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?

score 2 · Accepted Answer · answered Sep 14 '11 at 06:03

2

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

answered Sep 14 '11 at 06:03

eudoxos

Always use `"$@"` to relay arguments. – Jonathan Leffler Sep 14 '11 at 06:04
@Jonathan Leffler ,why is `""` necessary? – new_perl Sep 14 '11 at 06:06
@Jonathan Leffler: thanks, forgot the doublequotes, edited the answer. Explanation [here](http://wiki.bash-hackers.org/scripting/posparams#mass_usage) – eudoxos Sep 14 '11 at 06:08
Because you get different results with `$@` and `"$@"` when there are spaces inside the arguments. See [How to iterate over arguments in bash script](http://stackoverflow.com/questions/255898/how-to-iterate-over-arguments-in-bash-script/). – Jonathan Leffler Sep 14 '11 at 06:09

score 1 · Answer 2 · answered Sep 14 '11 at 06:03

1

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

exec a -additional "$@"

The exec is optional.

answered Sep 14 '11 at 06:03

Jonathan Leffler

manatwork · Answer 3 · 2011-09-14T06:30:07.370

0

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

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

edited Sep 14 '11 at 06:30

answered Sep 14 '11 at 06:23

manatwork

3 Answers3