batch parameters: everything after %1

不羁岁月 提交于 2019-12-21 03:10:06

问题


Duplicate:

  • Is there a way to indicate the last n parameters in a batch file?
  • how to get batch file parameters from Nth position on?

Clarification: I knew of the looping approach - this worked even before Command Extensions; I was hoping for something fun and undocumented like %~*1 or whatever - just like those documented at http://www.microsoft.com/resources/documentation/windows/xp/all/proddocs/en-us/percent.mspx?mfr=true.


In a Windows batch file (with the so called "Command Extensions" on), %1 is the first argument, %2 is the second, etc. %* is all arguments concatenated.

My question: is there a way to get everything AFTER %2, for example?

I couldn't find such a thing, and it would be helpful for something I'm working on.


回答1:


I am not sure if there is a direct command but you can always use a simple loop and shift to get the result in a variable. Something like:

@echo off
set RESTVAR=
shift
:loop1
if "%1"=="" goto after_loop
set RESTVAR=%RESTVAR% %1
shift
goto loop1

:after_loop
echo %RESTVAR%

Let me know if it helps!




回答2:


There is a shorter solution (one-liner) utilizing the tokenization capabilities of for loops:

:: all_but_first.bat
echo all: %*
for /f "tokens=1,* delims= " %%a in ("%*") do set ALL_BUT_FIRST=%%b
echo all but first: %ALL_BUT_FIRST%

output:

> all_but_first.bat foo bar baz
all: foo bar baz
all but first: bar baz



回答3:


The following will work for args with ", =, ' '. Based on Dmitry Sokolov answer. Fixed issue when second arg is the same as first arg.

@echo off
echo %*
set _tail=%*
call set _tail=%%_tail:*%1=%%
echo %_tail%



回答4:


The following will work for args with ", =, ' ' (as compared to @MaxTruxa answer)

echo %*
set _all=%*
call set _tail=%%_all:*%2=%%
set _tail=%2%_tail%
echo %_tail%

Test

> get_tail.cmd "first 1" --flag="other options" --verbose
"first 1" --flag="other options" --verbose
--flag="other options" --verbose



回答5:


You can use SHIFT for this. It removes %1 and shifts all other arguments one lower. This script outputs all the arguments after %2 (so it outputs %3, %4...) until one of them is empty (so it's the last one):

@echo off

SHIFT
SHIFT

:loop
if "%1" == "" goto end
echo %1
SHIFT
goto loop

:end

EDIT: Removed example using %* as this doesn't work - %* always outputs all of the parameters




回答6:


Building on schnaader's answer, I think this does it if you want everything after %1 concatenated.

@echo off

SHIFT

set after1=

:loop
if "%1" == "" goto end
set after1=%after1% %1
SHIFT
goto loop


:end

echo %after1%



回答7:


Sebi, here's the Syntax! There is a behavior, batch eating the equal signs which is not double quoted, it cause trouble in the scripts above. If you wan't to skip, i've made a modification, based on Raman Zhylich answer and strlen.cmd:

@ECHO OFF
SETLOCAL enableDelayedExpansion

SET _tail=%*
SET "_input="
SET /A _len=0

:again
SET "_param=%1"
SET "_input=%_input%%1"
FOR /L %%i in (0,1,8191) DO IF "!_param:~%%i,1!"=="" (
    REM skip param
    SET /A _len+=%%i
    REM _len can't be use in substring
    FOR /L %%j in (!_len!,1,!_len!) DO (
        REM skip param separator
        SET /A _len+=1
        IF "!_tail:~%%j,1!"=="=" (SET "_input=%_input%=" & SHIFT & goto :again)
    )
) & goto :next
:next
IF %_len% NEQ 0 SET _tail=!_tail:~%_len%!

ENDLOCAL & SET "_input=%_input%" & SET "_tail=%_tail%"


来源:https://stackoverflow.com/questions/935609/batch-parameters-everything-after-1

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