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日期在范围mssql之间添加分钟数

minute dateadd date-range sql-server

Kiel · 技术社区 · 8 年前

我想弄清楚一些事情有点困难,比如说我有约会和时间;

我想增加180分钟;

SELECT DATEADD(MINUTE,180,'2018-05-24 15:00')

这会给我一个“2018-05-24 18:00”的答案,但我想在一定范围内完成,所以如果你在09:00-17:00之间,就加上几分钟,就像这样;

SELECT DATEADD(MINUTES,180,'2018-05-24 15:00') WHERE '2018-05-24 15:00' BETWEEN '2018-05-24 09:00' AND '2018-05-24 17:00'

因此,答案是“2018-05-25 10:00”

3 回复 | 直到 8 年前

EzLo tumao kaixin 8 年前

很难,但这对你所有的案件都适用。此解决方案的工作时间为(正)分钟,结果始终在参数化小时内,并添加相应的天数。

DECLARE @RangeHourStart INT = 9
DECLARE @RangeHourEnd INT = 17

DECLARE @MinutesToAdd INT = 120

DECLARE @Date DATETIME = '2018-05-24 15:00'

SELECT
    FinalDate = CASE 
        WHEN -- When final hour exceeds the range hour
            DATEPART(HOUR, @Date) * 60 + 
            DATEPART(MINUTE, @Date) +
            @MinutesToAdd % ((@RangeHourEnd - @RangeHourStart) * 60) > @RangeHourEnd * 60 

        THEN
            DATEADD(HOUR, -1 * (@RangeHourStart - 1), 
                DATEADD(DAY, 1, 
                    DATEADD(MINUTE, @MinutesToAdd % ((@RangeHourEnd - @RangeHourStart) * 60), 
                        DATEADD(
                            DAY,
                            @MinutesToAdd / ((@RangeHourEnd - @RangeHourStart) * 60),
                            @Date)))) 

        ELSE
            DATEADD(MINUTE, @MinutesToAdd % ((@RangeHourEnd - @RangeHourStart) * 60), 
                DATEADD(
                    DAY,
                    @MinutesToAdd / ((@RangeHourEnd - @RangeHourStart) * 60),
                    @Date))
        END

我做到了,所以你不需要硬编码任何值。

Larnu 8 年前

不过,这看起来并不特别漂亮……

USE Sandbox;
GO

CREATE TABLE Times (DateNTime datetime2(0));
INSERT INTO Times
VALUES ('20180520 10:00:00'),
       ('20180520 15:20:00'),
       ('20180521 09:32:00'),
       ('20180521 14:17:00'),
       ('20180522 16:54:00'),
       ('20180523 12:46:00'),
       ('20180524 15:32:00');
GO


SELECT *
FROM Times;
GO

SELECT T.DateNTime,
       CASE WHEN CONVERT(time,T.DateNTime) <= '14:00' THEN DATEADD(MINUTE, 180,T.DateNTime)
            ELSE DATEADD(MINUTE, 180 - DATEDIFF(MINUTE,T.DateNTime,DATEADD(HOUR,17,DATEADD(DAY, DATEDIFF(DAY, 0, T.DateNTime),0))), DATEADD(HOUR,9,DATEADD(DAY, DATEDIFF(DAY, 0, T.DateNTime) + 1,0))) END
FROM Times T;

GO

DROP TABLE Times;

Larnu 8 年前

您可以尝试以下操作:

DECLARE @input DATETIME='2018-05-24 15:00'
    DECLARE @min INT=180

    SELECT CASE WHEN DATEADD(MINUTE,@min,@input)>DATEADD(HOUR, 17,DateAdd(Day, Datediff(Day,0, @input), 0)) 
        THEN  DATEADD(MINUTE,
                        DATEDIFF(MINUTE,
                                DATEADD(HOUR, 17,
                                    DATEADD(Day, 
                                        DATEDIFF(Day,0, @input),
                                    0)
                                ),
                                DATEADD(MINUTE,@min,@input)), 
                                    DATEADD(Hour,9,
                                        DATEADD(Day,1,
                                            DateAdd(Day, 
                                                Datediff(Day,0, @input),
                                                0)
                                        )
                                    )
                                )

        ELSE DATEADD(MINUTE,@min,@input)
        END