How to write a bash script to execute a different command with additional arguments?

Question

When called as my_cmd -a -b ... c,

the script will finally call program a with addition parameters:

a -additional -a -b ... c

How can I write such a bash script?

eudoxos · Accepted Answer · 2011-09-14 06:03:06Z

2

a -additional "$@" [pad to 30 characters]

answered Sep 14, 2011 at 6:03

eudoxos

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4 Comments

Jonathan Leffler Over a year ago

Always use "$@" to relay arguments.

new_perl Over a year ago

@Jonathan Leffler ,why is "" necessary?

eudoxos Over a year ago

@Jonathan Leffler: thanks, forgot the doublequotes, edited the answer. Explanation here

Jonathan Leffler Over a year ago

Because you get different results with $@ and "$@" when there are spaces inside the arguments. See How to iterate over arguments in bash script.

Jonathan Leffler · Accepted Answer · 2011-09-14 06:03:49Z

1

Given that the additional arguments come immediately after the command, it is trivial:

exec a -additional "$@"

The exec is optional.

answered Sep 14, 2011 at 6:03

Jonathan Leffler

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Comments

manatwork · Accepted Answer · 2011-09-14 06:30:07Z

0

If the command's name "a" has to be also taken from the my_cmd's parameters:

"${1:1}" -additional "$@"

edited Sep 14, 2011 at 6:30

answered Sep 14, 2011 at 6:23

manatwork

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