合并重叠日期间隔

Question

合并重叠日期间隔是否有更好的方法？
我提出的解决方案非常简单，现在我想知道是否有其他人更好地了解如何做到这一点。

/***** DATA EXAMPLE *****/ 
DECLARE @T TABLE (d1 DATETIME, d2 DATETIME) 
INSERT INTO @T (d1, d2) 
     SELECT '2010-01-01','2010-03-31' UNION SELECT '2010-04-01','2010-05-31' 
    UNION SELECT '2010-06-15','2010-06-25' UNION SELECT '2010-06-26','2010-07-10' 
    UNION SELECT '2010-08-01','2010-08-05' UNION SELECT '2010-08-01','2010-08-09' 
    UNION SELECT '2010-08-02','2010-08-07' UNION SELECT '2010-08-08','2010-08-08' 
    UNION SELECT '2010-08-09','2010-08-12' UNION SELECT '2010-07-04','2010-08-16' 
    UNION SELECT '2010-11-01','2010-12-31' UNION SELECT '2010-03-01','2010-06-13' 

/***** INTERVAL ANALYSIS *****/ 
WHILE (1=1) BEGIN 
    UPDATE t1 SET t1.d2 = t2.d2 
    FROM @T AS t1 INNER JOIN @T AS t2 ON 
      DATEADD(day, 1, t1.d2) BETWEEN t2.d1 AND t2.d2 
    IF @@ROWCOUNT = 0 BREAK 
END 

/***** RESULT *****/ 
SELECT StartDate = MIN(d1) , EndDate = d2 
FROM @T 
GROUP BY d2 
ORDER BY StartDate, EndDate 

/***** OUTPUT *****/ 
/***** 
StartDate EndDate 
2010-01-01 2010-06-13 
2010-06-15 2010-08-16 
2010-11-01 2010-12-31 
*****/ 

Answer 1

在此解决方案中，我创建了一个临时日历表，该表存储一个范围内每天的值。这种类型的表可以是静态的。另外，从2009-12-31开始，我只能存储400个奇怪的日期。很明显，如果你的日期跨度较大，你需要更多的价值。

此外，该解决方案将只使用SQL Server 2005+在我使用的是CTE。

With Calendar As 
    (
    Select DateAdd(d, ROW_NUMBER() OVER (ORDER BY s1.object_id), '1900-01-01') As [Date] 
    From sys.columns as s1 
     Cross Join sys.columns as s2 
    ) 
    , StopDates As 
    (
    Select C.[Date] 
    From Calendar As C 
     Left Join @T As T 
      On C.[Date] Between T.d1 And T.d2 
    Where C.[Date] >= (Select Min(T2.d1) From @T As T2) 
     And C.[Date] <= (Select Max(T2.d2) From @T As T2) 
     And T.d1 Is Null 
    ) 
    , StopDatesInUse As 
    (
    Select D1.[Date] 
    From StopDates As D1 
     Left Join StopDates As D2 
      On D1.[Date] = DateAdd(d,1,D2.Date) 
    Where D2.[Date] Is Null 
    ) 
    , DataWithEariestStopDate As 
    (
    Select * 
    , (Select Min(SD2.[Date]) 
     From StopDatesInUse As SD2 
     Where T.d2 < SD2.[Date]) As StopDate 
    From @T As T 
    ) 
Select Min(d1), Max(d2) 
From DataWithEariestStopDate 
Group By StopDate 
Order By Min(d1) 

编辑在2009年使用日期的问题无关，与最终的查询。问题是日历表不够大。我在2009-12-31开始了日程表。我已经在1900-01-01开始修改它。

Answer 2

试试这个

;WITH T1 AS 
(
    SELECT d1, d2, ROW_NUMBER() OVER(ORDER BY (SELECT 0)) AS R 
    FROM @T 
), NUMS AS 
(
    SELECT ROW_NUMBER() OVER(ORDER BY (SELECT 0)) AS R 
    FROM T1 A 
    CROSS JOIN T1 B 
    CROSS JOIN T1 C 
), ONERANGE AS 
(
    SELECT DISTINCT DATEADD(DAY, ROW_NUMBER() OVER(PARTITION BY T1.R ORDER BY (SELECT 0)) - 1, T1.D1) AS ELEMENT 
    FROM T1 
    CROSS JOIN NUMS 
    WHERE NUMS.R <= DATEDIFF(DAY, d1, d2) + 1 
), SEQUENCE AS 
(
    SELECT ELEMENT, DATEDIFF(DAY, '19000101', ELEMENT) - ROW_NUMBER() OVER(ORDER BY ELEMENT) AS rownum 
    FROM ONERANGE 
) 
SELECT MIN(ELEMENT) AS StartDate, MAX(ELEMENT) as EndDate 
FROM SEQUENCE 
GROUP BY rownum 

的基本思路是，首先展开的现有数据，所以你得到一个单独的行的每一天。这在ONERANGE

完成之后，确定之间的关系如何日期增量和行号做的方式。 现有范围/岛内的差异保持不变。只要你到一个新的数据孤岛，它们之间的差异增加，因为超过1日期增量，同时通过1

Answer 3

行数递增我一直在寻找同样的解决办法和整个这个职位上来到Combine overlapping datetime to return single overlapping range record。

上有Packing Date Intervals另一个线程。

我与不同日期范围，包括这里列出的进行了测试，它每次都正常工作。

---

SELECT 
     s1.StartDate, 
     --t1.EndDate 
     MIN(t1.EndDate) AS EndDate 
FROM @T s1 
INNER JOIN @T t1 ON s1.StartDate <= t1.EndDate 
    AND NOT EXISTS(SELECT * FROM @T t2 
       WHERE t1.EndDate >= t2.StartDate AND t1.EndDate < t2.EndDate) 
WHERE NOT EXISTS(SELECT * FROM @T s2 
       WHERE s1.StartDate > s2.StartDate AND s1.StartDate <= s2.EndDate) 
GROUP BY s1.StartDate 
ORDER BY s1.StartDate 

---

结果是：

StartDate | EndDate 
2010-01-01 | 2010-06-13 
2010-06-15 | 2010-06-25 
2010-06-26 | 2010-08-16 
2010-11-01 | 2010-12-31