More concise way to check to see if an array contains only numbers (integers)

Question

How do you verify an array contains only values that are integers?

I'd like to be able to check an array and end up with a boolean value of true if the array contains only integers and false if there are any other characters in the array. I know I can loop through the array and check each element individually and return true or false depending on the presence of non-numeric data:

For example:

$only_integers = array(1,2,3,4,5,6,7,8,9,10);
$letters_and_numbers = array('a',1,'b',2,'c',3);

function arrayHasOnlyInts($array)
{
    foreach ($array as $value)
    {
        if (!is_int($value)) // there are several ways to do this
        {
             return false;
        }
    }
    return true;
}

$has_only_ints = arrayHasOnlyInts($only_integers ); // true
$has_only_ints = arrayHasOnlyInts($letters_and_numbers ); // false

But is there a more concise way to do this using native PHP functionality that I haven't thought of?

Note: For my current task I will only need to verify one dimensional arrays. But if there is a solution that works recursively I'd be appreciative to see that to.

Answer 1

$only_integers === array_filter('is_int', $only_integers);             // true
$letters_and_numbers === array_filter('is_int', $letters_and_numbers); // false

It would help you in the future to define two helper, higher-order, functions:

/**
 * Tell whether all members of $array validate the $predicate.
 *
 * all(array(1, 2, 3),   'is_int'); -> true
 * all(array(1, 2, 'a'), 'is_int'); -> false
 */
function all($array, $predicate) {
    return array_filter($array, $predicate) === $array;
}

/**
 * Tell whether any member of $array validates the $predicate.
 *
 * any(array(1, 'a', 'b'),   'is_int'); -> true
 * any(array('a', 'b', 'c'), 'is_int'); -> false
 */
function any($array, $predicate) {
    return array_filter($array, $predicate) !== array();
}

Answer 2

function arrayHasOnlyInts($array) {
    return array_reduce(
        $array,
        function($result,$element) {
            return is_null($result) || $result && is_int($element);
        }
    );
}

returns true if array has only integers, false if at least one element is not an integer, and null if array is empty.

Answer 3

There's always array_reduce():

array_reduce($array, function($a, $b) { return $a && is_int($b); }, true);

But I would favor the fastest solution (which is what you supplied) over the most concise.

Answer 4

 <?php
 $only_integers = array(1,2,3,4,5,6,7,8,9,10);
 $letters_and_numbers = array('a',1,'b',2,'c',3);

 function arrayHasOnlyInts($array){
    $test = implode('',$array);
    return is_numeric($test);
 }

 echo "numbers:". $has_only_ints = arrayHasOnlyInts($only_integers )."<br />"; // true
 echo "letters:". $has_only_ints = arrayHasOnlyInts($letters_and_numbers )."<br />"; // false
 echo 'goodbye';
 ?>

Answer 5

Another alternative, though probably slower than other solutions posted here:

function arrayHasOnlyInts($arr) {
   $nonints = preg_grep('/\D/', $arr); // returns array of elements with non-ints
   return(count($nonints) == 0); // if array has 0 elements, there's no non-ints
}