Рассчитать разницу во времени в минутах в SQL Server

Мне нужна разница во времени между двумя временами в минутах. У меня есть время начала и время окончания, как показано ниже:

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

Мне нужен выход для первой строки как 45,60,15, что соответствует разнице во времени между 11:15 и 12:00, 12:00 и 13:00, 13:00 и 13:15 соответственно.

13
datetime difference sql sql-server sql-server-2008

4 ответов

следующие работы:

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);

чтобы получить 24 столбца, вы можете использовать 24 выражения case, что-то вроде:

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);

следующее также работает и может оказаться короче, чем повторение одного и того же выражения case снова и снова:

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;

он работает, присоединяясь к таблице из 24 чисел, поэтому выражение case не нужно повторять, а затем свертывать эти 24 числа обратно в столбцы с помощью PIVOT

12

использовать функция datediff С минутной разницей:

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

запрос, который может помочь вам: 

SELECT StartTime, EndTime, DATEDIFF(MINUTE, StartTime , EndTime) AS MinuteDiff 
FROM TableName
27

вы можете использовать DATEDIFF (это встроенная функция) и % (для вычисления масштаба) и CONCAT для получения результата только для одного столбца

select CONCAT('Month: ',MonthDiff,' Days: ' , DayDiff,' Minutes: ',MinuteDiff,' Seconds: ',SecondDiff) as T  from 
(SELECT DATEDIFF(MONTH, '2017-10-15 19:39:47' , '2017-12-31 23:59:59') % 12 as MonthDiff,
        DATEDIFF(DAY, '2017-10-15 19:39:47' , '2017-12-31 23:59:59') % 30 as DayDiff,
        DATEDIFF(HOUR, '2017-10-15 19:39:47' , '2017-12-31 23:59:59') % 24 as HourDiff,
        DATEDIFF(MINUTE, '2017-10-15 19:39:47' , '2017-12-31 23:59:59') % 60 AS MinuteDiff,
        DATEDIFF(SECOND, '2017-10-15 19:39:47' , '2017-12-31 23:59:59') % 60 AS SecondDiff) tbl
0

попробуйте это.. 

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

все, что вам нужно-это функция datediff..

-1