bash wrapper arguments problem ...

khanrockz · 06-14-2007, 06:58 PM

Hi !
I am trying to write a wrapper for ssh and it seems that it is not interpreting the arguments that are passed to it. The code for the wrapper goes like :

Code:

invoknam=`basename $0`
function make_command_line_args2 () {

    for (( i = 1; i <= $# ; i++ ))
    do
        eval VAR=\$$i
        countargs $VAR
        words=$?
        if [ "$words" -ge "2" ] ; then
                OPTIONS="$OPTIONS \"$VAR\""
        else
                OPTIONS="$OPTIONS $VAR"
        fi
    done
    echo $OPTIONS
}

cmd2=$(make_command_line_args2 "$@")
echo Command Options : $cmd2
exec /usr/bin/$invoknam $cmd2

but when I execute it it gives

khan /bin> ssh -x "-oForwardAgent no" khan@192.168.0.1
Command Options : -x "-oForwardAgent no" khan@192.168.0.1
ssh: "-oForwardAgent: Name or service not known

So it seems that i am getting the quotes right but they are not interpreted as I want them to. This command is a valid command with normal ssh executable.
Any help will be greatly appreciated ... how can I make this work ? Where am I going wrong ?

gnashley · 06-15-2007, 03:32 AM

This should get the second argument for you:
echo "$@" | awk '{ print $2}'

khanrockz · 06-15-2007, 10:44 AM

Quote:

This should get the second argument for you:
echo "$@" | awk '{ print $2}'

But how does this help in my code ? Can u plz be a little more specific. I am sorry but am a newbie

.

druuna · 06-15-2007, 11:38 AM

Hi,

Maybe the above is an exercise in programming (which I doubt) or maybe I'm missing the point, but the code below does what you want without all the extra de-assembling and re-assembling of the command line options:

Code:

#!/bin/bash

echo Command Options : "$@"
exec /usr/bin/ssh "$@"

An example run:

Code:

 $ ./ssh.sh -x "-oForwardAgent no" druuna@xxxx.yyyy.zzzz
Command Options : -x -oForwardAgent no druuna@xxxx.yyyy.zzzz
Password:

Hope this helps.