Question

我需要两分钟之间的时差。我有如下所示的开始时间和结束时间：

start time | End Time    
11:15:00   | 13:15:00    
10:45:00   | 18:59:00

我需要第一行的输出为45,60,15，分别对应于11:15和12:00,12：00和13：00,13：00和13:15之间的时差。

Answer 1

将DateDiff与MINUTE差异一起使用：

SELECT DATEDIFF(MINUTE, '11:10:10' , '11:20:00') AS MinuteDiff

可能对您有所帮助的查询：

SELECT StartTime, EndTime, DATEDIFF(MINUTE, StartTime , EndTime) AS MinuteDiff 
FROM TableName

Answer 2

以下按预期方式工作：

SELECT  Diff = CASE DATEDIFF(HOUR, StartTime, EndTime)
                    WHEN 0 THEN CAST(DATEDIFF(MINUTE, StartTime, EndTime) AS VARCHAR(10))
                    ELSE CAST(60 - DATEPART(MINUTE, StartTime) AS VARCHAR(10)) +
                        REPLICATE(',60', DATEDIFF(HOUR, StartTime, EndTime) - 1) + 
                        + ',' + CAST(DATEPART(MINUTE, EndTime) AS VARCHAR(10))
                END
FROM    (VALUES 
            (CAST('11:15' AS TIME), CAST('13:15' AS TIME)),
            (CAST('10:45' AS TIME), CAST('18:59' AS TIME)),
            (CAST('10:45' AS TIME), CAST('11:59' AS TIME))
        ) t (StartTime, EndTime);

要获得24列，您可以使用24个案例表达式，例如：

SELECT  [0] = CASE WHEN DATEDIFF(HOUR, StartTime, EndTime) = 0
                        THEN DATEDIFF(MINUTE, StartTime, EndTime)
                    ELSE 60 - DATEPART(MINUTE, StartTime)
                END,
        [1] = CASE WHEN DATEDIFF(HOUR, StartTime, EndTime) = 1 
                        THEN DATEPART(MINUTE, EndTime)
                    WHEN DATEDIFF(HOUR, StartTime, EndTime) > 1 THEN 60
                END,
        [2] = CASE WHEN DATEDIFF(HOUR, StartTime, EndTime) = 2
                        THEN DATEPART(MINUTE, EndTime)
                    WHEN DATEDIFF(HOUR, StartTime, EndTime) > 2 THEN 60
                END -- ETC
FROM    (VALUES 
            (CAST('11:15' AS TIME), CAST('13:15' AS TIME)),
            (CAST('10:45' AS TIME), CAST('18:59' AS TIME)),
            (CAST('10:45' AS TIME), CAST('11:59' AS TIME))
        ) t (StartTime, EndTime);

以下内容也有效，并且可能会比一遍又一遍地重复相同的case表达式更短：

WITH Numbers (Number) AS
(   SELECT  ROW_NUMBER() OVER(ORDER BY t1.N) - 1
    FROM    (VALUES (1), (1), (1), (1), (1), (1)) AS t1 (N)
            CROSS JOIN (VALUES (1), (1), (1), (1)) AS t2 (N)
), YourData AS
(   SELECT  StartTime, EndTime
    FROM    (VALUES 
                (CAST('11:15' AS TIME), CAST('13:15' AS TIME)),
                (CAST('09:45' AS TIME), CAST('18:59' AS TIME)),
                (CAST('10:45' AS TIME), CAST('11:59' AS TIME))
            ) AS t (StartTime, EndTime)
), PivotData AS
(   SELECT  t.StartTime,
            t.EndTime,
            n.Number,
            MinuteDiff = CASE WHEN n.Number = 0 AND DATEDIFF(HOUR, StartTime, EndTime) = 0 THEN DATEDIFF(MINUTE, StartTime, EndTime)
                                WHEN n.Number = 0 THEN 60 - DATEPART(MINUTE, StartTime)
                                WHEN DATEDIFF(HOUR, t.StartTime, t.EndTime) <= n.Number THEN DATEPART(MINUTE, EndTime)
                                ELSE 60
                            END
    FROM    YourData AS t
            INNER JOIN Numbers AS n
                ON n.Number <= DATEDIFF(HOUR, StartTime, EndTime)
)
SELECT  *
FROM    PivotData AS d
        PIVOT 
        (   MAX(MinuteDiff)
            FOR Number IN 
            (   [0], [1], [2], [3], [4], [5], 
                [6], [7], [8], [9], [10], [11],
                [12], [13], [14], [15], [16], [17], 
                [18], [19], [20], [21], [22], [23]
            ) 
        ) AS pvt;

它通过连接到一个包含24个数字的表来工作，因此不需要重复case表达式，然后使用PIVOT

将这24个数字回滚到列中

Answer 3

除了DATEDIFF，您还可以使用TIMEDIFF函数或TIMESTAMPDIFF。

示例

SET @date1 = '2010-10-11 12:15:35', @date2 = '2010-10-10 00:00:00';

SELECT 
TIMEDIFF(@date1, @date2) AS 'TIMEDIFF',
TIMESTAMPDIFF(hour, @date1, @date2) AS 'Hours',
TIMESTAMPDIFF(minute, @date1, @date2) AS 'Minutes',
TIMESTAMPDIFF(second, @date1, @date2) AS 'Seconds';

结果

TIMEDIFF : 36:15:35
Hours : -36
Minutes : -2175
Seconds : -130535

Answer 4

您可以使用DATEDIFF（它是一个内置函数）和％（用于比例计算）和CONCAT来将结果仅保留到一列

\return

Answer 5

请尝试以下操作以获取hh：mm：ss格式的时差

从[TableName]中选择StartTime，EndTime，CAST（（EndTime-StartTime）作为时间（0））“ TotalTime”

Answer 6

试试这个..

select starttime,endtime, case
  when DATEDIFF(minute,starttime,endtime) < 60  then DATEDIFF(minute,starttime,endtime) 
  when DATEDIFF(minute,starttime,endtime) >= 60
  then '60,'+ cast( (cast(DATEDIFF(minute,starttime,endtime) as int )-60) as nvarchar(50) )
end from TestTable123416

您需要的只是DateDiff ..

计算SQL Server中的时间差（以分钟为单位）

6 个答案: