I want to call a bash script like this
$ ./scriptName -o -p -t something path/to/file This is as far as I get
#!/bin/bash o=false p=false while getopts ":opt:" options do case $options in o ) opt1=true ;; p ) opt2=true ;; t ) opt3=$OPTARG ;; esac done but how do I get the path/to/file?
You can do something like:
shift $(($OPTIND - 1)) first_arg=$1 second_arg=$2 after the loop has run.
shift $((OPTIND - 1)) - i.e. losing the dollar sign inside the parentheses?
To capture all the remaining parameters after the getopts processing, a good solution is to shift (remove) all the processed parameters (variable $OPTIND) and assign the remaining parameters ($@) to a specific variable. Short answer:
shift $(($OPTIND - 1)) remaining_args="$@" Long example:
#!/bin/bash verbose=false function usage () { cat <<EOUSAGE $(basename $0) hvr:e: show usage EOUSAGE } while getopts :hvr:e: opt do case $opt in v) verbose=true ;; e) option_e="$OPTARG" ;; r) option_r="$option_r $OPTARG" ;; h) usage exit 1 ;; *) echo "Invalid option: -$OPTARG" >&2 usage exit 2 ;; esac done echo "Verbose is $verbose" echo "option_e is \"$option_e\"" echo "option_r is \"$option_r\"" echo "\$@ pre shift is \"$@\"" shift $((OPTIND - 1)) echo "\$@ post shift is \"$@\"" This will output
$ ./test-getopts.sh -r foo1 -v -e bla -r foo2 remain1 remain2 Verbose is true option_e is "bla" option_r is " foo1 foo2" $@ pre shift is "-r foo1 -v -e bla -r foo2 remain1 remain2" $@ post shift is "remain1 remain2" 