9

我目前遇到了这个问题。假设我有以下每日数据

    +-------------------------+-----+----+
    |          Date           | C1  | C2 |
    +-------------------------+-----+----+
    | 2012-08-01 00:00:00.000 |  44 | 44 |
    | 2012-08-02 00:00:00.000 |  51 | 49 |
    | 2012-08-03 00:00:00.000 |  60 | 59 |
    | 2012-08-04 00:00:00.000 |  68 | 67 |
    | 2012-08-05 00:00:00.000 |  82 | 78 |
    | 2012-08-06 00:00:00.000 |  62 | 59 |
    | 2012-08-07 00:00:00.000 |  58 | 53 |
    | 2012-08-08 00:00:00.000 |  69 | 65 |
    | 2012-08-09 00:00:00.000 |  82 | 72 |
    | 2012-08-10 00:00:00.000 |  70 | 68 |
    | 2012-08-11 00:00:00.000 |  75 | 71 |
    | 2012-08-12 00:00:00.000 |  64 | 64 |
    | 2012-08-13 00:00:00.000 |  74 | 69 |
    | 2012-08-14 00:00:00.000 |  60 | 56 |
    | 2012-08-15 00:00:00.000 |  66 | 60 |
    | 2012-08-16 00:00:00.000 |  57 | 51 |
    | 2012-08-17 00:00:00.000 |  52 | 49 |
    +-------------------------+-----+----+

我将如何对它进行分组，以便每周汇总 C1 和 C2？预期输出应该是

+---------------------------+------+----+
|          Date             |  C1  | C2 |
+---------------------------+------+----+
| 2012-08-06 to 2012-12-12  |  480 | 452|
| 2012-08-13 to 2012-08-19  |  430 | 394|
+---------------------------+------+----+

它从 2012-08-06 开始，因为周期应该是周一到周日。我已经尝试了大约一个小时左右的谷歌搜索，似乎没有结果适合我的问题，我希望有人可以帮助我。

谢谢！

4

2 回答 2

8

试试这个：

SET DATEFIRST 1 会将您的一周开始时间设置为星期一

SET DATEFIRST 1

    SELECT CAST(MIN( [DATE]) AS VARCHAR(20))+' TO '+CAST (MAX([DATE]) AS VARCHAR(20)) AS DATE,
           SUM(C1) AS GRU,
           SUM(C2) AS C1
    FROM   YOUR_TABLE
    GROUP BY DATEPART(WEEK,[DATE])
    HAVING COUNT(DISTINCT[DATE])=7

SET DATEFIRST 7
于 2012-09-18T06:31:43.620 回答
4

可能是这样的（记得接受答案）

declare @t table(Date datetime, C1 int, C2 int)
insert @t values('2012-08-01',44,44)
insert @t values('2012-08-02',51,49)
insert @t values('2012-08-03',60,59)
insert @t values('2012-08-04',68,67)
insert @t values('2012-08-05',82,78)
insert @t values('2012-08-06',62,59)
insert @t values('2012-08-07',58,53)
insert @t values('2012-08-08',69,65)
insert @t values('2012-08-09',82,72)
insert @t values('2012-08-10',70,68)
insert @t values('2012-08-11',75,71)
insert @t values('2012-08-12',64,64)
insert @t values('2012-08-13',74,69)
insert @t values('2012-08-14',60,56)
insert @t values('2012-08-15',66,60)
insert @t values('2012-08-16',57,51)
insert @t values('2012-08-17',52,49)

select 
convert(varchar(10), dateadd(week, datediff(week, 0, date-1),0), 120)+' to '+
convert(varchar(10), max(dateadd(week, datediff(week, 0,date-1),6)), 120) Date,
sum(C1) GRU, sum(C2) C1
from @t
group by dateadd(week, datediff(week, 0, date-1),0)
having datediff(day, 0, min(date)) %7 = 0
于 2012-09-18T07:34:04.353 回答