Passing directory name as a command line argument

adilwaheed · 03-10-2009, 05:33 AM

Hi!
I want to pass a directory name as a command-line argument and display attributes of all files in that directory.
Kindly tell me how to write script for this function.

pixellany · 03-10-2009, 05:40 AM

This looks like homework....

Regardless, you will need to supply some more detail as to what you are doing, and what you have tried.

For example, what attributes are you looking for? Something other than what is supplied by "ls -l"?

adilwaheed · 03-10-2009, 03:25 PM

I have tried

$ ls -l $(read dir) and
$ ls -l `read dir`

but it shows long listing of only root directory.

frieza · 03-10-2009, 03:41 PM

arguments for a shell script would be $X and up where X is a number >=1 (o is the command itself the rest are the arguments passed to that argument)

try creating a script script.sh as follows

echo $0
echo $1
echo $2
...
echo $x

then run that script with x number of agruments (any arbatrary numbers or words separated by spaces will work) and see what is output

adilwaheed · 03-12-2009, 03:01 PM

Thanks for your help. I've written the following script and it worked for me.

#! /bin/bash

if test -d "$1"
then
ls -l $1

elif test -z "$1"

then
echo Incomplete argument...
exit 0

else
echo Directory not found...
exit 0
fi