Bash script using getopts to store strings as an array-CodePudding

I am working on a Bash script that needs to take zero to multiple strings as an input but I am unsure how to do this because of the lack of a flag before the list.

The script usage:

script [ list ] [ -t <secs> ] [ -n <count> ]

The list takes zero, one, or multiple strings as input. When a space is encountered, that acts as the break between the strings in a case of two or more. These strings will eventually be input for a grep command, so my idea is to save them in an array of some kind. I currently have the -t and -n working correctly. I have tried looking up examples but have been unable to find anything that is similar to what I want to do. My other concern is how to ignore string input after a flag is set so no other strings are accepted.

My current script:

while getopts :t:n: arg; do
  case ${arg} in
    t)
      seconds=${OPTARG}
      if ! [[ $seconds =~ ^[1-9][0-9]*$ ]] ; then
        exit
      fi
      ;;
    n)
      count=${OPTARG}
      if ! [[ $count =~ ^[1-9][0-9]*$ ]] ; then
        exit
      fi
      ;;
    :)
      echo "$0: Must supply an argument to -$OPTARG" >&2
      exit
      ;;
    ?)
      echo "Invalid option: -${OPTARG}"
      exit
      ;;
  esac
done

Edit: This is for a homework assignment and am unsure if the order of arguments can change

CodePudding user response：

Would you please try the following:

#!/bin/bash

# parse the arguments before getopts
for i in "$@"; do
    if [[ $i = "-"* ]]; then
        break
    else                # append the arguments to "list" as long as it does not start with "-"
        list =("$1")
        shift
    fi
done

while getopts :t:n: arg; do
    : your "case" code here
done

# see if the variables are properly assigned
echo "seconds=$seconds" "count=$count"
echo "list=${list[@]}"

CodePudding user response：

Try:

#! /bin/bash -p

# Set defaults
count=10
seconds=20

args=( "$@" )
end_idx=$(($#-1))

# Check for '-n' option at the end
if [[ end_idx -gt 0 && ${args[end_idx-1]} == -n ]]; then
    count=${args[end_idx]}
    end_idx=$((end_idx-2))
fi

# Check for '-t' option at the (possibly new) end
if [[ end_idx -gt 0 && ${args[end_idx-1]} == -t ]]; then
    seconds=${args[end_idx]}
    end_idx=$((end_idx-2))
fi

# Take remaining arguments up to the (possibly new) end as the list of strings
strings=( "${args[@]:0:end_idx 1}" )

declare -p strings seconds count

The basic idea is to process the arguments right-to-left instead of left-to-right.
The code assumes that the only acceptable order of arguments is the one given in the question. In particular, it assumes that the -t and -n options must be at the end if they are present, and they must be in that order if both are present.
It makes no attempt to handle option arguments combined with options (e.g. -t5 instead of -t 5). That could be done fairly easily if required.
It's OK for strings in the list to begin with -.

CodePudding user response：

My shorter version

Some remarks:

Instead of loop over all argument**, then break if argument begin by -, I simply use a while loop.
From How do I test if a variable is a number in Bash?, added efficient is_int test function
As any output (echo) done in while getopts ... loop would be an error, redirection do STDERR (>&2) could be addressed to the whole loop instead of repeated on each echo line.
** Note doing a loop over all argument could be written for varname ;do. as $@ stand for default arguments, in "$@" are implicit in for loop.

#!/bin/bash

is_int() { case ${1#[- ]} in
               '' | *[!0-9]* ) echo "Argument '$1' is not a number"; exit 3;;
           esac ;}

while [[ ${1%%-*} ]];do
    args =("$1")
    shift
done

while getopts :t:n: arg; do
    case ${arg} in
        t ) is_int "${OPTARG}" ; seconds=${OPTARG} ;;
        n ) is_int "${OPTARG}" ; count=${OPTARG} ;;
        : ) echo "$0: Must supply an argument to -$OPTARG" ; exit 2;;
        ? ) echo "Invalid option: -${OPTARG}" ; exit 1;;
    esac
done >&2

declare -p seconds count args