTH,@sdate)+1) AS varchar(2))+ '-1' AS Date))
SET @sDate = CAST(Cast(DatePart(YEAR,@sdate) AS varchar(4)) +'-'
+ Cast(DatePart(MONTH,@sdate) AS varchar(2))+ '-1' AS Date)
DECLARE @startDayOfWeekOffSet int ;
SET @startDayOfWeekOffSet=
(SELECT DATEPART(DW,@sDate));
DECLARE @DaysToGetFirstSaturday int
SET @DaysToGetFirstSaturday= 7- @startDayOfWeekOffSet
DECLARE @firstEverStartOfWeekDay Date , @firstWeekEndDay Date
SET @firstEverStartOfWeekDay =@sDate;
IF @startDayOfWeekOffSet = 1 -- January
BEGIN
SET @firstEverStartOfWeekDay = @sDate
SET @DaysToGetFirstSaturday = 0
SET @firstWeekEndDay = @sDate END
ELSE BEGIN IF DATEPART(MONTH,@sDate) > DATEPART(MONTH,@firstEverStartOfWeekDay)
AND @startDayOfWeekOffSet = 7 --FirstSundayNewMonth
BEGIN
SET @firstEverStartOfWeekDay =@sDate
SET @firstWeekEndDay = @sDate END ELSE --NotASundayNewMonth
BEGIN
SET @firstWeekEndDay = DATEADD(DAY,@DaysToGetFirstSaturday,@firstEverStartOfWeekDay);
END END
SELECT @firstWeekEndDay AS firstWeekEndDay,
@firstEverStartOfWeekDay AS firstEverStartOfWeekDay,
@DaysToGetFirstSaturday AS DaysToGetFirstSaturday,
@startDayOfWeekOffSet AS startDayOfWeekOffSet ;
WITH cte AS
(SELECT 1 AS WeekNum,
@firstEverStartOfWeekDay StartDate,
@firstWeekEndDay EndDate
UNION ALL SELECT WeekNum + 1, --Case when @scenario = 'NewYear' Then
dateadd(DAY, 1, cte.EndDate) --Else dateadd(ww, 1, cte.StartDate) End as
StartDate,
CASE
WHEN StartDate = EndDate THEN DateAdd(DAY,-1,dateadd(DAY, 8, cte.EndDate))
ELSE dateadd(ww, 1, cte.EndDate)
END AS EndDate
FROM cte
WHERE dateadd(ww, 1, StartDate)<= @eDate )
SELECT WeekNum,
CASE
WHEN DatePart(YEAR, StartDate) < DATEPART(YEAR,@sDate)
THEN CAST(Cast(DatePart(YEAR,@sdate) AS varchar(4)) + '-1-1' AS Date)
WHEN DatePart(MONTH, StartDate) < DATEPART(MONTH,@sDate)
THEN CAST(Cast(DatePart(YEAR,@sdate) AS varchar(4)) + '-'
+ Cast(DatePart(MONTH,@sdate) AS varchar(2)) + '-1' AS Date)
ELSE StartDate
END AS StartDate,
CASE
WHEN DatePart(MONTH, EndDate) > DATEPART(MONTH,@eDate) THEN @eDate
ELSE EndDate
END AS EndDate
FROM cte
答案 3 :(得分：1)
在这种情况下使用数字/日期/计数表是非常聪明的。在许多情况下，这样的桌子非常漂亮！
在this answer中，我展示了如何创建和填充此类表格的方法。
创建此表后，查询就像：
一样简单SELECT n.CalendarDate AS StartOfWeek
,DATEADD(DAY,6,n.CalendarDate) AS EndOfWeek
FROM dbo.RunningNumbers AS n
WHERE n.CalendarDate>={d'2017-01-01'} AND n.CalendarDate<{d'2017-02-01'}
AND n.CalendarWeekDay=1;
提示这样的表可以很容易地扩展，只需通过手动将一些值添加到额外的列中来计算信息（例如，holydays， free-of-work 天）。 ..）
答案 4 :(得分：1)
如果您只想让一个函数返回一个月的一个月，那么这可以达到您想要的效果：
create function dbo.udf_weeks_of_month (@fromdate date)
returns table as return (
with n as (select n from (values(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) t(n))
, dates as (
select top (datediff(day, @fromdate, dateadd(month, datediff(month, 0, @fromdate )+1, 0)))
[DateValue]=convert(date,dateadd(day,row_number() over(order by (select 1))-1,@fromdate))
from n as deka cross join n as hecto
)
select
WeekOfMonth = row_number() over (order by datepart(week,DateValue))
, Week = datepart(week,DateValue)
, WeekStart = min(DateValue)
, WeekEnd = max(DateValue)
from dates
group by datepart(week,DateValue)
);
并像这样调用它：
set datefirst 1;
select * from dbo.udf_weeks_of_month('20170101');
返回：
+-------------+------+------------+------------+
| WeekOfMonth | Week | WeekStart | WeekEnd |
+-------------+------+------------+------------+
| 1 | 1 | 2017-01-01 | 2017-01-01 |
| 2 | 2 | 2017-01-02 | 2017-01-08 |
| 3 | 3 | 2017-01-09 | 2017-01-15 |
| 4 | 4 | 2017-01-16 | 2017-01-22 |
| 5 | 5 | 2017-01-23 | 2017-01-29 |
| 6 | 6 | 2017-01-30 | 2017-01-31 |
+-------------+------+------------+------------+
和这个电话：
select * from dbo.udf_weeks_of_month('february 2017');
返回：
+-------------+------+------------+------------+
| WeekOfMonth | Week | WeekStart | WeekEnd |
+-------------+------+------------+------------+
| 1 | 6 | 2017-02-01 | 2017-02-05 |
| 2 | 7 | 2017-02-06 | 2017-02-12 |
| 3 | 8 | 2017-02-13 | 2017-02-19 |
| 4 | 9 | 2017-02-20 | 2017-02-26 |
| 5 | 10 | 2017-02-27 | 2017-02-28 |
+-------------+------+------------+------------+
rextester演示：http://rextester.com/VKPQU7936（注意：rextester重新格式化日期）
答案 5 :(得分：1)
这个主题有很多变化。这个比其他答案略短。
declare @dt date = {fn current_date()};
declare @start_of_year date = datefromparts(year(@dt), 1, 1);
with digits(d) as (
select 0 union all select 1 union all select 2 union all select 3 union all
select 4 union all select 5 union all select 6 union all select 7
), wks as (
select
dateadd(week, d1.d * 8 + d0.d, dateadd(day, 1-datepart(weekday, @start_of_year), @start_of_year)) as week_start,
d1.d * 8 + d0.d as wk
from digits as d0 cross join digits as d1
)
select
case when year(week_start) < year(@dt)
then @start_of_year else week_start end as week_start,
case when year(dateadd(day, 6, week_start)) > year(@dt)
then datefromparts(year(@dt), 12, 31) else dateadd(day, 6, week_start) end as week_end
from wks
where wk between 0 and 53 and year(week_start) = year(@dt)
order by week_start;
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