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

Answer 2

Amplifying on baz's answer, if you need to enumerate the argument list with an index (such as to search for a specific word), you can do this without copying the list or mutating it.

Say you want to split an argument list at a double-dash ("--") and pass the arguments before the dashes to one command, and the arguments after the dashes to another:

 toolwrapper() {
   for i in $(seq 1 $#); do
     [[ "${!i}" == "--" ]] && break
   done || return $? # returns error status if we don't "break"

   echo "dashes at $i"
   echo "Before dashes: ${@:1:i-1}"
   echo "After dashes: ${@:i+1:$#}"
 }

Results should look like this:

 $ toolwrapper args for first tool -- and these are for the second
 dashes at 5
 Before dashes: args for first tool
 After dashes: and these are for the second

Answer 3

getopt Use command in your scripts to format any command line options or parameters.

#!/bin/bash
# Extract command line options & values with getopt
#
set -- $(getopt -q ab:cd "$@")
#
echo
while [ -n "$1" ]
do
case "$1" in
-a) echo "Found the -a option" ;;
-b) param="$2"
echo "Found the -b option, with parameter value $param"
shift ;;
-c) echo "Found the -c option" ;;
--) shift
break ;;
*) echo "$1 is not an option";;
esac
shift

Answer 4

You can also access them as an array elements, for example if you don't want to iterate through all of them

argc=$#
argv=("$@")

for (( j=0; j<argc; j++ )); do
    echo "${argv[j]}"
done

Answer 5

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

Answer 6

Note that Robert's answer is correct, and it works in sh as well. You can (portably) simplify it even further:

for i in "$@"

is equivalent to:

for i

I.e., you don't need anything!

Testing ($ is command prompt):

$ set a b "spaces here" d
$ for i; do echo "$i"; done
a
b
spaces here
d
$ for i in "$@"; do echo "$i"; done
a
b
spaces here
d

I first read about this in Unix Programming Environment by Kernighan and Pike.

In bash, help for documents this:

for NAME [in WORDS ... ;] do COMMANDS; done

If 'in WORDS ...;' is not present, then 'in "$@"' is assumed.