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

Use "$@" to represent all the arguments:

for var in "$@"
do
    echo "$var"
done

This will iterate over each argument and print it out on a separate line. $@ behaves like $* except that when quoted the arguments are broken up properly if there are spaces in them:

sh test.sh 1 2 '3 4'
1
2
3 4