SQL Server：日時単位のグループ

Question

SQL Server 2016 Expressを実行しています。

私は、サンプルデータを使用してこのテーブルを持っている：結果の私は、このようなものを持ちたい

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 ... 

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 ... 

我々は5またはX分のギャップを持っているときに、新しいグループが私が作成しなければなりません現在の行の日時値と次の行の日時の値の間の値。

私はこの種のクエリを持っていますが、5分以上のギャップがある場合には、行をグループ化する場所を特定できません。

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 

敬具、

私が生成したテストのためにまず

Answer 1

いくつかのランダムな日付：

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 

は今、それぞれの行と次の差分を計算し、当社グループは開始すべき場所を作業することができます：

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 

休憩の数を合計して各行にグループを割り当てます。

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 

最後に、あなたのグループ化された結果を選択：私はあなたがサブクエリを使用することができ、理解を容易にするためにテーブル変数を使用してきたが

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

を。

Answer 2

@apcのおかげです。参加の代わりに、ラグ機能を追加するためのコードを改訂し、行を削除した。ここで、[ブレイク]の<> 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]