running a command with args from a shell script

KM3 · 10-19-2006, 04:25 AM

Hi,
I'm sure this is easy, but it's been baffling me for an hour.

myscript.sh:

Code:

#!/bin/sh
cmd1="ls -l /home/me/ > /home/me/outfile"
$cmd1

but, cmd1 doesn't work as I want, which to behave exactly as if I had done this:

Code:

#!/bin/sh
ls -l /home/me/ > /home/me/outfile

thanks in advance!

greeklegend · 10-19-2006, 04:46 AM

Damn this one has me baffled. Note how echo $cmd1 yeilds exactly what it should...
Is this a bug or some expected behaviour?
EDIT: This works...

Code:

#!/bin/bash
cmd1="ls -l /home/me/"
$cmd1 > /home/me/outfile

KM3 · 10-19-2006, 05:28 AM

found the answer afterall:

Code:

#!/bin/sh
cmd1="ls -l /home/me/ > /home/me/outfile"
eval $cmd1

soggycornflake · 10-19-2006, 01:46 PM

You could also use an array, e.g.

Code:

#!/bin/sh
cmd1=(ls -l /home/me/ > /home/me/outfile)
${cmd1[*]}

This ${cmd1[*]} is slightly cumbersome, since bash only expands the first element given $cmd1. I use zsh, which expands $cmd1 to the entire array, which is a little more convenient.