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
The following script uses only "cmd.exe" and outputs the number of milliseconds from the time a pipeline is created to the time that the process preceding the script exits. i.e., Type your command, and pipe the to the script. Example: "timeout 3 | runtime.cmd" should yield something like "2990." If you need both the runtime output and the stdin output, redirect stdin before the pipe - ex: "dir /s 1>temp.txt | runtime.cmd" would dump the output of the "dir" command to "temp.txt" and would print the runtime to the console.
:: --- runtime.cmd ----
@echo off
setlocal enabledelayedexpansion
:: find target for recursive calls
if not "%1"=="" (
shift /1
goto :%1
exit /b
)
:: set pipeline initialization time
set t1=%time%
:: wait for stdin
more > nul
:: set time at which stdin was ready
set t2=!time!
::parse t1
set t1=!t1::= !
set t1=!t1:.= !
set t1=!t1: 0= !
:: parse t2
set t2=!t2::= !
set t2=!t2:.= !
set t2=!t2: 0= !
:: calc difference
pushd %~dp0
for /f %%i in ('%0 calc !t1!') do for /f %%j in ('%0 calc !t2!') do (
set /a t=%%j-%%i
echo !t!
)
popd
exit /b
goto :eof
:calc
set /a t=(%1*(3600*1000))+(%2*(60*1000))+(%3*1000)+(%4)
echo !t!
goto :eof
endlocal
其他回答
PowerShell为此提供了一个cmdlet，称为Measure-Command。您必须确保在运行PowerShell的机器上是可用的。
PS> Measure-Command { echo hi }
Days : 0
Hours : 0
Minutes : 0
Seconds : 0
Milliseconds : 0
Ticks : 1318
TotalDays : 1.52546296296296E-09
TotalHours : 3.66111111111111E-08
TotalMinutes : 2.19666666666667E-06
TotalSeconds : 0.0001318
TotalMilliseconds : 0.1318
Measure-Command捕获命令的输出。你可以使用Out-Default将输出重定向回控制台:
PS> Measure-Command { echo hi | Out-Default }
hi
Days : 0
...
正如Makotoe所评论的那样，Measure-Command返回一个TimeSpan对象，因此测量的时间被打印为一堆字段。你可以使用ToString()将对象格式化为时间戳字符串:
PS> (Measure-Command { echo hi | Out-Default }).ToString()
hi
00:00:00.0001318
如果Measure-Command中的命令改变了控制台文本的颜色，使用[console]::ResetColor()将其重置为正常。
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。
The following script uses only "cmd.exe" and outputs the number of milliseconds from the time a pipeline is created to the time that the process preceding the script exits. i.e., Type your command, and pipe the to the script. Example: "timeout 3 | runtime.cmd" should yield something like "2990." If you need both the runtime output and the stdin output, redirect stdin before the pipe - ex: "dir /s 1>temp.txt | runtime.cmd" would dump the output of the "dir" command to "temp.txt" and would print the runtime to the console.
:: --- runtime.cmd ----
@echo off
setlocal enabledelayedexpansion
:: find target for recursive calls
if not "%1"=="" (
shift /1
goto :%1
exit /b
)
:: set pipeline initialization time
set t1=%time%
:: wait for stdin
more > nul
:: set time at which stdin was ready
set t2=!time!
::parse t1
set t1=!t1::= !
set t1=!t1:.= !
set t1=!t1: 0= !
:: parse t2
set t2=!t2::= !
set t2=!t2:.= !
set t2=!t2: 0= !
:: calc difference
pushd %~dp0
for /f %%i in ('%0 calc !t1!') do for /f %%j in ('%0 calc !t2!') do (
set /a t=%%j-%%i
echo !t!
)
popd
exit /b
goto :eof
:calc
set /a t=(%1*(3600*1000))+(%2*(60*1000))+(%3*1000)+(%4)
echo !t!
goto :eof
endlocal
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
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;
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