计算两个连续行之间的日期差

Question

我有一个表，其中包含如下所示的日期时间行。

ID | DateTime

1  | 12:00
2  | 12:02
3  | 12:03
4  | 12:04
5  | 12:05
6  | 12:10

我想确定那些行之间有5分钟的“间隔”（例如，第5和6行）的那些行。

我知道我们需要使用DATEDIFF ，但是我如何只获取那些彼此连续的行呢？

Answer 1

您可以为此使用LAG ， LEAD窗口功能：

SELECT ID
FROM (
  SELECT ID, [DateTime], 
         DATEDIFF(mi, LAG([DateTime]) OVER (ORDER BY ID), [DateTime]) AS prev_diff,
         DATEDIFF(mi, [DateTime], LEAD([DateTime]) OVER (ORDER BY ID)) AS next_diff
  FROM mytable) AS t
WHERE prev_diff >= 5 OR next_diff >= 5

输出：

ID
==
5
6

注意：上面的查询假定顺序是由ID字段定义的。 您可以轻松地用其他指定表中顺序的其他字段替换该字段。

Answer 2

更新SS2012：使用LAG

DECLARE @tbl TABLE(ID INT, T TIME)
INSERT INTO @tbl VALUES
 (1,'12:00')
,(2,'12:02')
,(3,'12:03')
,(4,'12:04')
,(5,'12:05')
,(6,'12:10');

WITH TimesWithDifferenceToPrevious AS
(
    SELECT ID
          ,T
          ,LAG(T) OVER(ORDER BY T) AS prev
          ,DATEDIFF(MI,LAG(T) OVER(ORDER BY T),T) AS MinuteDiff
    FROM @tbl
)
SELECT *
FROM TimesWithDifferenceToPrevious
WHERE ABS(MinuteDiff) >=5

结果

6   12:10:00.0000000    12:05:00.0000000    5

Answer 3

您可以尝试一下（我不确定它是否真的很快）

SELECT current.datetime  AS current_datetime,
       previous.datetime AS previous_datetime,
       DATEDIFF(minute, previous.datetime, current.datetime) AS gap
FROM   my_table current
JOIN   my_table previous
  ON   previous.datetime < current.datetime
 AND   NOT EXISTS (SELECT *
                   FROM   my_table others
                   WHERE  others.datetime < current.datetime
                     AND  others.datetime > previous.datetime);