是否有办法在SQL Server中按月分组日期？
例如
Week 2: 05/07/2012 - 05/13/2012
Week 3: 05/14/2012 - 05/20/2012
但使用Sql server语句
我试过
SELECT SOMETHING,
datediff(wk, convert(varchar(6), getdate(), 112) + '01', getdate()) + 1 AS TIME_
FROM STATISTICS_
GROUP BY something, TIME_
ORDER BY TIME_
但它返回月份的周数。 （意思是3）
如何获得本周的那两天？
例如，现在我们处于第三（3）周并且我想要展示 05/14/2012 - 05/20/2012
我解决了 以某种方式 ：
SELECT DATEADD(ww, DATEDIFF(ww,0,<my_column_name>), 0)
select DATEADD(ww, DATEDIFF(ww,0,<my_column_name>), 0)+6
然后我会得到两天，我会在以后连接它们。
答案 0 :(得分：3)
然后我们可以生成不同周的范围，您可以使用它来加入其他表格。
SET DATEFIRST 7;
SET NOCOUNT ON;
DECLARE @today SMALLDATETIME, @fd SMALLDATETIME, @rc INT;
SELECT @today = DATEADD(DAY, DATEDIFF(DAY, 0, GETDATE()), 0), -- today
@fd = DATEADD(DAY, 1-DAY(@today), @today), -- first day of this month
@rc = DATEPART(DAY, DATEADD(DAY, -1, DATEADD(MONTH, 1, @fd)));-- days in month
DECLARE @thismonth TABLE (
[date] SMALLDATETIME,
[weekday] TINYINT,
[weeknumber] TINYINT
);
;WITH n(d) AS (
SELECT TOP (@rc+12) DATEADD(DAY, ROW_NUMBER() OVER
(ORDER BY [object_id]) - 7, @fd) FROM sys.all_objects
)
INSERT @thismonth([date], [weekday]) SELECT d, DATEPART(WEEKDAY, d) FROM n;
DELETE @thismonth WHERE [date] < (SELECT MIN([date]) FROM @thismonth WHERE [weekday] = 2)
OR [date] > (SELECT MAX([date]) FROM @thismonth WHERE [weekday] = 1);
;WITH x AS ( SELECT [date], weeknumber, rn = ((ROW_NUMBER() OVER
(ORDER BY [date])-1) / 7) + 1 FROM @thismonth ) UPDATE x SET weeknumber = rn;
- 现在，最后的查询给出了所有这些（我只是打破了这个以摆脱垂直滚动条）：
;WITH ranges(w,s,e) AS (
SELECT weeknumber, MIN([date]), MAX([date]) FROM @thismonth GROUP BY weeknumber
)
SELECT [week] = CONVERT(CHAR(10), r.s, 120) + ' - ' + CONVERT(CHAR(10), r.e, 120)
--, SOMETHING , other columns from STATISTICS_?
FROM ranges AS r
-- LEFT OUTER JOIN dbo.STATISTICS_ AS s
-- ON s.TIME_ >= r.s AND s.TIME_ < DATEADD(DAY, 1, r.e)
-- comment this out if you want all the weeks from this month:
WHERE w = (SELECT weeknumber FROM @thismonth WHERE [date] = @today)
GROUP BY r.s, r.e --, SOMETHING
ORDER BY [week];
WHERE子句的结果：
week
-----------------------
2012-05-14 - 2012-05-20
没有WHERE子句的结果：
week
-----------------------
2012-04-30 - 2012-05-06
2012-05-07 - 2012-05-13
2012-05-14 - 2012-05-20
2012-05-21 - 2012-05-27
2012-05-28 - 2012-06-03
请注意，我故意选择了YYYY-MM-DD。您应该避免像M / D / Y这样的区域格式，特别是输入和显示。无论你认为你的观众是多么有针对性，你总会有人认为05/07/2012是7月5日，而不是5月7日。 YYYY-MM-DD没有任何含糊之处。
答案 1 :(得分：2)
创建calendar table，然后您可以查询周数，特定周数和月份的第一天/最后一天等。您还可以加入查询以获取日期范围等。
答案 2 :(得分：0)
案例陈述怎么样？
case when datepart(day, mydatetime) between 1 and 7 then 1
when datepart(day, mydatetime) between 8 and 14 then 2
...
你还必须包括年份＆amp;月，除非你想要同一组中的所有第1周。
答案 3 :(得分：0)
您不清楚您是否要“按月份分组日期”，或者“选择某一周内的数据”
如果你的意思是“群组”，那么这个小片段应该可以让你获得“一周的一周”：
SELECT <stuff>
FROM CP_STATISTICS
WHERE Month(<YOUR DATE COL>) = 5 --april
GROUP BY Year(<YOUR DATE COL>),
Month(<YOUR DATE COL>),
DATEDIFF(week, DATEADD(MONTH, DATEDIFF(MONTH, 0, <YOUR DATE COL>), 0)
, <YOUR DATE COL>) +1
或者，如果您想要“4月第1周的销售额，按日期排序”，您可以做类似的事情。
DECLARE @targetDate datetime2 = '5/3/2012'
DECLARE @targetWeek int = DATEDIFF(week, DATEADD(MONTH,
DATEDIFF(MONTH, 0, @targetDate), 0), @targetDate) +1
SELECT <stuff>
FROM CP_STATISTICS
WHERE MONTH(@targetDate) = Month(myDateCol) AND
YEAR(@targetDate) = Year (myDateCol) AND
@targetWeek = DATEDIFF(week, DATEADD(MONTH,
DATEDIFF(MONTH, 0, myDateCol), 0), myDateCol) +1
ORDER BY myDateCol
请注意，如果您使用非标准周，或者想要在跨越月份边界的数周内达到前几个月的某些日子，事情会变得更加复杂。
编辑2
从查看“现在解决”部分开始。我认为你的问题是“如何在一周内从表格中获取数据？”
您的解决方案似乎是：
DECLARE @targetDate datetime2 = '5/1/2012'
DECLARE @startDate datetime2 = DATEADD(ww, DATEDIFF(ww,0,targetDate), 0)
DECLARE @endDate datetime2 = DATEADD(ww, DATEDIFF(ww,0,@now), 0)+6
SELECT <stuff>
FROM STATISTICS_
WHERE dateStamp >= @startDate AND dateStamp <= @endDate
请注意，如果日期为5/1，此解决方案将导致开始日期为“4/30/2012”。我指出这一点是因为你的解决方案跨越了月界限。这可能是也可能不是。