Pass text file as argument/variable [bash]

Question

I am trying to pass a text file as a specific argument and print its name and content...

#!/bin/bash

## set input args
while getopts "f" option; do
    case "${option}" in
        f)
          arg3=${OPTARG};;  
      esac
done

## script
echo $arg3
echo $(cat $arg3)

(for running it: sh myscript.sh -f filelist)

Something is really wrong because even the file name is not appearing! (and curiously, in bash everything goes well, why?).

It needs to be `getopts "f:"` option`. The `:` indicates that the option has an argument. — Barmar, Oct 05 '21 at 17:00

jgarces · Accepted Answer · 2021-10-06T08:02:33.673

According @Barmar's answer (thanks!) I forgot the colon with f... but, as I was trying to make this argument optional, this should be after. Based on this other question and @Barmar's point, the "final" code could be like that:

#!/bin/bash

## set input args
while getopts ":f" option; do
    case "${option}" in
        f)
          # Check next positional parameter
          eval nextopt=\${$OPTIND}
          # existing or starting with dash?
          if [[ -n $nextopt && $nextopt != -* ]] ; then
            OPTIND=$((OPTIND + 1))
            arg3=$nextopt
          else
            echo "Not filelist specified, closing..." && exit
          fi
        ;;
      esac
done

## script
echo $arg3
echo $(cat $arg3)

Pass text file as argument/variable [bash]

1 Answers1