Question

我有2列数据类型为Time。

实施例

08:00:00 - 12:00:00 = 4
08:30:00 - 23:00:00 = 15.5
20:00:00 - 00:00:00 = 4
22:00:00 - 06:00:00 = 8

Answer 1

所有建议使用ABS的答案似乎都在误解这里的情况。如果结果为负，则表示时间越过了日期边界，并且简单的补救措施是在结果中添加24小时以获得正确的值。

由于你只有时间，我们必须假设任何给定的一对代表一段不到24小时的时间（否则，没有办法区分代表2小时间隔的一对时间）一对代表26小时间隔的次数。）

因此，我们可以做的是始终将24小时添加到结果中，然后将结果模数为24小时。

以分钟计算：

SELECT (DATEDIFF(minute,FirstValue,SecondValue) + (24*60)) % (24*60)

Answer 2

使用declare @t1 as time = '08:30:00'; declare @t2 as time = '23:00:00'; select abs(cast((datediff(minute, @t1, @t2)) as float) / 60);

<强>查询

14.5

<强>结果

@t1

您可以将变量@t2和CASE更改为列名。

<强>更新

您可以使用select case when t1 > t2 then cast( datediff( minute, cast( cast( getdate() as date ) as varchar(50) ) + ' ' + cast( cast( left(t1, charindex('.', t1, 1) - 1) as varchar(50) ) as datetime ), cast( cast( getdate() + 1 as date ) as varchar(50) ) + ' ' + cast( cast( left(t2, charindex('.', t1, 1) - 1) as varchar(50) ) as datetime ) ) as float ) / 60 else cast((datediff(minute, t1, t2)) as float) / 60 end from [your_table_name];表达式来检查第一次比第二次列值更大的时间。如果它更大，那么第二个时间列应该在第二天。

<强>查询

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  {{$last := len $foo_srv | subtract 1}}
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    {{- with index $foo_srv $i}}{{.Address}}{{.Port}},{{end}}
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<强> there are some things to consider

Answer 3

DECLARE @start time;
DECLARE @end time;
DECLARE @null time;

SET @start = '09:06:00';
SET @end = '21:27:00';
SET @null = '00:00:00';

SELECT DATEADD(SECOND, - DATEDIFF(SECOND, @end, @start), @null)

Answer 4

DECLARE @time1 TIME ='22:00:00'
DECLARE @time2 TIME ='06:00:00'

SELECT CEILING(CAST(Diff as decimal)/(60 * 60)) FROM (SELECT  
CASE WHEN CAST(DATEDIFF(SECOND,@time1,@time2) AS INT) > 0 THEN  CAST(DATEDIFF(SECOND,@time1,@time2) AS INT)
ELSE (ABS(CAST(DATEDIFF(SECOND,CAST('23:59:59' AS TIME),@time1) AS INT)) + ABS(CAST(DATEDIFF(SECOND,@time2,CAST('00:00:00' as TIME)) AS INT)))
END AS Diff) AS A

<强>更新

上面的结果使用天花板功能将结果舍入到下一个小时，在片段下方使用小数部分

DECLARE @time1 TIME ='023:30:00'
DECLARE @time2 TIME ='06:00:00'

SELECT Round(CAST(Diff as decimal)/(60 * 60),2) FROM (SELECT  
CASE WHEN CAST(DATEDIFF(SECOND,@time1,@time2) AS INT) > 0 THEN  CAST(DATEDIFF(SECOND,@time1,@time2) AS INT)
ELSE (ABS(CAST(DATEDIFF(SECOND,CAST('23:59:59' AS TIME),@time1) AS INT)) + ABS(CAST(DATEDIFF(SECOND,@time2,CAST('00:00:00' as TIME)) AS INT)))
END AS Diff) AS A

Answer 5

使用DATEDIFF功能

作为下一个

select datediff(hour,'08:00:00','12:00:00')

结果： -

<强>更新

获取绝对值（正值）使用ABS函数作为下一个

select abs(datediff(hour,'22:00:00','06:00:00'))

结果： -

<强> UPDATE2： -

declare 
        @StartTime datetime = '22:00:00',
        @EndTime datetime = '06:00:00'


if (datepart(hour,@StartTime) >= 12)
set @StartTime = dateadd(day,-1,@StartTime)

select abs(datediff(hour,@StartTime,@EndTime))

结果： -

如何获得2次之间的时差

5 个答案: