更简洁的方法来检查数组是否只包含数字（整数）

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

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

 <?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 3

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
}

Answer 4

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 5

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 6

Why don't we give a go to Exceptions?

Take any built in array function that accepts a user callback ( array_filter() , array_walk() , even sorting functions like usort() etc.) and throw an Exception in the callback. Eg for multidimensional arrays:

function arrayHasOnlyInts($array)
{
    if ( ! count($array)) {
        return false;
    }

    try {
        array_walk_recursive($array, function ($value) {
            if ( ! is_int($value)) {
                throw new InvalidArgumentException('Not int');
            }

            return true;
        });
    } catch (InvalidArgumentException $e) {
        return false;
    }

    return true;
}

It's certainly not the most concise, but a versatile way.