如何编写脚本来计算完成脚本所需的时间?
我原以为会是这样的,但显然不是..
@echo off
set starttime=%time%
set endtime=%time%
REM do stuff here
set /a runtime=%endtime%-%starttime%
echo Script took %runtime% to complete
发布于 2009-04-11 06:28:41
这是一种切题,但它可能会对您有所帮助。
微软有一个timeit.exe程序,它的工作原理类似于unix 'time‘命令的增强版。它出现在Windows Server 2003 Resource Kit中,它几乎取代了我一直想要做的事情,就像你建议的那样。
这可能值得一看。
发布于 2012-10-31 21:37:48
很棒的小套路。但是,如果计算跨越两天,则计算是错误的。结果需要从24小时(8640000厘米秒)中减去。
同时删除相关行中的一个重复的'set‘命令。
请注意,区域设置会影响时间函数的格式,小数在UK中是句号,而不是逗号。
台词
rem我们可能已经测量了两天之间的时间
if %ENDTIME% LSS %STARTTIME% set set /A DURATION=%STARTTIME%-%ENDTIME%
需要更改为
rem我们可能已经测量了几天的时间
if %ENDTIME% LSS %STARTTIME% set /A DURATION=8640000 - (%STARTTIME% - %ENDTIME%)
发布于 2011-03-12 01:31:04
查看此脚本,它将通过WMI检索时间,因此它是独立于区域设置的。
@echo off
::::::::::::::::::::::::::::::::::::::::::
:: TimeDiff v1.00 by LEVENTE ROG ::
:: www.thesysadminhimself.com ::
::::::::::::::::::::::::::::::::::::::::::
::[ EULA ]:::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: Feel free to use this script. The code can be redistributed ::
:: and edited, but please keep the credits. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
::[ CHANGELOG ]::::::::::::::
:: v1.00 - First Version ::
:::::::::::::::::::::::::::::
FOR /F "skip=1 tokens=1-6" %%A IN ('WMIC Path Win32_LocalTime Get Day^,Hour^,Minute^,Second /Format:table ^| findstr /r "."') DO (
set Milisecond=%time:~9,2%
set Day=%%A
set Hour=%%B
set Minute=%%C
set Second=%%D
)
set /a Start=%Day%*8640000+%Hour%*360000+%Minute%*6000+%Second%*100+%Milisecond%
::
::
:: PUT COMMANDS HERE
ping www.thesysadminhimself.com
::
::
FOR /F "skip=1 tokens=1-6" %%A IN ('WMIC Path Win32_LocalTime Get Day^,Hour^,Minute^,Second /Format:table ^| findstr /r "."') DO (
set Day=%%A
set Hour=%%B
set Minute=%%C
set Second=%%D
)
set Milisecond=%time:~9,2%
set /a End=%Day%*8640000+%Hour%*360000+%Minute%*6000+%Second%*100+%Milisecond%
set /a Diff=%End%-%Start%
set /a DiffMS=%Diff%%%100
set /a Diff=(%Diff%-%DiffMS%)/100
set /a DiffSec=%Diff%%%60
set /a Diff=(%Diff%-%Diff%%%60)/60
set /a DiffMin=%Diff%%%60
set /a Diff=(%Diff%-%Diff%%%60)/60
set /a DiffHrs=%Diff%
:: format with leading zeroes
if %DiffMS% LSS 10 set DiffMS=0%DiffMS!%
if %DiffSec% LSS 10 set DiffMS=0%DiffSec%
if %DiffMin% LSS 10 set DiffMS=0%DiffMin%
if %DiffHrs% LSS 10 set DiffMS=0%DiffHrs%
echo %DiffHrs%:%DiffMin%:%DiffSec%.%DiffMS%
https://stackoverflow.com/questions/739606
复制相似问题