How to iterate over arguments in a Bash script

Question

I have a complex command that I\'d like to make a shell/bash script of. I can write it in terms of $1 easily:

foo $1 args -o $1.ext
         


        

Answer 1

For simple cases you can also use shift. It treats the argument list like a queue. Each shift throws the first argument out and the index of each of the remaining arguments is decremented.

#this prints all arguments
while test $# -gt 0
do
    echo "$1"
    shift
done