Check if a Bash array contains a value

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

Edit: With help from the answers and the comments, after some testing, I came up with this:

function contains() {
    local n=$#
    local value=${!n}
    for ((i=1;i < $#;i++)) {
        if [ "${!i}" == "${value}" ]; then

            echo "y"
            return 0
        fi
    }
    echo "n"
    return 1
}

A=("one" "two" "three four")
if [ $(contains "${A[@]}" "one") == "y" ]; then

    echo "contains one"
fi
if [ $(contains "${A[@]}" "three") == "y" ]; then
    echo "contains three"
fi

I'm not sure if it's the best solution, but it seems to work.

Answer

There is sample code that shows how to replace a substring from an array. You can make a copy of the array and try to remove the target value from the copy. If the copy and original are then different, then the target value exists in the original string.

The straightforward (but potentially more time-consuming) solution is to simply iterate through the entire array and check each item individually. This is what I typically do because it is easy to implement and you can wrap it in a function (see this info on passing an array to a function).