Noob variable question

[jamie1414]

new to linux and I'm trying to write a script where I have a while loop checking each argument but I can't use the shift option or else all those arguments go out the door for later use. So I'm trying to describe in my if statement the different variables. Basically what I'm trying to do is something like this variable:

$($args)

where $args = $# and is in a while loop that slowly counts down as it loops. The problem is I'm getting errors and I'm not sure if thats the proper format. The whole code looks like this:

args="$#"

while [ "args != "0"]
do
  if [ -d "$($args)" ]
  then
      x="0"
   else
      echo -e "usage: bla bla"
  fi
  args=`expr $args - 1`
done

Thanks in advance.

[barriehie]

So you want the CLI argument array; maybe something like this?

#!/bin/bash
##

declare -i how_many=$#
args=("$@")                          # CLI Argument Array

while [ $how_many != 0 ]
do
  if [ -d ${args[$how_many]} ]
  then
      x="0"
#      echo $how_many
      how_many=$how_many-1
  else
      echo -e "usage: bla bla"
  fi
#  args=`expr $args - 1`
done
exit 0

Short and simple:

#!/bin/sh
for i in $@;
do
echo $i
done

[utne2K]

Hi,
Newbie to bash.

Would a generic bash script work the same on SuSe as it would on Ubuntu?
Pardon me, if that is a goofy question.

By generic, I just mean like a 'hello world' and I guess I'm asking if, fundamentally,
bash works on all Linux flavours.

Thank you.

[barriehie]

Hi,
Newbie to bash.

Would a generic bash script work the same on SuSe as it would on Ubuntu?
Pardon me, if that is a goofy question.

By generic, I just mean like a 'hello world' and I guess I'm asking if, fundamentally,
bash works on all Linux flavours.

Thank you.

bash should work like bash irregardless of the OS it's installed on. Might be version differences. See what they say in the programmers section?