BASH: Best way to set variable from array

Question

Bash 4 on Linux ~
I have an array of possible values. I must restrict user input to these values.

Arr=(hello kitty goodbye quick fox)

User supplies value as argument to script:

bash myscript.sh -b var

Currently, I'm trying the following:

function func_exists () {
_var="$1"
for i in ${Arr[@]}
do
    if [ "$i" == "$_var" ]
    then
        echo hooray for "$_var"
        return 1
    fi
done
    return 0
}

func_exists $var
if [ $? -ne 1 ];then
    echo "Not a permitted value."
    func_help
    exit $E_OPTERROR
fi

Seems to work fine, are there better methods for testing user input against an array of allowed values?

UPDATE: I like John K's answer ...can someone clarify the use of $@? I understand that this represents all positional parameters -- so we shift the first param off the stack and $@ now represents all remaining params, those being the passed array ...is that correct? I hate blindly using code without understanding ...even if it works!

John Kugelman · Accepted Answer · 2011-06-02 02:55:25Z

1

Your solution is what I'd do. Maybe using a few more shell-isms, such as returning 0 for success and non-0 for failure like UNIX commands do in general.

# Tests if $1 is in the array ($2 $3 $4 ...).
is_in() {
    value=$1
    shift

    for i in "$@"; do
        [[ $i == $value ]] && return 0
    done

    return 1
}

if ! is_in "$var" "${Arr[@]}"; then
    echo "Not a permitted value." >&2
    func_help
    exit $E_OPTERROR
fi

Careful use of double quotes makes sure this will work even if the individual array entries contain spaces, which is allowed. This is a two element array: list=('hello world' 'foo bar').

edited Jun 2, 2011 at 2:55

answered Jun 2, 2011 at 2:50

John Kugelman

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2 Comments

Bubnoff Over a year ago

Nice improvements! Thanks! Barring further brilliance this will get my vote.

Bubnoff Over a year ago

My understanding of the $@ in your function: shift the first arg off the stack - $@ now represents the remainder. Is this correct?

Fritz G. Mehner · Accepted Answer · 2011-06-02 08:52:46Z

1

Another solution. is_in is just a variable:

Arr=(hello kitty goodbye quick fox)

var='quick'

string=" ${Arr[*]} "                            # array to string, framed with blanks
is_in=1                                         # false
# try to delete the variable inside the string; true if length differ 
[ "$string" != "${string/ ${var} /}" ] && is_in=0

echo -e "$is_in"

answered Jun 2, 2011 at 8:52

Fritz G. Mehner

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1 Comment

Bubnoff Over a year ago

Great solution. Thanks! I'll keep this one for later.

John Kugelman · Accepted Answer · 2011-06-02 12:57:28Z

1

function func_exists () {
  case "$1"
  in
    hello)
    kitty)
    goodbye) 
    quick)
    fox)
      return 1;;
    *)
      return 0;;
  esac
}

edited Jun 2, 2011 at 12:57

John Kugelman

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answered Jun 2, 2011 at 2:29

John Zwinck

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5 Comments

John Zwinck Over a year ago

To the down-voter: sorry, you saw my answer at the moment I accidentally submitted it due to hitting the wrong key. :)

Bubnoff Over a year ago

This is a nice solution but I am not sure how readable it would be if the allowable input array contained 50 - 100 members.

John Kugelman Over a year ago

Could also write it compactly in one case as hello|kitty|goodbye|quick|fox).

Bubnoff Over a year ago

Am thinking for versatility and readability, an array or an array pulled from a file would be appropriate. Not sure the most efficient way to test input against such an array. The solution should be able to deal efficiently and readably against arrays with 50 plus members.

Bubnoff Over a year ago

Another advantage to a separately defined array is that you could pass the array with the variable to test and have a reusable function.

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