是否有一种内置的方法来测量Windows命令行上命令的执行时间?
当前回答
下面的脚本模拟*nix纪元时间，但它是本地和区域性的。它应该处理日历边缘情况，包括闰年。如果Cygwin可用，则可以通过指定Cygwin选项来比较epoch值。
我在EST，报告的差异是4小时，这是相对正确的。有一些有趣的解决方案可以删除TZ和区域依赖，但我注意到没有什么微不足道的。
@ECHO off
SETLOCAL EnableDelayedExpansion
::
:: Emulates local epoch seconds
::
:: Call passing local date and time
CALL :SECONDS "%DATE%" "%TIME%"
IF !SECONDS! LEQ 0 GOTO END
:: Not testing - print and exit
IF NOT "%~1"=="cygwin" (
ECHO !SECONDS!
GOTO END
)
:: Call on Cygwin to get epoch time
FOR /F %%c IN ('C:\cygwin\bin\date +%%s') DO SET EPOCH=%%c
:: Show the results
ECHO Local Seconds: !SECONDS!
ECHO Epoch Seconds: !EPOCH!
:: Calculate difference between script and Cygwin
SET /A HOURS=(!EPOCH!-!SECONDS!)/3600
SET /A FRAC=(!EPOCH!-!SECONDS!)%%3600
:: Delta hours shown reflect TZ
ECHO Delta Hours: !HOURS! Remainder: !FRAC!
GOTO END
:SECONDS
SETLOCAL EnableDelayedExpansion
:: Expecting values from caller
SET DATE=%~1
SET TIME=%~2
:: Emulate Unix epoch time without considering TZ
SET "SINCE_YEAR=1970"
:: Regional constraint! Expecting date and time in the following formats:
:: Sun 03/08/2015 Day MM/DD/YYYY
:: 20:04:53.64 HH:MM:SS
SET VALID_DATE=0
ECHO !DATE! | FINDSTR /R /C:"^... [0-9 ][0-9]/[0-9 ][0-9]/[0-9][0-9][0-9][0-9]" > nul && SET VALID_DATE=1
SET VALID_TIME=0
ECHO !TIME! | FINDSTR /R /C:"^[0-9 ][0-9]:[0-9 ][0-9]:[0-9 ][0-9]" > nul && SET VALID_TIME=1
IF NOT "!VALID_DATE!!VALID_TIME!"=="11" (
IF !VALID_DATE! EQU 0 ECHO Unsupported Date value: !DATE! 1>&2
IF !VALID_TIME! EQU 0 ECHO Unsupported Time value: !TIME! 1>&2
SET SECONDS=0
GOTO SECONDS_END
)
:: Parse values
SET "YYYY=!DATE:~10,4!"
SET "MM=!DATE:~4,2!"
SET "DD=!DATE:~7,2!"
SET "HH=!TIME:~0,2!"
SET "NN=!TIME:~3,2!"
SET "SS=!TIME:~6,2!"
SET /A YEARS=!YYYY!-!SINCE_YEAR!
SET /A DAYS=!YEARS!*365
:: Bump year if after February - want leading zeroes for this test
IF "!MM!!DD!" GEQ "0301" SET /A YEARS+=1
:: Remove leading zeros that can cause octet probs for SET /A
FOR %%r IN (MM,DD,HH,NN,SS) DO (
SET "v=%%r"
SET "t=!%%r!"
SET /A N=!t:~0,1!0
IF 0 EQU !N! SET "!v!=!t:~1!"
)
:: Increase days according to number of leap years
SET /A DAYS+=(!YEARS!+3)/4-(!SINCE_YEAR!%%4+3)/4
:: Increase days by preceding months of current year
FOR %%n IN (31:1,28:2,31:3,30:4,31:5,30:6,31:7,31:8,30:9,31:10,30:11) DO (
SET "n=%%n"
IF !MM! GTR !n:~3! SET /A DAYS+=!n:~0,2!
)
:: Multiply and add it all together
SET /A SECONDS=(!DAYS!+!DD!-1)*86400+!HH!*3600+!NN!*60+!SS!
:SECONDS_END
ENDLOCAL & SET "SECONDS=%SECONDS%"
GOTO :EOF
:END
ENDLOCAL
其他回答
在程序所在的目录中，键入记事本mytimer.bat，单击“是”创建一个新文件。 粘贴下面的代码，用你的程序替换YourApp.exe，然后保存。 @echo掉 日期/ t 时间/ t YourApp.exe 日期/ t 时间/ t 在命令行中输入mytimer.bat，然后按Enter。
呵呵，最简单的解决办法可能是:
echo %time%
YourApp.exe
echo %time%
这适用于所有开箱即用的Windows。
在应用程序使用控制台输出的情况下，将开始时间存储在临时变量中可能会很方便:
set startTime=%time%
YourApp.exe
echo Start Time: %startTime%
echo Finish Time: %time%
这是凯西对答案的扩展。关于使用PowerShell中的Measure-Command:
您可以从标准的命令提示符调用PowerShell，如下所示: powershell -Command "Measure-Command {echo hi}" 这将消耗标准输出，但你可以通过从PowerShell添加| Out-Default来防止这种情况: 测量命令{echo hi | Out-Default} 或者从命令提示符: powershell -Command "Measure-Command {echo hi | Out-Default}"
当然，您可以自由地将其包装在脚本文件*中。Ps1或*.bat。
下面的脚本模拟*nix纪元时间，但它是本地和区域性的。它应该处理日历边缘情况，包括闰年。如果Cygwin可用，则可以通过指定Cygwin选项来比较epoch值。
我在EST，报告的差异是4小时，这是相对正确的。有一些有趣的解决方案可以删除TZ和区域依赖，但我注意到没有什么微不足道的。
@ECHO off
SETLOCAL EnableDelayedExpansion
::
:: Emulates local epoch seconds
::
:: Call passing local date and time
CALL :SECONDS "%DATE%" "%TIME%"
IF !SECONDS! LEQ 0 GOTO END
:: Not testing - print and exit
IF NOT "%~1"=="cygwin" (
ECHO !SECONDS!
GOTO END
)
:: Call on Cygwin to get epoch time
FOR /F %%c IN ('C:\cygwin\bin\date +%%s') DO SET EPOCH=%%c
:: Show the results
ECHO Local Seconds: !SECONDS!
ECHO Epoch Seconds: !EPOCH!
:: Calculate difference between script and Cygwin
SET /A HOURS=(!EPOCH!-!SECONDS!)/3600
SET /A FRAC=(!EPOCH!-!SECONDS!)%%3600
:: Delta hours shown reflect TZ
ECHO Delta Hours: !HOURS! Remainder: !FRAC!
GOTO END
:SECONDS
SETLOCAL EnableDelayedExpansion
:: Expecting values from caller
SET DATE=%~1
SET TIME=%~2
:: Emulate Unix epoch time without considering TZ
SET "SINCE_YEAR=1970"
:: Regional constraint! Expecting date and time in the following formats:
:: Sun 03/08/2015 Day MM/DD/YYYY
:: 20:04:53.64 HH:MM:SS
SET VALID_DATE=0
ECHO !DATE! | FINDSTR /R /C:"^... [0-9 ][0-9]/[0-9 ][0-9]/[0-9][0-9][0-9][0-9]" > nul && SET VALID_DATE=1
SET VALID_TIME=0
ECHO !TIME! | FINDSTR /R /C:"^[0-9 ][0-9]:[0-9 ][0-9]:[0-9 ][0-9]" > nul && SET VALID_TIME=1
IF NOT "!VALID_DATE!!VALID_TIME!"=="11" (
IF !VALID_DATE! EQU 0 ECHO Unsupported Date value: !DATE! 1>&2
IF !VALID_TIME! EQU 0 ECHO Unsupported Time value: !TIME! 1>&2
SET SECONDS=0
GOTO SECONDS_END
)
:: Parse values
SET "YYYY=!DATE:~10,4!"
SET "MM=!DATE:~4,2!"
SET "DD=!DATE:~7,2!"
SET "HH=!TIME:~0,2!"
SET "NN=!TIME:~3,2!"
SET "SS=!TIME:~6,2!"
SET /A YEARS=!YYYY!-!SINCE_YEAR!
SET /A DAYS=!YEARS!*365
:: Bump year if after February - want leading zeroes for this test
IF "!MM!!DD!" GEQ "0301" SET /A YEARS+=1
:: Remove leading zeros that can cause octet probs for SET /A
FOR %%r IN (MM,DD,HH,NN,SS) DO (
SET "v=%%r"
SET "t=!%%r!"
SET /A N=!t:~0,1!0
IF 0 EQU !N! SET "!v!=!t:~1!"
)
:: Increase days according to number of leap years
SET /A DAYS+=(!YEARS!+3)/4-(!SINCE_YEAR!%%4+3)/4
:: Increase days by preceding months of current year
FOR %%n IN (31:1,28:2,31:3,30:4,31:5,30:6,31:7,31:8,30:9,31:10,30:11) DO (
SET "n=%%n"
IF !MM! GTR !n:~3! SET /A DAYS+=!n:~0,2!
)
:: Multiply and add it all together
SET /A SECONDS=(!DAYS!+!DD!-1)*86400+!HH!*3600+!NN!*60+!SS!
:SECONDS_END
ENDLOCAL & SET "SECONDS=%SECONDS%"
GOTO :EOF
:END
ENDLOCAL
我在Windows Server 2008 R2中使用的一行程序是:
cmd /v:on /c "echo !TIME! & *mycommand* & echo !TIME!"
只要mycommand不需要引号(这会影响cmd的引号处理)。on允许两个不同的TIME值独立计算，而不是在执行命令时一次计算。
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