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%

一个解决方案使用纯PHP为cmd和一个env。变量:

@echo off
setlocal enableextensions

REM set start time env var
FOR /F "tokens=* USEBACKQ" %%F IN (`php -r "echo microtime(true);"`) DO ( SET start_time=%%F )

## PUT_HERE_THE_COMMAND_TO_RUN ##

REM echo elapsed time
php -r "echo 'elapsed: ' . (round(microtime(true) - trim(getenv('start_time')), 2)) . ' seconds' . mb_convert_encoding('
', 'UTF-8', 'HTML-ENTITIES');"

不需要cygwin或不受信任的实用程序。当PHP在本地可用时非常有用 精度和输出格式可以很容易地调整 同样的想法也可以移植到PowerShell

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。

不像Unix上的一些功能那样优雅，但是创建一个cmd文件，看起来像这样:

@echo off
time < nul
yourexecutable.exe > c:\temp\output.txt
time < nul
rem on newer windows system you can try time /T

将显示开始和停止时间，如下所示:

The current time is: 10:31:57.92
Enter the new time:
The current time is: 10:32:05.94
Enter the new time:

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

如果你愿意

测量执行时间精确到(hh:mm:ss)的百分之一秒。ff格式) 不需要下载和安装资源包 看起来像一个巨大的DOS书呆子(谁不是呢)

尝试将以下脚本复制到一个新的批处理文件(例如timecmd.bat):

@echo off
@setlocal

set start=%time%

:: Runs your command
cmd /c %*

set end=%time%
set options="tokens=1-4 delims=:.,"
for /f %options% %%a in ("%start%") do set start_h=%%a&set /a start_m=100%%b %% 100&set /a start_s=100%%c %% 100&set /a start_ms=100%%d %% 100
for /f %options% %%a in ("%end%") do set end_h=%%a&set /a end_m=100%%b %% 100&set /a end_s=100%%c %% 100&set /a end_ms=100%%d %% 100

set /a hours=%end_h%-%start_h%
set /a mins=%end_m%-%start_m%
set /a secs=%end_s%-%start_s%
set /a ms=%end_ms%-%start_ms%
if %ms% lss 0 set /a secs = %secs% - 1 & set /a ms = 100%ms%
if %secs% lss 0 set /a mins = %mins% - 1 & set /a secs = 60%secs%
if %mins% lss 0 set /a hours = %hours% - 1 & set /a mins = 60%mins%
if %hours% lss 0 set /a hours = 24%hours%
if 1%ms% lss 100 set ms=0%ms%

:: Mission accomplished
set /a totalsecs = %hours%*3600 + %mins%*60 + %secs%
echo command took %hours%:%mins%:%secs%.%ms% (%totalsecs%.%ms%s total)

使用

如果你把timecmd.bat放在你路径中的一个目录中，你可以在任何地方调用它，像这样:

timecmd [your command]

E.g.

C:\>timecmd pause
Press any key to continue . . .
command took 0:0:1.18

如果你想做输出重定向，你可以这样引用命令:

timecmd "dir c:\windows /s > nul"

这应该可以处理从午夜之前到午夜之后运行的命令，但是如果您的命令运行了24小时或更长时间，则输出将是错误的。