Check if a variable exists in a list in Bash

Question

I am trying to write a script in bash that check the validity of a user input.
I want to match the input (say variable x) to a list of valid values.

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Answer 1

An alternative solution inspired by the accepted response, but that uses an inverted logic:

MODE="${1}"

echo "<${MODE}>"
[[ "${MODE}" =~ ^(preview|live|both)$ ]] && echo "OK" || echo "Uh?"

Here, the input ($MODE) must be one of the options in the regular expression ('preview', 'live', or 'both'), contrary to matching the whole options list to the user input. Of course, you do not expect the regular expression to change.

Answer 2

If the list is fixed in the script, I like the following the best:

validate() {
    grep -F -q -x "$1" <<EOF
item 1
item 2
item 3
EOF
}

Then use validate "$x" to test if $x is allowed.

If you want a one-liner, and don't care about whitespace in item names, you can use this (notice -w instead of -x):

validate() { echo "11 22 33" | grep -F -q -w "$1"; }

Notes:

• This is POSIX sh compliant.
• validate does not accept substrings (remove the -x option to grep if you want that).
• validate interprets its argument as a fixed string, not a regular expression (remove the -F option to grep if you want that).

Sample code to exercise the function:

for x in "item 1" "item2" "item 3" "3" "*"; do
    echo -n "'$x' is "
    validate "$x" && echo "valid" || echo "invalid"
done

Answer 3

Prior answers don't use tr which I found to be useful with grep. Assuming that the items in the list are space delimited, to check for an exact match:

echo $mylist | tr ' ' '\n' | grep -F -x -q "$myitem"

This will return exit code 0 if the item is in the list, or exit code 1 if it isn't.

It's best to use it as a function:

_contains () {  # Check if space-separated list $1 contains line $2
  echo "$1" | tr ' ' '\n' | grep -F -x -q "$2"
}

mylist="aa bb cc"

# Positive check
if _contains "${mylist}" "${myitem}"; then
  echo "in list"
fi

# Negative check
if ! _contains "${mylist}" "${myitem}"; then
  echo "not in list"
fi

Answer 4

[[ $list =~ (^|[[:space:]])$x($|[[:space:]]) ]] && echo 'yes' || echo 'no'

or create a function:

contains() {
    [[ $1 =~ (^|[[:space:]])$2($|[[:space:]]) ]] && exit(0) || exit(1)
}

to use it:

contains aList anItem
echo $? # 0： match, 1: failed

Answer 5

Consider exploiting the keys of associative arrays. I would presume this outperforms both regex/pattern matching and looping, although I haven't profiled it.

declare -A list=( [one]=1 [two]=two [three]='any non-empty value' )
for value in one two three four
do
    echo -n "$value is "
    # a missing key expands to the null string, 
    # and we've set each interesting key to a non-empty value
    [[ -z "${list[$value]}" ]] && echo -n '*not* '
    echo "a member of ( ${!list[*]} )"
done

Output:

one is a member of ( one two three )
two is a member of ( one two three )
three is a member of ( one two three )
four is *not* a member of ( one two three )