Shell Script: command line argument with quotes

Question

I have a script which takes command line arguments and after some processing, it sends these arguments to the java application.

Some of these arguments can be quoted and I want to send those arguments as it is (in the quote). I am using following script:

$ARGS=""
for a in $@; do
    ARGS = ${ARGS} ${a};
done
exec $JAR $ARGS

where $JAR contains command to run the jar.
But, if I run the script with following options:

script x y "a b"

The script treats the three parameters as x, y and a b.
And I want it to consider these parameters as: x, y and "a b".

The user will provide the command line arguments. So, I need to do the processing in the script itself. — sachinpkale, Jul 23 '12 at 05:22
With the example arguments to the script, how should the `exec` line inside the script look like? `exec $JAR x y "a b"`? — Some programmer dude, Jul 23 '12 at 07:04
yes. but it is preferable to have it as three arguments viz. x, y and "a b" — sachinpkale, Jul 23 '12 at 09:04

Ignacio Vazquez-Abrams · Accepted Answer · 2012-07-23T05:28:58.297

2

$ARGS=()
for a in "$@"; do
    ARGS+=("${a}")
done
exec "$JAR" "${ARGS[@]}"

edited Jul 23 '12 at 05:28

answered Jul 23 '12 at 05:19

It will not work for me. I need to process some of the arguments differently (which are not shown in the example above). So, the loop is necessary in this case. Sorry! I was not clear initially. – sachinpkale Jul 23 '12 at 05:27
Main answer still doesn't change. – Ignacio Vazquez-Abrams Jul 23 '12 at 05:29

1 Answers1