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请有人可以在这里帮助我，我正在尝试获取日期范围之间的小时表。

例如：

if starttime == 2012-02-06 23:59:00'
then endtime = '2012-02-29 10:26:17'

我应该有一个表（变量表），内容如下：

Start                    End                      Seconds
2012-02-06 23:59:00.000  2012-02-06 23:59:00.000  0
2012-02-29 09:00:00.000  2012-02-29 09:59:00.000  3540
2012-02-29 10:00:00.000  2012-02-29 10:26:17.000  1577
4

4 回答 4

1

second通过查看您的预期表，您在两次之间得到了差异。

SELECT DATEDIFF(SECOND, '2012-02-29 09:00:00.000', '2012-02-29 09:59:00.000')只给出 3540 秒的差异。

要获得小时差，您可以尝试以下代码：

SELECT  DATEDIFF(hour, '2012-02-06 23:59:00.000', '2012-02-06 23:59:00.000')
于 2012-07-13T10:21:59.353 回答
0

感谢大家的建议和意见。我终于得到了一个解决可能问题的方法。

以下是我提出的解决方案的脚本：

DECLARE @start_date datetime = CONVERT(DATETIME,'2012-02-06 23:59:01.000',20);
DECLARE @end_date datetime = CONVERT(DATETIME,'2012-12-08 23:59:17.000',20);
DECLARE @org datetime  ;
DECLARE @end datetime  ;
DECLARE @datetable TABLE (h_start datetime, h_end datetime,h_sesc int);

WHILE (dateadd(second, -1, dateadd(hour, datediff(hour, 0, @start_date)+1, 0))) < @end_date
BEGIN
SET @org = null;
SET @org = @start_date;
SET @end = (dateadd(second, -1, dateadd(hour, datediff(hour, 0, @org)+1, 0)));
INSERT INTO @datetable (h_start, h_end,h_sesc)
VALUES(dateadd(second, 0,@org), @end,DATEDIFF(second, @org,@end));

SET @start_date = dateadd(second, 1,@end);

END;


INSERT INTO @datetable (h_start, h_end,h_sesc)
VALUES(dateadd(second, 0,@start_date), @end_date,DATEDIFF(second, dateadd(second, 0,@start_date),@end_date));

SELECT * FROM @datetable;

以上将给出以下结果：

h_start                 h_end                   h_sesc
2012-02-06 23:59:01.000 2012-02-06 23:59:59.000 58
2012-02-07 00:00:00.000 2012-02-07 00:59:59.000 3599
2012-02-07 01:00:00.000 2012-02-07 01:59:59.000 3599
2012-02-07 02:00:00.000 2012-02-07 02:59:59.000 3599
2012-02-07 03:00:00.000 2012-02-07 03:59:59.000 3599
2012-02-07 04:00:00.000 2012-02-07 04:59:59.000 3599
2012-02-07 05:00:00.000 2012-02-07 05:59:59.000 3599

......

2012-12-08 18:00:00.000 2012-12-08 18:59:59.000 3599
2012-12-08 19:00:00.000 2012-12-08 19:59:59.000 3599
2012-12-08 20:00:00.000 2012-12-08 20:59:59.000 3599
2012-12-08 21:00:00.000 2012-12-08 21:59:59.000 3599
2012-12-08 22:00:00.000 2012-12-08 22:59:59.000 3599
2012-12-08 23:00:00.000 2012-12-08 23:59:17.000 3557

希望有人会发现它有用。

于 2012-07-17T11:04:09.640 回答
0 
DECLARE @start_date datetime = CONVERT(DATETIME,'2012-02-06 23:59:01.000', 20);
DECLARE @end_date datetime = CONVERT(DATETIME,'2012-12-08 23:59:17.000', 20);

-- Using a recursive query after round down to full hours avoids a loop. The resulting intervals should be easier to use that way, e.g. you can but don't have to insert them into a table variable.
SET @start_date = DATEADD(HOUR, DATEPART(HOUR, @start_date), CAST(FLOOR(CAST(@start_date AS float)) AS datetime));
SET @end_date = DATEADD(HOUR, DATEPART(HOUR, @end_date), CAST(FLOOR(CAST(@end_date AS float)) AS datetime));
WITH RecursiveTimeIntervals AS
(SELECT @start_date Interval
UNION ALL SELECT DATEADD(HOUR, 1, r.Interval)
FROM RecursiveTimeIntervals r
WHERE r.Interval < @end_date)
SELECT * 
FROM RecursiveTimeIntervals
ORDER BY 1
OPTION (MAXRECURSION 0)
于 2017-06-12T09:46:25.010 回答
-1 
DATEDIFF(hour, [start], [end])

请参阅w3schools 上的说明。

于 2012-07-13T10:23:16.663 回答