In array operator in bash

Question

Is there a way to test whether an array contains a specified element?

e.g., something like:

array=(one two three)

if [ \"one\" in ${array} ]; then
...
f         


        

Answer 1

Try this:

array=(one two three)
if [[ "${array[*]}" =~ "one" ]]; then
  echo "'one' is found"
fi

Answer 2

if you just want to check whether an element is in array, another approach

case "${array[@]/one/}" in 
 "${array[@]}" ) echo "not in there";;
 *) echo "found ";;
esac

Answer 3

A for loop will do the trick.

array=(one two three)

for i in "${array[@]}"; do
  if [[ "$i" = "one" ]]; then
    ...
    break
  fi
done

Answer 4

I got an function 'contains' in my .bashrc-file:

contains () 
{ 
    param=$1;
    shift;
    for elem in "$@";
    do
        [[ "$param" = "$elem" ]] && return 0;
    done;
    return 1
}

It works well with an array:

contains on $array && echo hit || echo miss
  miss
contains one $array && echo hit || echo miss
  hit
contains onex $array && echo hit || echo miss
  miss

But doesn't need an array:

contains one four two one zero && echo hit || echo miss
  hit

Answer 5

OPTIONS=('-q','-Q','-s','-S')

find="$(grep "\-q" <<< "${OPTIONS[@]}")"
if [ "$find" = "${OPTIONS[@]}" ];
then
    echo "arr contains -q"
fi

Answer 6

I like using grep for this:

if echo ${array[@]} | grep -qw one; then
  # "one" is in the array
  ...
fi

(Note that both -q and -w are non-standard options to grep: -w tells it to work on whole words only, and -q ("quiet") suppresses all output.)