Is there a way to indicate the last n parameters in a batch file?

user_28757 0
Viewed 34453

In the following example, I want to call a child batch file from a parent batch file and pass all of the remaining parameters to the child.

C:\> parent.cmd child1 foo bar
C:\> parent.cmd child2 baz zoop
C:\> parent.cmd child3 a b c d e f g h i j k l m n o p q r s t u v w x y z

Inside parent.cmd, I need to strip %1 off the list of parameters and only pass the remaining parameters to the child script.

set CMD=%1
%CMD% <WHAT DO I PUT HERE>

I've investigated using SHIFT with %*, but that doesn't work. While SHIFT will move the positional parameters down by 1, %* still refers to the original parameters.

Anyone have any ideas? Should I just give up and install Linux?

cmd parameters shell
8 Answers

%* will always expand to all original parameters, sadly. But you can use the following snippet of code to build a variable containing all but the first parameter:

rem throw the first parameter away
shift
set params=%1
:loop
shift
if [%1]==[] goto afterloop
set params=%params% %1
goto loop
:afterloop

I think it can be done shorter, though ... I don't write these sort of things very often :)

Should work, though.

user_730700

I came across this question while I was facing a similar problem, but I solved it using a different approach. Instead of re-creating the input line using a loop (as in the answer by @Joey) I simply removed the first parameter from the input string.

set _input=%*
set params=!_input:%1 =!
call OTHER_BATCH_FILE.cmd %params%

Of course, this assumes that all the parameters are different.

user_86680670
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