[英]T-SQL DATEDIFF giving returning result 1 hours when difference is only 45 minutes
Facing an issue with T-sql Datediff function, I am calculating date difference in minutes between two dates, and then in hours. 面对T-sql Datediff函数的问题,我正在以两个日期之间的分钟为单位,然后以小时为单位,计算日期差。
Minute is giving me correct result 分钟给我正确的结果
SELECT DATEDIFF(minute,'2018-01-22 23:59:00.000','2018-01-23 00:44:00.000')
Result 45 minutes
But when I am trying to calculate hours it's giving me incorrect results for days that are almost over and new day begins, So if the time parameter is '23:59:00' and the second parameter is '00:44:00' it returns 1 hour difference when its only 45 minutes. 但是,当我尝试计算小时数时,它给我几乎结束了几天并开始新的一天的不正确结果,所以如果time参数为'23:59:00'而第二个参数为'00:44:00',当只有45分钟时,返回1小时时差。
SELECT DATEDIFF(HOUR,'2018-01-22 23:59:00.000','2018-01-23 00:44:00.000')
Result 1 Hour --Incorrect
I am expecting this result to be zero 我期望这个结果为零
SELECT DATEDIFF(HOUR,'2018-01-22 23:59:00.000','2018-01-23 00:44:00.000')
Result 0 Hour -- This is the result expected
Update: Posting my Function here if anyone needs to Calculate difference between two dates in format as Day:Hour:Minute 更新:如果有人需要以日期:小时:分钟的格式计算两个日期之间的差,请在此处发布我的函数
ALTER FUNCTION [dbo].[UDF_Fedex_CalculateDeliveryOverdue]
(
-- Add the parameters for the function here
@requiredDate VARCHAR(50),
@deliveryStamp VARCHAR(50)
)
RETURNS VARCHAR(25)
AS
BEGIN
DECLARE @ResultVar VARCHAR(25)
SET @ResultVar = ( SELECT CASE WHEN a.Days = 0 AND a.Hours = 0 THEN CAST(a.Minutes AS VARCHAR(10)) + ' Minutes'
WHEN a.Days = 0 THEN CAST(a.Hours AS VARCHAR(10)) + ' Hours ' + CAST(a.Minutes AS VARCHAR(10)) + ' Minutes'
ELSE CAST(a.Days AS VARCHAR(10)) +' Day ' + CAST(a.Hours AS VARCHAR(10)) +' Hours ' + CAST(a.Minutes AS VARCHAR(10)) + ' Minutes'
END
FROM ( SELECT DATEDIFF(hh, @requiredDate,@deliveryStamp)/24 AS 'Days'
,(DATEDIFF(MI, @requiredDate,@deliveryStamp)/60) -
(DATEDIFF(hh, @requiredDate,@deliveryStamp)/24)*24 AS 'Hours'
,DATEDIFF(mi, @requiredDate,@deliveryStamp) -
(DATEDIFF(mi, @requiredDate,@deliveryStamp)/60)*60 AS 'Minutes'
) a)
-- Return the result of the function
RETURN @ResultVar
END
To get value 0 you need to get the result in minutes, and convert to hours: 要获得值0,您需要以分钟为单位的结果,然后转换为小时:
SELECT DATEDIFF(minute,'2018-01-22 23:59:00.000','2018-01-23 00:44:00.000')/60
For more precision: 为了获得更高的精度:
SELECT DATEDIFF(minute,'2018-01-22 23:59:00.000','2018-01-23 00:44:00.000')/60.0
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