来源日期：
CREATE TABLE #Temp (ID INT Identity(1,1) Primary Key, BeginDate datetime, EndDate datetime, GroupBy INT)
INSERT INTO #Temp
SELECT '2015-06-05 00:00:00.000','2015-06-12 00:00:00.000',7
UNION
SELECT '2015-06-05 00:00:00.000', '2015-06-08 00:00:00.000',7
UNION
SELECT '2015-10-22 00:00:00.000', '2015-10-31 00:00:00.000',7
SELECT *, DATEDIFF(DAY,BeginDate, EndDate) TotalDays FROM #Temp
DROP TABLE #Temp
ID BeginDate EndDate GroupBy TotalDays
1 6/5/15 0:00 6/8/15 0:00 7 3
2 6/5/15 0:00 6/12/15 0:00 7 7
3 10/22/15 0:00 10/31/15 0:00 7 9
期望输出：
ID BeginDate EndDate GroupBy TotalDays GroupCnt GroupNum
1 6/5/15 0:00 6/8/15 0:00 7 3 1 1
2 6/5/15 0:00 6/12/15 0:00 7 7 1 1
3 10/22/15 0:00 10/29/15 0:00 7 9 2 1
3 10/29/15 0:00 10/31/15 0:00 7 9 2 2
目标：
根据ID/BeginDate/EndDate对记录进行分组。
基于GroupBy 号 (# of days)和 TotalDays (days diff)，
如果GroupBy =＆gt; TotalDays，保持单组记录
否则将组记录（每GroupBy计数1条记录）乘以TotalDays限制。
道歉，如果它令人困惑，但基本上，在上面的例子中，每个组(ID/BeginDate/EndDate)应该有一条记录，其中记录天差异b/w Begin/End date = 7 or less（GroupBy）。
如果天数差异超过7天，则创建另一条记录（每隔7天差异）。
因此，由于前两个记录的天差为7天或更短，因此只有一个记录。
第3条记录的天差为9 (7 + 2)。因此，应该有2条记录（前7天为第1天，另外2天为第2条）。
GroupCNT = how many records there're of the grouped records after applying the above records.
GroupNum基本上是该组的row number。
每个记录的GroupBy＃可以不同。数据集非常庞大，因此性能非常重要。
我能够弄清楚的一个模式与模数b / w GroupBy和days diff有关。
当GroupBy value is < days diff时，模数总是小于GroupBy。当GroupBy value = days diff时，模数始终为0.当GroupBy value > days diff时，模数总是等于GroupBy。我不确定是否/如何使用它来分组/乘以记录以满足要求。
SELECT DISTINCT
ID
, BeginDate
, EndDate
, GroupBy
, DATEDIFF(DAY,BeginDate, EndDate) TotalDays
, CAST(GroupBy as decimal(18,6))%CAST(DATEDIFF(DAY,BeginDate, EndDate) AS decimal(18,6)) Modulus
, CASE WHEN DATEDIFF(DAY,BeginDate, EndDate) <= GroupBy THEN BeginDate END NewBeginDate
, CASE WHEN DATEDIFF(DAY,BeginDate, EndDate) <= GroupBy THEN EndDate END NewEndDate
FROM #Temp
更新： 忘记提及/包括开始/结束，当记录成倍增加时，将相应地改变。换句话说，开始/结束日期将反映GroupBy - 期望的输出在第3和第4条记录中更清楚地显示我的意思。 此外，GroupCnt / GroupNum的计算不如分组/乘以记录那么重要。
答案 0 :(得分：3)
你可以使用递归CTE做这样的事情。
;WITH cte AS (
SELECT ID,
BeginDate,
EndDate,
GroupBy,
DATEDIFF(DAY, BeginDate, EndDate) AS TotalDays,
1 AS GroupNum
FROM #Temp
UNION ALL
SELECT ID,
BeginDate,
EndDate,
GroupBy,
TotalDays,
GroupNum + 1
FROM cte
WHERE GroupNum * GroupBy < TotalDays
)
SELECT ID,
BeginDate = CASE WHEN GroupNum = 1 THEN BeginDate
ELSE DATEADD(DAY, GroupBy * (GroupNum - 1), BeginDate)
END ,
EndDate = CASE WHEN TotalDays <= GroupBy THEN EndDate
WHEN DATEADD(DAY, GroupBy * GroupNum, BeginDate) > EndDate THEN EndDate
ELSE DATEADD(DAY, GroupBy * GroupNum, BeginDate)
END ,
GroupBy,
TotalDays,
COUNT(*) OVER (PARTITION BY ID) GroupCnt,
GroupNum
FROM cte
OPTION (MAXRECURSION 0)
cte构建了一个像这样的记录集。
ID BeginDate EndDate GroupBy TotalDays GroupNum
----------- ----------------------- ----------------------- ----------- ----------- -----------
1 2015-06-05 00:00:00.000 2015-06-08 00:00:00.000 7 3 1
2 2015-06-05 00:00:00.000 2015-06-12 00:00:00.000 7 7 1
3 2015-10-22 00:00:00.000 2015-10-31 00:00:00.000 7 9 1
3 2015-10-22 00:00:00.000 2015-10-31 00:00:00.000 7 9 2
然后你只需要这个并使用一些case语句来确定开始和结束日期应该是什么。
你最终应该
ID BeginDate EndDate GroupBy TotalDays GroupCnt GroupNum
----------- ----------------------- ----------------------- ----------- ----------- ----------- -----------
1 2015-06-05 00:00:00.000 2015-06-08 00:00:00.000 7 3 1 1
2 2015-06-05 00:00:00.000 2015-06-12 00:00:00.000 7 7 1 1
3 2015-10-22 00:00:00.000 2015-10-29 00:00:00.000 7 9 2 1
3 2015-10-29 00:00:00.000 2015-10-31 00:00:00.000 7 9 2 2
由于您使用的是SQL 2012，因此您还可以在最终查询中使用LAG和LEAD函数。
;WITH cte AS (
SELECT ID,
BeginDate,
EndDate,
GroupBy,
DATEDIFF(DAY, BeginDate, EndDate) AS TotalDays,
1 AS GroupNum
FROM #Temp
UNION ALL
SELECT ID,
BeginDate,
EndDate,
GroupBy,
TotalDays,
GroupNum + 1
FROM cte
WHERE GroupNum * GroupBy < TotalDays
)
SELECT ID,
BeginDate = COALESCE(LAG(BeginDate) OVER (PARTITION BY ID ORDER BY GroupNum) + GroupBy * (GroupNum - 1), BeginDate),
EndDate = COALESCE(LEAD(BeginDate) OVER (PARTITION BY ID ORDER BY GroupNum) + GroupBy * GroupNum, EndDate),
GroupBy,
TotalDays,
COUNT(*) OVER (PARTITION BY ID) GroupCnt,
GroupNum
FROM cte
OPTION (MAXRECURSION 0)
答案 1 :(得分：1)
CREATE TABLE dim_number (id INT);
INSERT INTO dim_number VALUES ((0), (1), (2), (3)); -- Populate this to a large number
SELECT
#Temp.Id,
CASE WHEN dim_number.id = 0
THEN #Temp.BeginDate
ELSE DATEADD(DAY, dim_number.id * #Temp.GroupBy, #Temp.BeginDate)
END AS BeginDate,
CASE WHEN dim_number.id = parts.count
THEN #Temp.EndDate
ELSE DATEADD(DAY, (dim_number.id + 1) * #Temp.GroupBy, #Temp.BeginDate)
END AS EndDate,
#Temp.GroupBy AS GroupBy,
DATEDIFF(DAY, #Temp.BeginDate, #Temp.EndDate) AS TotalDays,
parts.count + 1 AS GroupCnt,
dim_number.id + 1 AS GroupNum
FROM
#Temp
CROSS APPLY
(SELECT DATEDIFF(DAY, #Temp.BeginDate, #Temp.EndDate) / #Temp.GroupBy AS count) AS parts
INNER JOIN
dim_number
ON dim_number.id >= 0
AND dim_number.id <= parts.count