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

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%

如果您使用的是Windows 2003(注意不支持Windows server 2008及更高版本)，您可以使用Windows server 2003资源工具包，其中包含显示详细执行统计信息的time .exe。这里有一个例子，计时命令“timeit -?”：

C:\>timeit timeit -?
Invalid switch -?
Usage: TIMEIT [-f filename] [-a] [-c] [-i] [-d] [-s] [-t] [-k keyname | -r keyname] [-m mask] [commandline...]
where:        -f specifies the name of the database file where TIMEIT
                 keeps a history of previous timings.  Default is .\timeit.dat
              -k specifies the keyname to use for this timing run
              -r specifies the keyname to remove from the database.  If
                 keyname is followed by a comma and a number then it will
                 remove the slowest (positive number) or fastest (negative)
                 times for that keyname.
              -a specifies that timeit should display average of all timings
                 for the specified key.
              -i specifies to ignore non-zero return codes from program
              -d specifies to show detail for average
              -s specifies to suppress system wide counters
              -t specifies to tabular output
              -c specifies to force a resort of the data base
              -m specifies the processor affinity mask

Version Number:   Windows NT 5.2 (Build 3790)
Exit Time:        7:38 am, Wednesday, April 15 2009
Elapsed Time:     0:00:00.000
Process Time:     0:00:00.015
System Calls:     731
Context Switches: 299
Page Faults:      515
Bytes Read:       0
Bytes Written:    0
Bytes Other:      298

您可以在Windows 2003资源工具包中获得TimeIt。它不能从微软下载中心直接下载，但仍然可以从archive.org - Windows Server 2003资源工具包工具中获得。

在Perl安装了可用的雇佣解决方案后，运行:

C:\BATCH>time.pl "echo Fine result"
0.01063
Fine result

STDERR出现在被测量的秒之前

#!/usr/bin/perl -w

use Time::HiRes qw();
my $T0 = [ Time::HiRes::gettimeofday ];

my $stdout = `@ARGV`;

my $time_elapsed = Time::HiRes::tv_interval( $T0 );

print $time_elapsed, "\n";
print $stdout;

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

因为其他人建议安装像免费软件和PowerShell这样的东西，你也可以安装Cygwin，它可以让你访问许多基本的Unix命令，比如time:

abe@abe-PC:~$ time sleep 5

real    0m5.012s
user    0m0.000s
sys 0m0.000s

不知道Cygwin增加了多少开销。