pass partial list of arguments from the parent script to the child script

Question

Suppose my parent script is called with n arguments, and I want to pass all arguments from argument #3 to the child script.

How do I do that in bash script?

I know $@ would get you the entire argument list, which is not what I want.

rici · Accepted Answer · 2013-07-19 18:37:06Z

7

Like this:

child_script "${@:3}"

If you wanted to pass four arguments starting at argument 3:

child_script "${@:3:4}"

answered Jul 19, 2013 at 18:37

rici

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Comments

Barmar · Accepted Answer · 2013-07-19 18:25:07Z

4

Use shift to remove the first two arguments:

shift 2
child_script "$@"

If you need to use $1 and $2, you can save them in variables first.

Another option is to assign the arguments to an array:

args=("$@")

remove the first two elements of the array:

unset args[0]
unset args[1]

Then call the child script with this array:

child_script "${args[@]}"

answered Jul 19, 2013 at 18:13

Barmar

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But what if I want to use $1 and $2 later in the file? SHould I just save their values in a separate variable?

That's what I would probably do, but I added another solution to the answer.

pmoosh · Accepted Answer · 2013-07-19 21:23:47Z

0

As for the discussion of keeping the original values: You could do the shift in the child script.

answered Jul 19, 2013 at 21:23

pmoosh

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