Take multiple arguments in batch file

Question

I have a batch file test.bat I understand we can give it multiple arguments and take those values using %1, %2, and so on. But I do not know how many arguments will be given. I thought of constructing an array and for loop to decide. But the point I am failing is constructing the for loop.

what I did is:

set count=0
// for loop until we have arguments left
set list[%count%]=%var%
set /A count=count+1

I know little bit of for loop in batch files, but I do not quite understand how to use in this scenario. Number of arguments are unknown. Any direction would be appreciated?

Mofi · Accepted Answer · 2025-10-23 16:59:41Z

3

%* references all arguments. You could use

for %%a in (%*) do echo %%a

or use shift which advances %1 to the next argument and so on.

:loop
echo %1
shift
if not "%~1" == "" goto :loop

edited Oct 23 at 16:59

Mofi

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answered Aug 21, 2014 at 20:54

ths

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Mofi · Accepted Answer · 2025-10-23 16:56:19Z

0

You need to add double quotes around %1 to get the whole argument if it is a string.

@echo off
:loop
echo %1
shift
if not "%~1" == "" goto :loop

When you run it, you will see the result like this:

D:\2\test2>arg1 abc "abc def"
abc
"abc def"

edited Oct 23 at 16:56

Mofi

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answered Oct 19 at 22:42

Ben Mar

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