Sql server 如何在SQL中计算时间

我有一张像这样的桌子

ID              DTTM
123456789   2017-10-05 08:00:00.000
123456789   2017-10-05 08:05:00.000
123456789   2017-10-05 08:07:00.000
123456789   2017-10-05 08:15:00.000
123456789   2017-10-05 08:25:00.000
123456789   2017-10-05 09:00:00.000
123456789   2017-10-05 09:01:00.000
123456789   2017-10-05 09:02:00.000
123456789   2017-10-05 09:03:00.000
123456789   2017-10-05 11:00:00.000

我需要根据时间间隔创建一个标志（日期无关紧要） 必须每隔15分钟将标志设置为1 因此，在本例中，第1、4、6、10行将是标志，该ID的总数将为4 每次如果

间隔>=15分钟

将再次启动，直到下一次 我试过这样的东西； 以我的头为例

(
Select        top 100 percent 
            ID,
            DTTM,                
            LEAD(DTTM,1) over (partition by ID order by DTTM) as NextDTTM
From        Example
Order by    ID
), myCount
AS
(
Select      Top 100 percent
            ID,
            DTTM,
            NextDTTM,
            DateDiff("MINUTE",DTTM,NextDTTM) as Interval
from        myLead
)
Select      ID,
            DTTM,
            NextDTTM,
            Interval,
            Case When Interval >= 15 then 1 else 0 END as CountFlag
From        myCount
Where       Interval is not NULL
Order by    ID

但它并没有像预期的那样工作？我该如何解决这个问题

谢谢，

---

Oded Dror

如果您使用MySQL，那么您可以尝试使用

TIMEDIFF（param1，param2）

---

*param1和param2可以是times或datetimes值

如果使用MySQL，则可以尝试使用

TIMEDIFF（param1，param2）

---

*param1和param2可以是times或datetimes值

我不确定您需要标志的外观，但使用SQL Server 2012它们是两个棘手的部分。首先，要获取下一行和上一行之间的时间差：

DECLARE @DataSource TABLE
(
    [ID] BIGINT
   ,[DTTM] DATETIME2
);

INSERT INTO @DataSource ([ID], [DTTM])
VALUES   (123456789, '2017-10-05 08:00:00.000')
        ,(123456789, '2017-10-05 08:05:00.000')
        ,(123456789, '2017-10-05 08:07:00.000')
        ,(123456789, '2017-10-05 08:15:00.000')
        ,(123456789, '2017-10-05 08:25:00.000')
        ,(123456789, '2017-10-05 09:00:00.000')
        ,(123456789, '2017-10-05 09:01:00.000')
        ,(123456789, '2017-10-05 09:02:00.000')
        ,(123456789, '2017-10-05 09:03:00.000')
        ,(123456789, '2017-10-05 11:00:00.000');

SELECT *  
      ,DATEDIFF(MINUTE,CONVERT(TIME, [DTTM]),LEAD(CONVERT(TIME, [DTTM]), 1, NULL) OVER (PARTITION BY [ID] ORDER BY [DTTM])) AS [Minutes]
FROM @DataSource
ORDER BY [DTTM];

然后，我们需要

求和

从超间隔乞讨到当前行的分钟数，并将它们除以

15

：

WITH DataSource AS
(
    SELECT *  
          ,DATEDIFF(MINUTE,LAG(CONVERT(TIME, [DTTM]), 1, NULL) OVER (PARTITION BY [ID] ORDER BY [DTTM]), CONVERT(TIME, [DTTM])) AS [Minutes]
    FROM @DataSource
)
SELECT *
      ,SUM(ISNULL([Minutes], 0)) OVER (PARTITION BY [ID] ORDER BY [DTTM] ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) / 15 AS [MinutesRank]
FROM DataSource;

有了这个，你可以做你需要做的任何事情，就像我们现在知道每一排还有多少分钟一样。例如：

WITH DataSource AS
(
    SELECT *  
          ,DATEDIFF(MINUTE,LAG(CONVERT(TIME, [DTTM]), 1, NULL) OVER (PARTITION BY [ID] ORDER BY [DTTM]), CONVERT(TIME, [DTTM])) AS [Minutes]
    FROM @DataSource
)
, DataSourceMinutesTotal AS
(
    SELECT *
          ,SUM(ISNULL([Minutes], 0)) OVER (PARTITION BY [ID] ORDER BY [DTTM] ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) / 15 AS [MinutesRank]
    FROM DataSource
)
SELECT [ID], [DTTM]
      ,DENSE_RANK() OVER (PARTITION BY [ID] ORDER BY ISNULL([MinutesRank], 0)) AS [IntervalID]
FROM DataSourceMinutesTotal
ORDER BY [DTTM];

---

我不确定您需要标志的外观，但是使用

SQLServer2012

它们是两个棘手的部分。首先，要获取下一行和上一行之间的时间差：

DECLARE @DataSource TABLE
(
    [ID] BIGINT
   ,[DTTM] DATETIME2
);

INSERT INTO @DataSource ([ID], [DTTM])
VALUES   (123456789, '2017-10-05 08:00:00.000')
        ,(123456789, '2017-10-05 08:05:00.000')
        ,(123456789, '2017-10-05 08:07:00.000')
        ,(123456789, '2017-10-05 08:15:00.000')
        ,(123456789, '2017-10-05 08:25:00.000')
        ,(123456789, '2017-10-05 09:00:00.000')
        ,(123456789, '2017-10-05 09:01:00.000')
        ,(123456789, '2017-10-05 09:02:00.000')
        ,(123456789, '2017-10-05 09:03:00.000')
        ,(123456789, '2017-10-05 11:00:00.000');

SELECT *  
      ,DATEDIFF(MINUTE,CONVERT(TIME, [DTTM]),LEAD(CONVERT(TIME, [DTTM]), 1, NULL) OVER (PARTITION BY [ID] ORDER BY [DTTM])) AS [Minutes]
FROM @DataSource
ORDER BY [DTTM];

然后，我们需要

求和

从超间隔乞讨到当前行的分钟数，并将它们除以

15

：

WITH DataSource AS
(
    SELECT *  
          ,DATEDIFF(MINUTE,LAG(CONVERT(TIME, [DTTM]), 1, NULL) OVER (PARTITION BY [ID] ORDER BY [DTTM]), CONVERT(TIME, [DTTM])) AS [Minutes]
    FROM @DataSource
)
SELECT *
      ,SUM(ISNULL([Minutes], 0)) OVER (PARTITION BY [ID] ORDER BY [DTTM] ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) / 15 AS [MinutesRank]
FROM DataSource;

有了这个，你可以做你需要做的任何事情，就像我们现在知道每一排还有多少分钟一样。例如：

WITH DataSource AS
(
    SELECT *  
          ,DATEDIFF(MINUTE,LAG(CONVERT(TIME, [DTTM]), 1, NULL) OVER (PARTITION BY [ID] ORDER BY [DTTM]), CONVERT(TIME, [DTTM])) AS [Minutes]
    FROM @DataSource
)
, DataSourceMinutesTotal AS
(
    SELECT *
          ,SUM(ISNULL([Minutes], 0)) OVER (PARTITION BY [ID] ORDER BY [DTTM] ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) / 15 AS [MinutesRank]
    FROM DataSource
)
SELECT [ID], [DTTM]
      ,DENSE_RANK() OVER (PARTITION BY [ID] ORDER BY ISNULL([MinutesRank], 0)) AS [IntervalID]
FROM DataSourceMinutesTotal
ORDER BY [DTTM];

---

谢谢您联系我，问题解决了。你的方法没有解决这个问题。谢谢你联系我，问题解决了。你的方法并没有解决问题。Gotqn，谢谢你的帮助，问题已经解决了。这个解决方案不会给出我想要的结果。Gotqn，谢谢你的帮助，问题已经解决了。这个解决方案不会给出我想要的结果。