sql-server - 在sql server中获取精确计算的总小时和分钟

Question

再会。我使用 ms sql server 2005 express 并继承了我的表结构：

      User  |        Log_Date         |     Log_In_Time         |       Log_Out_Time      |
-----------------------------------------------------------------------------------------
    xxx     | 2013-10-02 23:00:31.623 | 2013-10-02 23:07:31.623 | 2013-10-03 01:08:31.623 |

笔记：

LOG_OUT_TIME是在其他日子，即10 月 3 日上午 1:08，LOG_IN_TIME是10 月 2 日晚上 11:00

现在我想添加列： TOTAL

它看起来像这样：

      User  |        Log_Date         |     Log_In_Time         |       Log_Out_Time      | TOTAL  |
---------------------------------------------------------------------------------------------------
    xxx     | 2013-10-02 23:00:31.623 | 2013-10-02 23:07:31.623 | 2013-10-03 01:08:31.623 | 2.08   |

有人能知道路吗？

先感谢您！

Answer 1

很简单的方法：

CAST(Log_Out_Time - Log_In_Time AS TIME)

给出：

02:01:00.0000000

编辑：

准确查询：

DECLARE @ST DATETIME = '2013-10-02 23:07:31.623'
DECLARE @ET DATETIME = '2013-10-03 01:08:31.623'

SELECT CAST(@ET - @ST AS TIME)

Answer 2

这将起作用：

SELECT CAST(DATEDIFF(hour, Log_In_Time, Log_Out_Time) AS varchar) + '.' + 
    CAST(DATEDIFF(mi, Log_In_Time, Log_Out_Time) / (60 * DATEDIFF(hour, Log_In_Time, Log_Out_Time) ) AS varchar)
FROM tableName

我已经分别计算了小时和分钟。

Answer 3

You could use DateDiff @Unknownymous, what I would do is seconds comparison, cast this to a float and the divide by 60 as necessary. This will give you minutes, hours, etc scaling up to the precision you need.

DECLARE @diff float
DECLARE @diffMin float
DECLARE @diffHour float
DECLARE @diffDay float

SELECT @diff = cast(datediff(ss,Log_in_Time,Log_Out_Time) as FLOAT)
SELECT @diffMin = @diff/60
SELECT @diffHour = @diffMin/60
SELECT @diffDay = @diffHour/24

print @diff
print @diffmin
print @diffhour
print @diffday

Answer 4

select 
cast(datepart(hour,Log_Out_Time - Log_Date) as varchar) + '.' +  
right('00' + cast(datepart(minute,Log_Out_Time - Log_Date) as varchar),2)