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

This approach has the advantage of not needing to loop over all the elements (at least not explicitly). But since array_to_string_internal() in array.c still loops over array elements and concatenates them into a string, it's probably not more efficient than the looping solutions proposed, but it's more readable.

if [[ " ${array[*]} " =~ " ${value} " ]]; then     # whatever you want to do when array contains value fi  if [[ ! " ${array[*]} " =~ " ${value} " ]]; then     # whatever you want to do when array doesn't contain value fi 

Note that in cases where the value you are searching for is one of the words in an array element with spaces, it will give false positives. For example

array=("Jack Brown") value="Jack" 

The regex will see "Jack" as being in the array even though it isn't. So you'll have to change IFS and the separator characters on your regex if you want still to use this solution, like this

IFS="|" array=("Jack Brown${IFS}Jack Smith") value="Jack"  if [[ "${IFS}${array[*]}${IFS}" =~ "${IFS}${value}${IFS}" ]]; then     echo "true" else     echo "false" fi  unset IFS # or set back to original IFS if previously set 

This will print "false".

Obviously this can also be used as a test statement, allowing it to be expressed as a one-liner

[[ " ${array[*]} " =~ " ${value} " ]] && echo "true" || echo "false"