如何在 DOS 提示符下为脚本计时？
*NIX 等价物是：
$ time myscript
real 0m0.886s
user 0m0.846s
sys 0m0.031s
$
有与此等效的DOS吗？谢谢
skcubrats
问问题
14913 次
4 回答
5
您可以使用Windows Server 2003 Resource Kit中的 timeit.exe 。您还可以获得大量其他有用的工具和 Unix 命令端口。
C:\Documents and Settings\Rob>timeit sleep 3
Version Number: Windows NT 5.1 (Build 2600)
Exit Time: 2:40 pm, Thursday, December 10 2009
Elapsed Time: 0:00:03.296
Process Time: 0:00:00.015
System Calls: 7576
Context Switches: 1974
Page Faults: 1072
Bytes Read: 6172
Bytes Written: 11086
Bytes Other: 144432
于 2009-12-10T20:33:28.513 回答
5
这是一个 DOS 批处理脚本，它将为您计算时间差，而不仅仅是echo %time%.
只要您的“定时用户部分”运行时间不超过 24 小时，以下代码就可以正常工作。请原谅或给我建议，这是我花了大约一个小时完成的工作。
将“PAUSE”替换为对脚本的调用，以便为您自己的代码计时。
DOS BAT 文件如下：
注意：已编辑以解决前导零是八进制而不是十进制的问题。
注意：经过编辑以处理“午夜”小时环绕运行并在输出中放置前导零。
@echo off
@rem --------------------------------------------
setlocal ENABLEEXTENSIONS
set start_time=%time%
echo Beginning at: %start_time%
echo Running Timed Batch File
echo.
@rem echo %* <-- Extension for all args -- easier than following line
@rem echo %0 %1 %2 %3 %4 %5 %6 %7 %8 %9 ... ...
@rem DO YOUR WORK HERE -- see "call" / PAUSE is here to "fake" work
@rem call userscript.bat %*
PAUSE
set stop_time=%time%
echo.
echo Timed Batch File Completed
echo Start time: %start_time%
echo Stop time : %stop_time%
set TEMPRESULT=%start_time:~0,2%
call:FN_REMOVELEADINGZEROS
set start_hour=%TEMPRESULT%
@rem
set TEMPRESULT=%start_time:~3,2%
call:FN_REMOVELEADINGZEROS
set start_min=%TEMPRESULT%
@rem
set TEMPRESULT=%start_time:~6,2%
call:FN_REMOVELEADINGZEROS
set start_sec=%TEMPRESULT%
@rem
set TEMPRESULT=%start_time:~9,2%
call:FN_REMOVELEADINGZEROS
set start_hundredths=%TEMPRESULT%
set TEMPRESULT=%stop_time:~0,2%
call:FN_REMOVELEADINGZEROS
set stop_hour=%TEMPRESULT%
@rem
set TEMPRESULT=%stop_time:~3,2%
call:FN_REMOVELEADINGZEROS
set stop_min=%TEMPRESULT%
@rem
set TEMPRESULT=%stop_time:~6,2%
call:FN_REMOVELEADINGZEROS
set stop_sec=%TEMPRESULT%
@rem
set TEMPRESULT=%stop_time:~9,2%
call:FN_REMOVELEADINGZEROS
set stop_hundredths=%TEMPRESULT%
set /A start_total=(((((%start_hour%*60)+%start_min%)*60)+%start_sec%)*100)+%start_hundredths%
set /A stop_total=(((((%stop_hour%*60)+%stop_min%)*60)+%stop_sec%)*100)+%stop_hundredths%
set /A total_time=%stop_total% - %start_total%
set /A total_hundredths=%total_time% %% 100
set total_hundredths=00%total_hundredths%
set total_hundredths=%total_hundredths:~-2%
set /A total_time=%total_time% / 100
set /A total_sec="%total_time% %% 60"
set total_sec=00%total_sec%
set total_sec=%total_sec:~-2%
set /A total_time=%total_time% / 60
set /A total_min="%total_time% %% 60"
set total_min=00%total_min%
set total_min=%total_min:~-2%
set /A total_time=%total_time% / 60
set /A total_hour="%total_time% %% 60"
@rem Handle if it wrapped around over midnight
if "%total_hour:~0,1%"=="-" set /A total_hour=%total_hour% + 24
echo Total time: %total_hour%:%total_min%:%total_sec%.%total_hundredths%
@rem --------------------------------------------
@rem Exit the BAT Program
endlocal
goto END
@rem --------------------------------------------
@rem FN_REMOVELEADINGZEROS function
@rem Used to remove leading zeros from Decimal
@rem numbers so they are not treated as Octal.
:FN_REMOVELEADINGZEROS
if "%TEMPRESULT%"=="0" goto END
if "%TEMPRESULT:~0,1%" NEQ "0" goto END
set TEMPRESULT=%TEMPRESULT:~1%
goto FN_REMOVELEADINGZEROS
@rem --------------------------------------------
@rem BAT PROGRAM / FUNCTION FILE EXIT
:END
于 2009-12-10T20:28:21.947 回答
5
您可以将其放在批次的开头和结尾：
echo %time%
于 2009-09-22T19:49:54.333 回答