Executing commands through shell variable

shriyer · 07-23-2009, 04:21 PM

Hey there,
I am trying to pass a command as argument to a function.
The command shows up in $1.
Now I want to execute this command, but if I do a $1

./sample
"bla/blaprintf: warning: ignoring excess arguments, starting with `bla/bla'

The code is :

#!/bin/ksh
fn()
{
$1
}

fn 'printf "bla/bla bla/bla bla bla bla bla\\n"'

Any ideas how to get it work please ?
Thanks,
Shrikant

choogendyk · 07-23-2009, 06:27 PM

It is an issue with shell substitution. I couldn't get it to work with a single argument, no matter what combination of quoting characters or escape characters I tried. it was breaking on the first space. So the error you got was discarding beginning with the second bla/bla. Playing around a bit, I got the following to work both in bash and ksh on my Mac.

Code:

#!/bin/bash
fn()
{
echo "==${1}==";
echo "==${2}==";
$1 "$2"
}

fn printf 'whataver thingy\n'

I added in the echo statements to get the feedback on what was going in to the function. Obviously, you can strip them out.

thinknix · 07-23-2009, 07:03 PM

Here is an easier solution:

Code:

#!/bin/sh
fn()
{
eval "$1" 
}

fn 'printf "bla/bla bla/bla bla bla bla bla\n"'