Php, in_array with no exactly match

Question

I want to do the following:

$a = array();
$a[] = array(1,2);
$a[] = array(2,5);
$a[] = array(3,4);
var_dump (in_array(array(2,5), $a));

this returns OK, as it expected, but if the source array is not fully matched:

$a = array();
$a[] = array(1,2, 'f' => array());
$a[] = array(2,5, 'f' => array());
$a[] = array(3,4, 'f' => array());
var_dump (in_array(array(2,5), $a));

it returns false. Is there a way to do it with the built-in way, or I have to code it?

that wont take the orders into account. Would be true for 2,5 and 5,2 — John Smith
– John Smith, Commented Jun 5, 2014 at 9:37
that will not work at all. First approach checks if there are array as element in $a, and if it is with values 2, 5. But separately 2 and 5 are not into $a because $a is represented as $a = array(1 = > array(2, 5, 'f')) — Ivan Yonkov
– Ivan Yonkov, Commented Jun 5, 2014 at 9:39
in_array won't do it. It's simply not in the scope of what it does. You'll have to loop yourself. — deceze
– deceze ♦, Commented Jun 5, 2014 at 9:41

Alma Do · Accepted Answer · 2014-06-05 09:47:08Z

2

in_array() is just not the thing that you should use for this issue. Because it will compare values with type casting, if that's needed. Instead, you may use plain loop or something like:

function in_array_array(array $what, array $where)
{
   return count(array_filter($where, function($x) use ($what)
   {
      return $x===$what;
   }))>0;
}

So then

var_dump(in_array_array(array(2, 5), $a)); //true

answered Jun 5, 2014 at 9:47

Alma Do

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deceze · Accepted Answer · 2014-06-05 10:10:01Z

1

$needle = array(2, 5);
$found = array_reduce($a, function ($found, array $array) use ($needle) {
    return $found || !array_diff($needle, $array);
});

This does an actual test of whether the needle is a subset of an array.

function subset_in_array(array $needle, array $haystack) {
    return array_reduce($haystack, function ($found, array $array) use ($needle) {
        return $found || !array_diff($needle, $array);
    });
}

if (subset_in_array(array(2, 5), $a)) ...

answered Jun 5, 2014 at 10:10

deceze♦

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