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

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。

其他回答

driblio的答案可以短一点(虽然可读性不高)

@echo off

:: Calculate the start timestamp
set _time=%time%
set /a _hours=100%_time:~0,2%%%100,_min=100%_time:~3,2%%%100,_sec=100%_time:~6,2%%%100,_cs=%_time:~9,2%
set /a _started=_hours*60*60*100+_min*60*100+_sec*100+_cs


:: yourCommandHere


:: Calculate the difference in cSeconds
set _time=%time%
set /a _hours=100%_time:~0,2%%%100,_min=100%_time:~3,2%%%100,_sec=100%_time:~6,2%%%100,_cs=%_time:~9,2%
set /a _duration=_hours*60*60*100+_min*60*100+_sec*100+_cs-_started

:: Populate variables for rendering (100+ needed for padding)
set /a _hours=_duration/60/60/100,_min=100+_duration/60/100%%60,_sec=100+(_duration/100%%60%%60),_cs=100+_duration%%100

echo Done at: %_time% took : %_hours%:%_min:~-2%:%_sec:~-2%.%_cs:~-2%

::prints something like:
::Done at: 12:37:53,70 took: 0:02:03.55

根据Luke Sampson的评论,这个版本是八进制安全的,尽管任务应该在24小时内完成。

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。

如果有人来这里寻找这个问题的答案,有一个Windows API函数叫做GetProcessTimes()。编写一个小的C程序来启动这个命令、执行这个调用并返回进程时间,看起来并没有太多的工作。

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

如果你打开了一个命令窗口并手动调用命令,你可以在每个提示符上显示时间戳,例如:

prompt $d $t $_$P$G

它会给你这样的东西:

23.03.2009 15:45:50 77 C: \ >

如果你有一个小的批处理脚本来执行你的命令,在每个命令之前有一个空行,例如。

(空行) myCommand.exe (下一行空) myCommand2.exe

根据提示符中的时间信息,可以计算出每条命令的执行时间。最好的可能是管道输出到一个文本文件进行进一步分析:

MyBatchFile.bat > output.txt