Check if a Bash array contains a value

Question

In Bash, what is the simplest way to test if an array contains a certain value?

Answer 1

Here's my take on this.

I'd rather not use a bash for loop if I can avoid it, as that takes time to run. If something has to loop, let it be something that was written in a lower level language than a shell script.

function array_contains { # arrayname value
  local -A _arr=()
  local IFS=
  eval _arr=( $(eval printf '[%q]="1"\ ' "\${$1[@]}") )
  return $(( 1 - 0${_arr[$2]} ))
}

This works by creating a temporary associative array, _arr, whose indices are derived from the values of the input array. (Note that associative arrays are available in bash 4 and above, so this function won't work in earlier versions of bash.) We set $IFS to avoid word splitting on whitespace.

The function contains no explicit loops, though internally bash steps through the input array in order to populate printf. The printf format uses %q to ensure that input data are escaped such that they can safely be used as array keys.

$ a=("one two" three four)
$ array_contains a three && echo BOOYA
BOOYA
$ array_contains a two && echo FAIL
$

Note that everything this function uses is a built-in to bash, so there are no external pipes dragging you down, even in the command expansion.

And if you don't like using eval ... well, you're free to use another approach. :-)