dannyman.toldme.com : Argument Parsing in Shell Scripts

June 22, 2005
Technical

Argument Parsing in Shell Scripts

So, say you are writing a shell script (sh or bash) that needs to take arguments like so:

./script.sh start
./script.sh -v start
./script.sh -c foo.conf start
./script.sh -vc foo.conf start

This took me a bit of doing. First, I tried getopt but apparently all that does is disambiguate -v from -c. That, and the FreeBSD man page claims to have improved the example by making it completely inscrutable.

Eventually I found a reference that explained how to use getopts, which worked a lot better, but all the examples I could find didn’t tell you how to reset $@.

Eventually, I figured out that that last bit is accomplished with shift.

Here’s example code:

#!/bin/sh
#

conf=default.conf
while getopts c:hv o
do case "$o" in
    c)  conf="$OPTARG";;
    h)  echo "Usage: $0 -hv -c <config> [start|stop|restart|status]" && exit 1;;
    v)  verbose="yes";;
    \?)  echo "Usage: $0 -hv -c <config> [start|stop|restart|status]" && exit 1;;
esac
done

# Reset $@
shift `echo $OPTIND-1 | bc`

echo "conf: $conf"
echo "command: $1"

Notes:

The while loop iterates $o through $@ where it conforms to the getopts configuration.
getopts c:hv indicates that -h and -v are boolean arguments, and -c is followed by a string argument.
$OPTIND is set at the end of getopt, and indicates how many arguments through $@ it parsed.
Oh, and the shift operation resets $@ by $OPTIND-1, so that $1 ends up pointing at the start command, or whatever.
If you’re programming in ksh, you can just set $OPTIND-1. But if you’re programming in ksh you are far beyond needing to check my blog for argument parsing.

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while [ $# != 0 ]; do flag="$1" case "$flag" in -a) echo "You provided the -a flag, which takes no arguments" ;; -b) if [ $# -gt 1 ]; then arg="$2" shift else echo "You did not provide an argument for the -b flag" fi echo "You supplied an argument for the -b flag: $arg" ;; *) echo "Unrecognized flag or argument: $flag" ;; esac shift done

http://www.zeuscat.com/andrew/ Andrew Ho

My standard sh command line parsing loop looks like this:

If you are using bash instead of sh, you can accept both -fvalue and -f value variants:

           -b*) arg="${arg##-b}"
                echo "You attached an argument to the -b flag: $arg"
                ;;

My code doesn’t reset $@. I have a README file in my homedir that reminds me of the other neat bash-ism that I like to use: ${1+"$@"} expands to $@, but if $@ is empty, it expands to an empty list (different from “$@” which would expand to the single empty string “”).

an_agent

FAI…

shift $((OPTIND - 1))

works in bash (not sh) without having to pipe to bc.
Ted Henry

Thanks for this post! This is exactly what I needed :-)
Alexander Brankov

Hi guys,

These are very good examples! Thank you to all of you.

It is comprehensive and easy but here is a different case:

myscript.sh --first-argument=23 --second-argument="some string" unnamed arguments

I could not find anything to cook up a good example for this case.

Your competent help would be much appreciated!

Many thanks,

Alex
http://dannyman.toldme.com/ dannyman
Alexander,

The code in my post handles spaces just fine:
```
0-11:45 dannhowa@ops-dev ~> ./test.sh -c 'foo foo' start
conf: foo foo
command: start
```
http://AlexanderBrankov Alexander Brankov

Hi Guys,

It is the equal sign that I am worried about. In this case I can have something like:

./my_script.sh –option_1=”Some Text!” –option-2=10

I struggle to handle this case when I am converting some tools that use this way of calling the commands.

Kind regards,

Alex
http://AlexanderBrankov Alexander Brankov

Again the two dashes in front of the arguments names ware merged.

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Argument Parsing in Shell Scripts

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