如何在SQL Server中查找小时、分钟和秒（所有结果）的时差

如何找到所有条件下的时差，如

如果差别只有几分钟

 49 mintus

如果时差

 1 hour 20 minutes

我正在努力

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

但这只在几分钟内播出

我也试过这个

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

---

但我想说的是2小时27分钟16秒

如果您的SQL server版本高于2012，您可以尝试使用函数

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')

问题1：

:

注

反斜杠\可以从hr、min、s…等格式中转义关键字，然后将其显示为原始单词。

如果获取hh.mm.ss的日期时间功能较差，则可以使用apply：

 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

---

此答案不会返回正确的结果。这将返回1小时，即使只有1分钟：选择datediffhh，'11:59'，'12:00'，您需要使用相同的DATEDIFF来计算所有时间成分并输入答案。

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

 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

DECLARE @StartTime  datetime = '2018-08-09 12:16:49.000',
        @EndTime    datetime = '2018-08-09 12:44:05.000'

select  [hour] + [minute] + [second] as TimeDiffere
from    (
            select diff_sec = datediff(second, @StartTime, @EndTime)
        ) t
        cross apply
        (
            select  [hour] = isnull(convert(varchar(10), nullif(diff_sec / 60 / 60, 0)) 
                           + ' hours ', '')
        ) hr
        cross apply
        (
            select  [minute] = isnull(convert(varchar(10), nullif(diff_sec / 60 % 60, 0)) 
                             + ' mintues ', '')
        ) mn
        cross apply
        (
            select  [second] = isnull(convert(varchar(10), nullif(diff_sec % 60, 0)) 
                             + ' seconds', '')
        ) sc

/*  RESULT

27 mintues 16 seconds

*/

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);