正确计算 2 个日期范围之间的时间

Question

我正在使用 Datetime 列，由于时间重叠，我无法计算时间。

这是我的示例数据

    Declare @T table (ID Int, InTime Datetime,OutTime Datetime)
    Insert into @T values (1,'2020-08-23 09:26:07.000','2020-08-23 09:57:55.000')
    Insert into @T values (1,'2020-08-23 14:09:08.000','2020-08-26 08:13:45.000')
    Insert into @T values (1,'2020-08-24 11:14:37.000','2020-08-24 18:07:25.000')
    Insert into @T values (1,'2020-08-25 09:19:38.000','2020-08-25 19:19:29.000')
    Insert into @T values (1,'2020-08-26 08:13:50.000','2020-08-28 08:39:23.000')
    Insert into @T values (1,'2020-08-27 08:42:16.000','2020-08-27 11:38:06.000')
    Insert into @T values (1,'2020-08-27 09:51:14.000','2020-08-28 18:23:06.000')
    Insert into @T values (1,'2020-08-29 09:51:14.000','2020-08-30 18:23:06.000')

我的预期输出：

+----+-------------------------+--------------------------+
| ID |         InTime          |         OutTime          |
+----+-------------------------+--------------------------+
|  1 | 2020-08-23 09:26:07.000 | 2020-08-23 09:57:55.000  |
|  1 | 2020-08-23 14:09:08.000 | 2020-08-26 08:13:45.000  |
|  1 | 2020-08-26 08:13:50.000 | 2020-08-28 08:39:23.000  |
|  1 | 2020-08-29 09:51:14.000 | 2020-08-30 18:23:06.000  |
+----+-------------------------+--------------------------+

如果您在表格中看到第二个条目，其中我的 Intime 是 2020-08-23 14:09:08.000，outTime 是 2020-08-26 08:13:45.000。因此，如果表中有 23 到 26 之间的任何条目，我们应该跳过该条目。所以在输出表中，我们需要跳过第 24 和第 25 条目。

有人可以帮我解决这个问题吗？任何帮助将不胜感激

我尝试了这个链接但无法理解逻辑 Link

Answer 1

这是“聚合区间”的经典案例。 Snodgrass 给出了一个经典的查询来做到这一点：

WITH  
T0 AS  
(SELECT PRE.id,
        PRE.intime AS D1, PRE.outtime AS F1,  
        DER.intime AS D2, DER.outtime AS F2  
 FROM   @T PRE  
        INNER JOIN @T DER  
              ON PRE.intime <= DER.outtime
                 AND PRE.id = DER.id)  
SELECT DISTINCT id, D1 AS intime, F2 AS outtime
FROM   T0 AS I  
WHERE  NOT EXISTS (SELECT *  
                   FROM   @T SI1  
                   WHERE  (SI1.intime < I.D1  
                           AND I.D1 <= SI1.outtime
                           AND I.id = SI1.id )  
                      OR  (SI1.intime <= I.F2  
                           AND I.F2 < SI1.outtime
                           AND I.id = SI1.id))  
AND NOT EXISTS (SELECT *  
                FROM   @T SI2
                WHERE  D1 < SI2.intime  
                  AND  SI2.intime <= F2
                  AND  I.id = SI2.id  
                  AND  NOT EXISTS (SELECT *  
                                   FROM   @T SI3  
                                   WHERE  SI3.intime < SI2.intime  
                                     AND  SI2.intime <= SI3.outtime
                                     AND  SI2.id = SI3.id ));

Chris Date 给出另一个版本：

WITH T  
AS (SELECT F.intime, L.outtime, F.id
    FROM   @T AS F  
           JOIN @T AS L  
                ON F.outtime <= L.outtime
                   AND F.id = L.id
           INNER JOIN @T AS E      
                 ON F.id = E.id  
    GROUP  BY F.intime, L.outtime,  F.id  
    HAVING COUNT(CASE  
                    WHEN (E.intime < F.intime AND F.intime <= E.outtime)  
                          OR (E.intime <= L.outtime AND L.outtime < E.outtime)
                    THEN 1  
                 END) = 0)  
SELECT id, intime, MIN(outtime) AS outtime  
FROM   T  
GROUP  BY id, intime;

最后是我写的：

WITH  
T0 AS -- suprime les périodes incluses
(SELECT DISTINCT Tout.id, Tout.intime, Tout.outtime
 FROM   @T  AS Tout  
 WHERE  NOT EXISTS(SELECT *  
                   FROM   @T  AS Tin  
                   WHERE  Tout.intime >= Tin.intime  
                     AND  Tout.outtime < Tin.outtime
                     AND Tout.id = Tin.id)),  
T1 AS -- ancres : périodes de tête...  
(SELECT Ta.*, 1 AS ITERATION  
 FROM   T0 AS Ta  
 WHERE  NOT EXISTS(SELECT *  
                   FROM   T0 AS Tb  
                   WHERE  Tb.outtime >= Ta.intime  
                      AND Tb.outtime  < Ta.outtime
                      AND Tb.id = Ta.id)  
 UNION  ALL -- itération sur période dont le debut est inclus dans les bornes de la période ancre  
 SELECT T1.id, T1.intime, T0.outtime, T1.ITERATION + 1
 FROM   T1  
        INNER JOIN T0  
              ON T1.intime < T0.intime  
                 AND T1.outtime >= T0.intime  
                 AND T1.outtime < T0.outtime
                 AND T1.id = T0.id),  
T2 AS  
(SELECT *, ROW_NUMBER() OVER(PARTITION BY id, intime ORDER BY DATEDIFF(s, intime, outtime) DESC) AS N1,
           ROW_NUMBER() OVER(PARTITION BY id, outtime ORDER BY DATEDIFF(s, intime, outtime) DESC) AS N2
 FROM   T1)  
SELECT id, intime, outtime
FROM   T2  
WHERE  N1 = 1 AND N2 = 1;

递归查询

Itzik Ben Gan 完成了一些更复杂且性能更高的查询...