Question

如何在所有条件下查找时差

如果仅相差几分钟

 49 mintus

如果小时数不同

 1 hour 20 minutes

我正在尝试

SELECT DATEDIFF(MINUTE,'2018-08-09 10:16:49.000','2018-08-09 11:14:40.000') AS Diff

但这仅在几分钟内显示

我也尝试过这个

DECLARE @StartTime datetime = '2018-08-09 10:16:49.000',
        @EndTime datetime = '2018-08-09 12:44:05.000'
SELECT CAST(@EndTime - @StartTime as Time) As TimeDiffere

输出：02:27:16.0000000

但是我想2hr 27mins 16Sec

Answer 1

如果您的SQL Server版本高于2012，则可以尝试起诉format函数。

CREATE TABLE T(
 StartTime datetime,
  EndTime datetime
);

insert into t values ('2018-08-09 10:16:49.000','2018-08-09 12:44:05.000')

查询1 ：

SELECT format(EndTime - StartTime ,'hh\hr mm\min\s ss\Sec') As TimeDiffere
FROM T

Results ：

|       TimeDiffere |
|-------------------|
| 02hr 27mins 16Sec |

注意

反斜杠\可以像format，hr，min ...这样从s逃逸关键字，然后将其显示为原始单词。 / p>

Answer 2

 DECLARE @StartTime datetime = '2018-08-09 10:01:15.000',
        @EndTime datetime = '2018-08-09 12:50:00.000'
SELECT CONCAT((DATEDIFF(second,@StartTime,@EndTime )/60)/60,'hr ',(DATEDIFF(second,@StartTime,@EndTime)/60)%60,'mins ',DATEDIFF(second,@StartTime,@EndTime)%60,'Sec') 
As TimeDiffere

Answer 3

original_pic = Photo.objects.filter(id=photo_id).values_list('original_pic', flat=True)[0]
image = Image.open(original_pic).convert('RGBA')

Answer 4

如果您的datetime功能差，无法获得hh.mm.ss，则可以使用apply：

select t.*, concat( (dtdiff / 3600), 'hr ', (dtdiff / 60) % 60, 'mins ', dtdiff % 60, 'sec')
from table t cross apply
     ( values (datediff(second, startdate, enddate)) 
     ) tt(dtdiff);

如何在SQL Server中查找时，分，秒的时差（所有结果）

4 个答案: