Checking if array is multidimensional or not?

Question

Use count() twice; one time in default mode and one time in recursive mode. If the values match, the array is not multidimensional, as a multidimensional array would have a higher recursive count.

if (count($array) == count($array, COUNT_RECURSIVE)) 
{
  echo 'array is not multidimensional';
}
else
{
  echo 'array is multidimensional';
}

This option second value mode was added in PHP 4.2.0. From the PHP Docs:

If the optional mode parameter is set to COUNT_RECURSIVE (or 1), count() will recursively count the array. This is particularly useful for counting all the elements of a multidimensional array. count() does not detect infinite recursion.

However this method does not detect array(array()).

The short answer is no you can't do it without at least looping implicitly if the 'second dimension' could be anywhere. If it has to be in the first item, you'd just do

is_array($arr[0]);

But, the most efficient general way I could find is to use a foreach loop on the array, shortcircuiting whenever a hit is found (at least the implicit loop is better than the straight for()):

$ more multi.php
<?php

$a = array(1 => 'a',2 => 'b',3 => array(1,2,3));
$b = array(1 => 'a',2 => 'b');
$c = array(1 => 'a',2 => 'b','foo' => array(1,array(2)));

function is_multi($a) {
    $rv = array_filter($a,'is_array');
    if(count($rv)>0) return true;
    return false;
}

function is_multi2($a) {
    foreach ($a as $v) {
        if (is_array($v)) return true;
    }
    return false;
}

function is_multi3($a) {
    $c = count($a);
    for ($i=0;$i<$c;$i++) {
        if (is_array($a[$i])) return true;
    }
    return false;
}
$iters = 500000;
$time = microtime(true);
for ($i = 0; $i < $iters; $i++) {
    is_multi($a);
    is_multi($b);
    is_multi($c);
}
$end = microtime(true);
echo "is_multi  took ".($end-$time)." seconds in $iters times\n";

$time = microtime(true);
for ($i = 0; $i < $iters; $i++) {
    is_multi2($a);
    is_multi2($b);
    is_multi2($c);
}
$end = microtime(true);
echo "is_multi2 took ".($end-$time)." seconds in $iters times\n";
$time = microtime(true);
for ($i = 0; $i < $iters; $i++) {
    is_multi3($a);
    is_multi3($b);
    is_multi3($c);
}
$end = microtime(true);
echo "is_multi3 took ".($end-$time)." seconds in $iters times\n";
?>

$ php multi.php
is_multi  took 7.53565130424 seconds in 500000 times
is_multi2 took 4.56964588165 seconds in 500000 times
is_multi3 took 9.01706600189 seconds in 500000 times

Implicit looping, but we can't shortcircuit as soon as a match is found...

$ more multi.php
<?php

$a = array(1 => 'a',2 => 'b',3 => array(1,2,3));
$b = array(1 => 'a',2 => 'b');

function is_multi($a) {
    $rv = array_filter($a,'is_array');
    if(count($rv)>0) return true;
    return false;
}

var_dump(is_multi($a));
var_dump(is_multi($b));
?>

$ php multi.php
bool(true)
bool(false)

For PHP 4.2.0 or newer:

function is_multi($array) {
    return (count($array) != count($array, 1));
}

I think this is the most straight forward way and it's state-of-the-art:

function is_multidimensional(array $array) {
    return count($array) !== count($array, COUNT_RECURSIVE);
}

You can simply execute this:

if (count($myarray) !== count($myarray, COUNT_RECURSIVE)) return true;
else return false;

If the optional mode parameter is set to COUNT_RECURSIVE (or 1), count() will recursively count the array. This is particularly useful for counting all the elements of a multidimensional array.

If it's the same, means there are no sublevels anywhere. Easy and fast!

You could look check is_array() on the first element, under the assumption that if the first element of an array is an array, then the rest of them are too.

After PHP 7 you could simply do:

public function is_multi(array $array):bool
{
    return is_array($array[array_key_first($array)]);
}