Command line arguments in Bash

I want to write a bash script which takes different arguments. It should be used like normal linux console programs:

my_bash_script -p 2 -l 5 -t 20

So the value 2 should be saved in a variable called pages and the parameter l should be saved in a variable called length and the value 20 should be saved in a variable time.

What is the best way to do this?

bash
command-line-arguments

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小开

Something along the lines of

pages=
length=
time=


while test $# -gt 0
do
case $1 in
-p)
pages=$2
shift
;;
-l)
length=$2
shift
;;
-t)
time=$2
shift
;;
*)
echo >&2 "Invalid argument: $1"
;;
esac
shift
done

小开

最佳答案

Use the getopts builtin:
here's a tutorial

pages=  length=  time=


while getopts p:l:t: opt; do
case $opt in
p)
pages=$OPTARG
;;
l)
length=$OPTARG
;;
t)
time=$OPTARG
;;
esac
done


shift $((OPTIND - 1))

shift $((OPTIND - 1)) shifts the command line parameters so that you can access possible arguments to your script, i.e. $1, $2, ...