SQL Server：按日期时间间隔分组

Question

运行SQL Server 2016 Express。

我有这个表的样本数据：

ID Datetime2    other columns 
------------------------------------------ 
1  2017-01-14 11:00:00 ... 
1  2017-01-14 11:01:00 ... 
1  2017-01-14 11:02:00 ... 
1  2017-01-14 11:03:00 ... 
1  2017-01-14 11:10:00 ... --> 7 minutes gap 
1  2017-01-14 11:11:00 ... 
1  2017-01-14 11:20:00 ... --> 9 minutes gap 
1  2017-01-14 11:22:00 ... 
1  2017-01-14 11:24:00 ... 

我想有这种结果

ID start    end     other columns 
----------------------------------------------------------- 
1 2017-01-14 11:00 2017-01-14 11:03 ... 
1 2017-01-14 11:10 2017-01-14 11:11 ... 
1 2017-01-14 11:20 2017-01-14 11:24 ... 

一定要我创建一个新的组时，我们有5或X分钟的间隔或更多当前行日期时间值和下一行日期时间值之间。

我有这种类型的查询，但我不知道在5分钟或更长的时间间隔内将行分组在哪里。

WITH groups(DateTimeField, grp) AS 
(
    SELECT DISTINCT 
     DateTimeField, 
     DATEDIFF(MINUTE, DateTimeField, lag(DateTimeField) OVER (ORDER BY DateTimeField DESC)) grp 
    FROM 
     MyTable 
    WHERE 
     ID = 1 
) 
SELECT 
    COUNT(*) AS consecutiveDates, 
    MIN(DateTimeField) AS minDate, 
    MAX(DateTimeField) AS maxDate 
FROM 
    groups 
GROUP BY 
    grp 
ORDER BY 
    1 DESC, 2 DESC 

最好的问候，

Answer 1

首先因为我已经生成测试一些随机日期：

DECLARE @DatesTables TABLE (ID INT, [DateTime2] DATETIME) 

DECLARE @ID INT 
DECLARE @Date DATETIME 
SET @Date = GETDATE() 

WHILE (SELECT COUNT(*) FROM @DatesTables) < 50 
BEGIN 
    SET @ID = (SELECT COUNT(*) FROM @DatesTables) + 1 
    SET @Date = DATEADD(MINUTE, rand() * 7 + 1,@Date) 
    INSERT INTO @DatesTables (ID, [DateTime2]) VALUES (@ID, @Date) 
END 

现在让我们来计算每一行和下一之间的差异，制定出在那里我们组应该开始：

DECLARE @DatesTables2 TABLE ([NewID] INT, [DateA] DATETIME, [DateB] DATETIME, DiffMin INT, [Break] INT) 

INSERT INTO @DatesTables2 
SELECT 
    ROW_NUMBER() OVER (ORDER BY a.ID), 
    a.DateTime2 AS DateA, 
    b.DateTime2 AS DateB, 
    DATEDIFF(MINUTE, a.[DateTime2], b.[DateTime2]) AS DiffMin, 
    CASE WHEN DATEDIFF(MINUTE, a.[DateTime2], b.[DateTime2]) > 5 THEN 1 ELSE 0 END AS [Break] 
FROM @DatesTables a JOIN @DatesTables b ON a.id = b.ID - 1 

通过累加断点数来为每行分配一组：

DECLARE @Groups TABLE (DateA DATETIME, DateB DATETIME, DiffMin INT, [Break] INT, [Group] INT) 

INSERT INTO @Groups 
SELECT a.DateA, a.DateB,a.DiffMin, a.[Break], SUM(b.[Break]) + a.[Break] AS [Group] FROM @DatesTables2 a JOIN @DatesTables2 b 
ON b.newid < a.newid 
GROUP BY a.DateA, a.DateB, a.[Break],a.DiffMin 

最后选择你的分组结果：

SELECT [Group], MIN(DateA) AS Start, MAX(DateB) AS [End] FROM @Groups GROUP BY [Group] ORDER BY [Group] 

虽然我已经使用表变量更容易理解，你可以使用子查询。

Answer 2

感谢@apc。修改后的代码添加滞后函数，而不是加入和删除的行，其中的[BREAK] <> 0

//.. 
--use of the lag function 

INSERT INTO @DatesTables2 
SELECT 
    ROW_NUMBER() OVER (ORDER BY a.DateTime2 DESC), 
    a.DateTime2 AS DateA, 
    lag(DateTime2) OVER (ORDER BY DateTime2 DESC) AS DateB, 
    DATEDIFF(MINUTE, a.DateTime2 , lag(DateTime2) OVER (ORDER BY DateTime2 DESC)) AS DiffMin, 
    CASE WHEN DATEDIFF(MINUTE, a.DateTime2 , lag(DateTime2) OVER (ORDER BY DateTime2 DESC)) > 5 THEN 1 ELSE 0 END AS [Break] 
FROM DatesTables a 

//.. 
--Query only where [Break=0] to discard the end value where the gap is more than 5 minutes. 

INSERT INTO @Groups 
SELECT a.DateA, a.DateB, a.DiffMin, a.[Break], SUM(b.[Break]) + a.[Break] AS [Group] 
FROM @DatesTables2 a JOIN @DatesTables2 b ON b.newid < a.newid 
WHERE a.[Break] = 0 
GROUP BY a.DateA, a.DateB, a.[Break], a.DiffMin 

//.. 
--Finally get the datetime diff between min and max. 

SELECT [Group], MIN(DateA) AS Start, MAX(DateB) AS [End], DATEDIFF(MINUTE, MIN(DateA), MAX(DateB)) 
FROM @Groups 
GROUP BY [Group] 
ORDER BY [Group]